![](\"http://i.imgur.com/A4ikuJh.png\")
\n\nWe could also represent $\\bP$ by [factoring](https://arbital.com/p/factoring_probability), so using $\\bP(A,B) = \\bP(A)\\; \\bP(B \\mid A)$ we'd make this picture:\n\n![](\"http://i.imgur.com/QaVKTX4.png\")
\n\n\nVisualizing probabilities in a square is neat because we can draw simple pictures that highlight interesting facts about our probability distribution.\n\nBelow are some pictures illustrating:\n\n* [independent events](https://arbital.com/p/4cf) (What happens if the columns and the rows in our square *both* line up?)\n\n* [marginal probabilities](https://arbital.com/p/marginal_probability) (If we're looking at a square of probabilities, where's the probability $\\bP(A)$ of $A$ or the probability $\\bP(\\neg B)$?)\n\n* [conditional probabilities](https://arbital.com/p/1rj) (Can we find in the square the probability $\\bP(B \\mid A)$ of $B$ if we condition on seeing $A$? What about the conditional probability $\\bP(A \\mid B)$?)\n\n* [factoring a distribution](https://arbital.com/p/factoring_probability) (Can we always write $\\bP$ as a square? Why do the columns line up but not the rows?)\n\n* the process of computing [joint probabilities](https://arbital.com/p/1rh) from [factored probabilities](https://arbital.com/p/factoring_probability)\n\nIndependent events\n===\n\nHere's a picture of the joint distribution of [two independent events](https://arbital.com/p/4cf) $A$ and $B$:\n\n![](\"http://i.imgur.com/0STax7m.png\")
\n\nHere's the probability $\\bP(\\neg B)$ of $\\neg B$:\n\n![](\"http://i.imgur.com/uNGdze2.png\")
\n\nIn these pictures we're dividing by the area of the whole square. Since the probability of anything at all happening is 1, we could just leave it out, but it'll be helpful for comparison while we think about conditionals next.\n\n\n\nConditional probabilities\n===\n\nWe can start with some probability $\\bP(B)$, and then *assume* that $A$ is true to get a [conditional probability](https://arbital.com/p/1rj) $\\bP(B \\mid A)$ of $B$. Conditioning on $A$ being true is like restricting our whole attention to just the possible worlds where $A$ happens:\n\n\n![](\"http://i.imgur.com/ubybRoH.png\")
\n\nThen the conditional probability of $B$ given $A$ is the proportion of these $A$ worlds where $B$ also happens:\n\n![](\"http://i.imgur.com/bFCHiZ8.png\")
\n\nIf instead we condition on $\\neg A$, we get:\n\n![](\"http://i.imgur.com/LHkGLVl.png\")
\n\n\nSo our square visualization gives a nice way to see, at a glance, the conditional probabilities of $B$ given $A$ or given $\\neg A$:\n\n![](\"http://i.imgur.com/G6Sja4f.png\")
\n\n\nMore on this next.\n\nFactoring a distribution\n===\n\nRecall the square showing our joint distribution $\\bP$:\n\n![](\"http://i.imgur.com/4oCyg8q.png\")
\n\n\nand for the conditional probability $\\bP(A \\mid \\neg B)$:\n\n\n![](\"http://i.imgur.com/pK7o56J.png\")
\n\n\n\nComputing joint probabilities from factored probabilities \n===\n\nLet's say we know the factored probabilities for $A$ and $B$, factoring by $A$. That is, we know $\\bP(A = \\true)$, and we also know $\\bP(B = \\true \\mid A = \\true)$ and $\\bP(B = \\true \\mid A = \\false)$. How can we recover the joint probability $\\bP(A = t_A, B = t_B)$ that $A = t_A$ is the case and also $B = t_B$ is the case?\n\n\nSince \n\n$$\\bP(B = \\false \\mid A = \\true) = \\frac{\\bP(A = \\true, B = \\false)}{\\bP(A = \\true)}\\ ,$$ \n\nwe can multiply the prior $\\bP(A)$ by the conditional $\\bP(\\neg B \\mid A)$ to get the joint $\\bP(A, \\neg B)$:\n\n$$\\bP(A = \\true)\\; \\bP(B = \\false \\mid A = \\true) = \\bP(A = \\true, B = \\false)\\ .$$ \n\nIf we do this at the same time for all the possible truth values $t_A$ and $t_B$, we get back the full joint distribution:\n\n![](\"http://i.imgur.com/7OnspSN.png\")
", "date_published": "2016-06-19T08:37:06Z", "authors": ["Tsvi BT", "Eric Rogstad", "Team Arbital"], "summaries": [], "tags": [], "alias": "496"}
{"id": "363519aa663c58390f7c6602890a0db0", "title": "Symmetric group", "url": "https://arbital.com/p/symmetric_group", "source": "arbital", "source_type": "text", "text": "The notion that group theory captures the idea of \"symmetry\" derives from the notion of the symmetric group, and the very important theorem due to Cayley that every group is a subgroup of a symmetric group.\n\n# Definition\n\nLet $X$ be a [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz). A [bijection](https://arbital.com/p/499) $f: X \\to X$ is a *permutation* of $X$.\nWrite $\\mathrm{Sym}(X)$ for the set of permutations of the set $X$ (so its elements are functions).\n\nThen $\\mathrm{Sym}(X)$ is a group under the operation of composition of functions; it is the *symmetric group on $X$*.\n(It is also written $\\mathrm{Aut}(X)$, for the *automorphism group*.)\n\nWe write $S_n$ for $\\mathrm{Sym}(\\{ 1,2, \\dots, n\\})$, the *symmetric group on $n$ elements*.\n\n# Elements of $S_n$\n\nWe can represent a permutation of $\\{1,2,\\dots, n\\}$ in two different ways, each of which is useful in different situations.\n\n## Double-row notation\n\nLet $\\sigma \\in S_n$, so $\\sigma$ is a function $\\{1,2,\\dots,n\\} \\to \\{1,2,\\dots,n\\}$.\nThen we write $$\\begin{pmatrix}1 & 2 & \\dots & n \\\\ \\sigma(1) & \\sigma(2) & \\dots & \\sigma(n) \\\\ \\end{pmatrix}$$\nfor $\\sigma$.\nThis has the advantage that it is immediately clear where every element goes, but the disadvantage that it is quite hard to see the properties of an element when it is written in double-row notation (for example, \"$\\sigma$ cycles round five elements\" is hard to spot at a glance), and it is not very compact.\n\n## Cycle notation\n\n[Cycle notation](https://arbital.com/p/49f) is a different notation, which has the advantage that it is easy to determine an element's order and to get a general sense of what the element does.\nEvery element of $S_n$ [can be expressed in ](https://arbital.com/p/49k).\n\n## Product of transpositions\n\nIt is a useful fact that every permutation in a (finite) symmetric group [may be expressed](https://arbital.com/p/4cp) as a product of [transpositions](https://arbital.com/p/4cn).\n\n# Examples\n\n- The group $S_1$ is the group of permutations of a one-point set. It contains the identity only, so $S_1$ is the trivial group.\n- The group $S_2$ is isomorphic to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) of order $2$. It contains the identity map and the map which interchanges $1$ and $2$.\n\nThose are the only two [abelian](https://arbital.com/p/3h2) symmetric groups.\nIndeed, in cycle notation, $(123)$ and $(12)$ do not commute in $S_n$ for $n \\geq 3$, because $(123)(12) = (13)$ while $(12)(123) = (23)$.\n\n- The group $S_3$ contains the following six elements: the identity, $(12), (23), (13), (123), (132)$. It is isomorphic to the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) $D_6$ on three vertices. ([Proof.](https://arbital.com/p/group_s3_isomorphic_to_d6))\n\n# Why we care about the symmetric groups\n\nA very important (and rather basic) result is [Cayley's Theorem](https://arbital.com/p/49b), which states the link between group theory and symmetry.\n\n%%%knows-requisite([https://arbital.com/p/4bj](https://arbital.com/p/4bj)):\n# Conjugacy classes of $S_n$\n\nIt is a useful fact that the conjugacy class of an element in $S_n$ is precisely the set of elements which share its [cycle type](https://arbital.com/p/4cg). ([Proof.](https://arbital.com/p/4bh))\nWe can therefore [list the conjugacy classes](https://arbital.com/p/4bk) of $S_5$ and their sizes.\n%%%\n\n# Relationship to the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf)\n\nThe [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$ is defined as the collection of elements of $S_n$ which can be made by an even number of [transpositions](https://arbital.com/p/4cn). This does form a group ([proof](https://arbital.com/p/4hg)).\n\n%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):\nIn fact $A_n$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $S_n$, obtained by taking the quotient by the [sign homomorphism](https://arbital.com/p/4hk).\n%%%", "date_published": "2016-06-17T13:13:31Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Needs summary"], "alias": "497"}
{"id": "fdd553b3eef91d9ca4bf364e3b4e331c", "title": "Bijective function", "url": "https://arbital.com/p/bijective_function", "source": "arbital", "source_type": "text", "text": "A bijective function is a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) which has an inverse. Equivalently, it is both [injective](https://arbital.com/p/injective_function) and [surjective](https://arbital.com/p/surjective_function).", "date_published": "2016-06-14T13:11:21Z", "authors": ["Eric Rogstad", "Mark Chimes", "Patrick Stevens"], "summaries": [], "tags": ["Definition", "Stub"], "alias": "499"}
{"id": "c34c416412b8dba17dc1fefed02e88a1", "title": "Cayley's Theorem on symmetric groups", "url": "https://arbital.com/p/cayley_theorem_symmetric_groups", "source": "arbital", "source_type": "text", "text": "Cayley's Theorem states that every group $G$ appears as a certain subgroup of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $\\mathrm{Sym}(G)$ on the [https://arbital.com/p/-3gz](https://arbital.com/p/-3gz) of $G$.\n\n# Formal statement\n\nLet $G$ be a group.\nThen $G$ is [isomorphic](https://arbital.com/p/49x) to a subgroup of $\\mathrm{Sym}(G)$.\n\n# Proof\n\nConsider the left regular [action](https://arbital.com/p/3t9) of $G$ on $G$: that is, the function $G \\times G \\to G$ given by $(g, h) \\mapsto gh$.\nThis [induces a homomorphism](https://arbital.com/p/49c) $\\Phi: G \\to \\mathrm{Sym}(G)$ given by [https://arbital.com/p/-currying](https://arbital.com/p/-currying): $g \\mapsto (h \\mapsto gh)$.\n\nNow the following are equivalent:\n\n- $g \\in \\mathrm{ker}(\\Phi)$ the [kernel](https://arbital.com/p/49y) of $\\Phi$\n- $(h \\mapsto gh)$ is the identity map\n- $gh = h$ for all $h$\n- $g$ is the identity of $G$\n\nTherefore the kernel of the homomorphism is trivial, so it is injective.\nIt is therefore bijective onto its [image](https://arbital.com/p/3lh), and hence an isomorphism onto its image.\n\nSince [the image of a group under a homomorphism is a subgroup of the codomain of the homomorphism](https://arbital.com/p/4b4), we have shown that $G$ is isomorphic to a subgroup of $\\mathrm{Sym}(G)$.", "date_published": "2016-06-15T07:29:23Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "49b"}
{"id": "9320a016b71508f1653f365bd159d043", "title": "Group action induces homomorphism to the symmetric group", "url": "https://arbital.com/p/group_action_induces_homomorphism", "source": "arbital", "source_type": "text", "text": "Just as we can [curry](https://arbital.com/p/currying) functions, so we can \"curry\" [homomorphisms](https://arbital.com/p/47t) and [actions](https://arbital.com/p/3t9).\n\nGiven an action $\\rho: G \\times X \\to X$ of group $G$ on set $X$, we can consider what happens if we fix the first argument to $\\rho$. Writing $\\rho(g)$ for the induced map $X \\to X$ given by $x \\mapsto \\rho(g, x)$, we can see that $\\rho(g)$ is a [bijection](https://arbital.com/p/499).\n\nIndeed, we claim that $\\rho(g^{-1})$ is an inverse map to $\\rho(g)$.\nConsider $\\rho(g^{-1})(\\rho(g)(x))$.\nThis is precisely $\\rho(g^{-1})(\\rho(g, x))$, which is precisely $\\rho(g^{-1}, \\rho(g, x))$.\nBy the definition of an action, this is just $\\rho(g^{-1} g, x) = \\rho(e, x) = x$, where $e$ is the group's identity.\n\nWe omit the proof that $\\rho(g)(\\rho(g^{-1})(x)) = x$, because it is nearly identical.\n\nThat is, we have proved that $\\rho(g)$ is in $\\mathrm{Sym}(X)$, where $\\mathrm{Sym}$ is the [https://arbital.com/p/-497](https://arbital.com/p/-497); equivalently, we can view $\\rho$ as mapping elements of $G$ into $\\mathrm{Sym}(X)$, as well as our original definition of mapping elements of $G \\times X$ into $X$.\n\n# $\\rho$ is a homomorphism in this new sense\n\nIt turns out that $\\rho: G \\to \\mathrm{Sym}(X)$ is a homomorphism.\nIt suffices to show that $\\rho(gh) = \\rho(g) \\rho(h)$, where recall that the operation in $\\mathrm{Sym}(X)$ is composition of permutations.\n\nBut this is true: $\\rho(gh)(x) = \\rho(gh, x)$ by definition of $\\rho(gh)$; that is $\\rho(g, \\rho(h, x))$ because $\\rho$ is a group action; that is $\\rho(g)(\\rho(h, x))$ by definition of $\\rho(g)$; and that is $\\rho(g)(\\rho(h)(x))$ by definition of $\\rho(h)$ as required.", "date_published": "2016-06-14T15:05:26Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "49c"}
{"id": "6efa7fb8e0b518dd8ca9194423356a97", "title": "Cycle notation in symmetric groups", "url": "https://arbital.com/p/cycle_notation_symmetric_group", "source": "arbital", "source_type": "text", "text": "There is a convenient way to represent the elements of a [https://arbital.com/p/-497](https://arbital.com/p/-497) on a finite set.\n\n# $k$-cycle\n\nA $k$-cycle is a member of $S_n$ which moves $k$ elements to each other cyclically.\nThat is, letting $a_1, \\dots, a_k$ be distinct in $\\{1,2,\\dots,n\\}$, a $k$-cycle $\\sigma$ is such that $\\sigma(a_i) = a_{i+1}$ for $1 \\leq i < k$, and $\\sigma(a_k) = a_1$, and $\\sigma(x) = x$ for any $x \\not \\in \\{a_1, \\dots, a_k \\}$.\n\nWe have a much more compact notation for $\\sigma$ in this case: we write $\\sigma = (a_1 a_2 \\dots a_k)$.\n(If spacing is ambiguous, we put in commas: $\\sigma = (a_1, a_2, \\dots, a_k)$.)\nNote that there are several ways to write this: $(a_1 a_2 \\dots a_k) = (a_2 a_3 \\dots a_k a_1)$, for example.\nIt is conventional to put the smallest $a_i$ at the start.\n\nNote also that a cycle's inverse is extremely easy to find: the inverse of $(a_1 a_2 \\dots a_k)$ is $(a_k a_{k-1} \\dots a_1)$.\n\nFor example, the double-row notation $$\\begin{pmatrix}1 & 2 & 3 \\\\ 2 & 3 & 1 \\\\ \\end{pmatrix}$$\nis written as $(123)$ or $(231)$ or $(312)$ in cycle notation.\n\nHowever, it is unclear without context which symmetric group $(123)$ lies in: it could be $S_n$ for any $n \\geq 3$.\nSimilarly, $(145)$ could be in $S_n$ for any $n \\geq 5$.\n\n# General elements, not just cycles\n\nNot every element of $S_n$ is a cycle. For example, the following element of $S_4$ has [order](https://arbital.com/p/4cq) $2$ so could only be a $2$-cycle, but it moves all four elements:\n$$\\begin{pmatrix}1 & 2 & 3 & 4 \\\\ 2 & 1 & 4 & 3 \\\\ \\end{pmatrix}$$\n\nHowever, it may be written as the composition of the two cycles $(12)$ and $(34)$: it is the result of applying one and then the other.\nNote that since the cycles are disjoint (having no elements in common), [it doesn't matter in which order we perform them](https://arbital.com/p/49g).\nIt is a very important fact that [every permutation may be written as the product of disjoint cycles](https://arbital.com/p/49k).\nIf $\\sigma$ is a permutation obtained by first doing cycle $c_1 = (a_1 a_2 \\dots a_k)$, then by doing cycle $c_2$, then cycle $c_3$, we write $\\sigma = c_3 c_2 c_1$; this is by analogy with function composition, indicating that the first permutation to apply is on the rightmost end of the expression.\n(Be aware that some authors differ on this.)\n\n## Order of an element\n\nFirstly, a cycle has [order](https://arbital.com/p/4cq) equal to its length.\nIndeed, the cycle $(a_1 a_2 \\dots a_k)$ has the effect of rotating $a_1 \\mapsto a_2 \\mapsto a_3 \\dots \\mapsto a_k \\mapsto a_1$, and if we do this $k$ times we get back to where we started.\n(And if we do it fewer times - say $i$ times - we can't get back to where we started: $a_1 \\mapsto a_{i+1}$.)\n\nNow, suppose we have an element in disjoint cycle notation: $(a_1 a_2 a_3)(a_4 a_5)$, say, where all the $a_i$ are different.\nThen the order of this element is $3 \\times 2 = 6$, because: \n\n- $(a_1 a_2 a_3)$ and $(a_4 a_5)$ are disjoint and hence commute, so $[https://arbital.com/p/(](https://arbital.com/p/()^n = (a_1 a_2 a_3)^n (a_4 a_5)^n$\n- $(a_1 a_2 a_3)^n (a_4 a_5)^n$ is the identity if and only if $(a_1 a_2 a_3)^n = (a_4 a_5)^n = e$ the identity, because otherwise (for instance, if $(a_1 a_2 a_3)^n$ is not the identity) it would move $a_1$.\n- $(a_1 a_2 a_3)^n$ is the identity if and only if $n$ is divisible by $3$, since $(a_1 a_2 a_3)$'s order is $3$.\n- $(a_4 a_5)^n$ is the identity if and only if $n$ is divisible by $2$.\n\nThis reasoning generalises: the order of an element in disjoint cycle notation is equal to the [https://arbital.com/p/-least_common_multiple](https://arbital.com/p/-least_common_multiple) of the lengths of the cycles.\n\n# Examples\n\n- The element $\\sigma$ of $S_5$ given by first performing $(123)$ and then $(345)$ is $(345)(123) = (12453)$. Indeed, the first application takes $1$ to $2$ and the second application does not affect the resulting $2$, so $\\sigma$ takes $1$ to $2$; the first application takes $2$ to $3$ and the second application takes the resulting $3$ to $4$, so $\\sigma$ takes $2$ to $4$; the first application does not affect $4$ and the second application takes $4$ to $5$, so $\\sigma$ takes $4$ to $5$; and so on.\n\nThis example suggests a general procedure for expressing a permutation which is already in cycle form, in *disjoint* cycle form. It turns out that [this can be done in an essentially unique way](https://arbital.com/p/49k).\n\n## Cycle type\n\nThe [https://arbital.com/p/-4cg](https://arbital.com/p/-4cg) is given by taking the list of lengths of the cycles in the disjoint cycle form.", "date_published": "2016-06-15T08:17:16Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "49f"}
{"id": "b20eef2d1a9ea3a5fe7d87f941754f13", "title": "Disjoint cycles commute in symmetric groups", "url": "https://arbital.com/p/disjoint_cycles_commute_symmetric_group", "source": "arbital", "source_type": "text", "text": "Consider two [cycles](https://arbital.com/p/49f) $(a_1 a_2 \\dots a_k)$ and $(b_1 b_2 \\dots b_m)$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, where all the $a_i, b_j$ are distinct.\n\nThen it is the case that the following two elements of $S_n$ are equal:\n\n- $\\sigma$, which is obtained by first performing the permutation notated by $(a_1 a_2 \\dots a_k)$ and then by performing the permutation notated by $(b_1 b_2 \\dots b_m)$\n- $\\tau$, which is obtained by first performing the permutation notated by $(b_1 b_2 \\dots b_m)$ and then by performing the permutation notated by $(a_1 a_2 \\dots a_k)$\n\nIndeed, $\\sigma(a_i) = (b_1 b_2 \\dots b_m)[https://arbital.com/p/(](https://arbital.com/p/() = (b_1 b_2 \\dots b_m)(a_{i+1}) = a_{i+1}$ (taking $a_{k+1}$ to be $a_1$), while $\\tau(a_i) = (a_1 a_2 \\dots a_k)[https://arbital.com/p/(](https://arbital.com/p/() = (a_1 a_2 \\dots a_k)(a_i) = a_{i+1}$, so they agree on elements of $(a_1 a_2 \\dots a_k)$.\nSimilarly they agree on elements of $(b_1 b_2 \\dots b_m)$; and they both do not move anything which is not an $a_i$ or a $b_j$.\nHence they are the same permutation: they act in the same way on all elements of $\\{1,2,\\dots, n\\}$.\n\nThis reasoning generalises to more than two disjoint cycles, to show that disjoint cycles commute.", "date_published": "2016-06-14T14:53:51Z", "authors": ["Patrick Stevens"], "summaries": ["In a symmetric group, if we are applying a collection of permutations which are each disjoint cycles, we get the same result no matter the order in which we perform the cycles."], "tags": [], "alias": "49g"}
{"id": "e6db08fdda06bc0155b06b59630af9c9", "title": "Disjoint cycle notation is unique", "url": "https://arbital.com/p/disjoint_cycle_notation_is_unique", "source": "arbital", "source_type": "text", "text": "", "date_published": "2016-06-14T14:34:15Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Stub"], "alias": "49k"}
{"id": "85920bd6b70bccbf2e87f7a9fb71da82", "title": "Pi", "url": "https://arbital.com/p/pi", "source": "arbital", "source_type": "text", "text": "Pi, usually written $π$, is a number equal to the ratio of a circle's [https://arbital.com/p/-circumference](https://arbital.com/p/-circumference) to its [https://arbital.com/p/-diameter](https://arbital.com/p/-diameter). The value of $π$ is approximately $3.141593$.\n\n\n\nIf the length of a curve seems like an ill-defined concept to you (maybe you only understand how lines could have lengths), consider bigger and bigger [regular polygons](https://arbital.com/p/-regular_polygon) that make better and better approximations of the circle. As the number of sides $N$ of the polygon goes to $∞$, the perimeter will approach a length of $π$ times the diameter.\n\nOne could also define $π$ to be the area of a circle divided the area of a square, whose edge is the radius of the circle.\n\n## What Kind of Number It Is ##\n\nIt's not an [https://arbital.com/p/48l](https://arbital.com/p/48l).\n\n\n\nIf the diameter here is 1, then the perimeter of the hexagon is 3, the perimeter of the square is 4, and the circumference of the circle is in between. There are no integers between 3 and 4.\n\nIt's [not rational](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational), [nor is it algebraic](https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem#Transcendence_of_e_and_.CF.80). It's transcendental.", "date_published": "2016-07-21T16:02:45Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Michael Cohen"], "summaries": [], "tags": [], "alias": "49r"}
{"id": "37d26663af702784762e2d3397557c75", "title": "Complexity theory", "url": "https://arbital.com/p/complexity_theory", "source": "arbital", "source_type": "text", "text": "**Complexity theory** is the study of the efficiency of algorithms with respect to several metrics, usually time and memory usage. Complexity theorists aim to classify different problems into [classes of difficulty](https://arbital.com/p/complexity_class) and study the relations that hold between the classes.\n\nWhen studying [computability](https://arbital.com/p/), we are concerned with the identification of that which is or not computable in an ideal sense, without worrying about time or memory limitations.\n\nHowever, often in practice we have to be more pragmatic. A program which takes a [googol](https://arbital.com/p/42m) years to run is not going to see much use. If you need more [GbBytes](https://arbital.com/p/) to solve a computational problem that atoms exist in the universe, you may as well go ahead and declare the problem unsolvable for all practical purposes.\n\n**Complexity theory** raises the standards of computability in drawing the boundary between that which you can do with a computer and that which you cannot. It concerns the study of the [asymptotic behavior](https://arbital.com/p/oh_notation) of programs when fed inputs of growing size, in terms of the resources they consume. The kind of resources with which complexity theorists work more often are the *time* a program takes to finish and the highest *memory usage* %%note:the memory usage is frequently called **space complexity**%% in any given point of the execution.\n\nComplexity theory allows us to have a deeper understanding of what makes an algorithm efficient, which in turn allows us to develop better and faster algorithms. Surprisingly enough, it turns out that proving that some computational problems are hard to solve has incredibly important practical applications in [https://arbital.com/p/-cryptography](https://arbital.com/p/-cryptography).\n\n----\n\nThe abstract framework in which we develop this theory are [Turing machines](https://arbital.com/p/) and [decision problems](https://arbital.com/p/). In this context, the [time complexity](https://arbital.com/p/) is associated with the number of steps a TM takes to halt and return an output, while the [space complexity](https://arbital.com/p/) corresponds to the length of the tape we would need for the TM to never fall off when moving left or right.\n\nOne may worry that the complexity measures are highly dependent on the choice of computational framework used. After all, if we allow our TM to skip two spaces per step each time it moves it is going to take potentially half the time to compute something. However, the asymptotic behavior of complexity is [surprisingly robust](https://arbital.com/p/robustness_of_tm), though there are [some caveats](https://arbital.com/p/failure_of_strong_CT).\n\nThe most interesting characterization of complexity comes in the form of [complexity classes](https://arbital.com/p/), which break down the family of [decision problems](https://arbital.com/p/) into sets of problems which can be solved with particular constraints.\n\nPerhaps the most important complexity class is [$P$](https://arbital.com/p/), the class of decision problems which can be efficiently computed %%note:For example, checking whether a graph is connected or not%%. The second best known class is [$NP$](https://arbital.com/p/), the class of problems whose solutions can be easily checked %%note:An example is factoring a number: it is hard to factor $221$, but easy to multiply $13$ and $17$ and check that $13 \\cdot 17 = 221$%% . It is an open problem whether those two classes are [one and the same](https://arbital.com/p/4bd); that is, that every problem whose solutions are easy to check is also easy to solve.\n\nThere are many more important complexity classes, and it can be daunting to contemplate the sheer variety with which complexity theory deals. Feel free to take a guided tour though the [complexity zoo](https://arbital.com/p/4b9) if you want an overview of some of the most relevant.\n\nAn important concept is that of a [reduction](https://arbital.com/p/). Some complexity classes have problems such that if you were able to solve them efficiently you could translate other problems in the class to this one and solve them efficiently. Those are called [complete problems of a complexity class](https://arbital.com/p/).\n\n-----\n\nThis page is meant to be a starting point to learn complexity theory from an entry level. If there is any concept which feels mysterious to you, try exploring the greenlinks in their order of appearance. If you feel like the concepts presented are too basic, try a different lens.", "date_published": "2016-10-08T15:56:04Z", "authors": ["Eric Bruylant", "Adom Hartell", "Eric Rogstad", "Jaime Sevilla Molina"], "summaries": ["Study of the computational resources needed to solve a problem, usually time and memory"], "tags": [], "alias": "49w"}
{"id": "0e4add28e6c0c6e4aac88a2d6b57a6cf", "title": "Group isomorphism", "url": "https://arbital.com/p/group_isomorphism", "source": "arbital", "source_type": "text", "text": "A group isomorphism is a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) which is [bijective](https://arbital.com/p/499).\nWe say that two groups are *isomorphic* if there is an isomorphism between them.\n\nIt turns out that isomorphism is a much more useful concept than true equality of [groups](https://arbital.com/p/-3gd), and it captures the idea that \"these two objects are the same group\": the isomorphism shows us how to relabel the elements to see that they are indeed the same group.\n\nFor example, the trivial group is in some sense \"the only group with one element\", but it can be instantiated in many different ways: as $(\\{ a \\}, +_a)$, or $(\\{ b \\}, +_b)$, and so on (where $+_x$ is the [https://arbital.com/p/-3kb](https://arbital.com/p/-3kb) taking $(x, x)$ to $x$).\nThey all behave in exactly the same ways for the purpose of group theory, but they are not literally identical.\nThey are all isomorphic, though: the map $\\{a \\} \\to \\{ b \\}$ given by $a \\mapsto b$ is an isomorphism of the respective groups.\n\nTwo groups are isomorphic if and only if they have the same [Cayley table](https://arbital.com/p/cayley_table), possibly with rearrangement of rows/columns and with relabelling of elements.", "date_published": "2016-06-15T06:30:30Z", "authors": ["Eric Bruylant", "Mark Chimes", "Patrick Stevens"], "summaries": [], "tags": ["Math 2", "C-Class", "Proposed B-Class"], "alias": "49x"}
{"id": "45ad5caf675bfd06f5dfc337c978ab68", "title": "Kernel of group homomorphism", "url": "https://arbital.com/p/kernel_of_group_homomorphism", "source": "arbital", "source_type": "text", "text": "The kernel of a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \\to H$ is the collection of all elements $g$ in $G$ such that $f(g) = e_H$ the identity of $H$.\n\nIt is important to note that the kernel of any group homomorphism $G \\to H$ is always a subgroup of $G$.\nIndeed:\n\n- if $f(g_1) = e_H$ and $f(g_2) = e_H$ then $e_H = f(g_1) f(g_2) = f(g_1 g_2)$, so the kernel is closed under $G$'s operation;\n- if $f(x) = e_H$ then $e_H = f(e_G) = f(x^{-1} x) = f(x^{-1}) f(x) = f(x^{-1})$ (where we have used that [the image of the identity is the identity](https://arbital.com/p/49z)), so inverses are contained in the putative subgroup;\n- $f(e_G) = e_H$ because the image of the identity is the identity, so the identity is contained in the putative subgroup.\n\nIt turns out that the notion of \"[https://arbital.com/p/-4h6](https://arbital.com/p/-4h6)\" coincides exactly with the notion of \"kernel of homomorphism\". ([Proof.](https://arbital.com/p/4h7))\nThe \"kernel of homomorphism\" viewpoint of normal subgroups is much more strongly motivated from the point of view of [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7); Timothy Gowers [considers this to be the correct way](https://gowers.wordpress.com/2011/11/20/normal-subgroups-and-quotient-groups/) to introduce the teaching of normal subgroups in the first place.", "date_published": "2016-06-17T13:06:36Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Needs clickbait", "Definition"], "alias": "49y"}
{"id": "1b5adddc1c243bd446e35420a846b442", "title": "Image of the identity under a group homomorphism is the identity", "url": "https://arbital.com/p/image_of_identity_under_group_homomorphism", "source": "arbital", "source_type": "text", "text": "For any [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \\to H$, we have $f(e_G) = e_H$ where $e_G$ is the identity of $G$ and $e_H$ the identity of $H$.\n\nIndeed, $f(e_G) f(e_G) = f(e_G e_G) = f(e_G)$, so premultiplying by $f(e_G)^{-1}$ we obtain $f(e_G) = e_H$.", "date_published": "2016-06-14T17:21:36Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "49z"}
{"id": "0796e7704b28a88161136960efce6b00", "title": "Under a group homomorphism, the image of the inverse is the inverse of the image", "url": "https://arbital.com/p/group_homomorphism_image_of_inverse", "source": "arbital", "source_type": "text", "text": "For any [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \\to H$, we have $f(g^{-1}) = f(g)^{-1}$.\n\nIndeed, $f(g^{-1}) f(g) = f(g^{-1} g) = f(e_G) = e_H$, and similarly for multiplication on the left.", "date_published": "2016-06-14T17:26:14Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4b1"}
{"id": "876b303b22ac6f89751ca041999ef0ee", "title": "The image of a group under a homomorphism is a subgroup of the codomain", "url": "https://arbital.com/p/image_of_group_under_homomorphism_is_subgroup", "source": "arbital", "source_type": "text", "text": "Let $f: G \\to H$ be a [https://arbital.com/p/-47t](https://arbital.com/p/-47t), and write $f(G)$ for the set $\\{ f(g) : g \\in G \\}$.\nThen $f(G)$ is a group under the operation inherited from $H$.\n\n# Proof\n\nTo prove this, we must verify the group axioms.\nLet $f: G \\to H$ be a group homomorphism, and let $e_G, e_H$ be the identities of $G$ and of $H$ respectively.\nWrite $f(G)$ for the image of $G$.\n\nThen $f(G)$ is closed under the operation of $H$: since $f(g) f(h) = f(gh)$, so the result of $H$-multiplying two elements of $f(G)$ is also in $f(G)$.\n\n$e_H$ is the identity for $f(G)$: it is $f(e_G)$, so it does lie in the image, while it acts as the identity because $f(e_G) f(g) = f(e_G g) = f(g)$, and likewise for multiplication on the right.\n\nInverses exist, by \"the inverse of the image is the image of the inverse\".\n\nThe operation remains associative: this is inherited from $H$.\n\nTherefore, $f(G)$ is a group, and indeed is a subgroup of $H$.", "date_published": "2016-06-14T17:30:27Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4b4"}
{"id": "4a4fe38fe888bf3d3093737625ad39fc", "title": "Expected value", "url": "https://arbital.com/p/expected_value", "source": "arbital", "source_type": "text", "text": "The expected value of an action is the [https://arbital.com/p/-mean](https://arbital.com/p/-mean) numerical outcome of the possible results weighted by their [https://arbital.com/p/-1rf](https://arbital.com/p/-1rf). It may actually be impossible to get the expected value, for example, if a coin toss decides between you getting \\$0 and \\$10, then we say you get \"\\$5 in expectation\" even though there is no way for you to get \\$5.\n\nThe expectation of V (often shortened to \"the expected V\") is how much V you expect to get on average. For example, the expectation of a payoff, or an expected payoff, is how much money you will get on average; the expectation of the duration of a speech, or an expected duration, is how long the speech will last \"on average.\"\n\nSuppose V has discrete possible values, say $V = x_{1},$ or $V = x_{2}, ..., $ or $V = x_{k}$. Let $P(x_{i})$ refer to the probability that $V = x_{i}$. Then the expectation of V is given by:\n\n$$\\sum_{i=1}^{k}x_{i}P(x_{i})$$\n\nSuppose V has continuous possible values, x. For instance, let $x \\in \\mathbb{R}$. Let $P(x)$ be the continuous probability distribution, or $\\lim_{dx \\to 0}$ of the probability that $x![](\"http://i.imgur.com/0off1db.png\")
\n\n\n\n\n\nThis is what independence looks like, using the [square visualization](https://arbital.com/p/496) of probabilities:\n\n\n\nWe can see that the [events](https://arbital.com/p/event_probability) $A$ and $B$ don't interact; we say that $A$ and $B$ are *independent*. Whether we look at the whole square, or just the red part of\nthe square where $A$ is true, the probability of $B$ stays the same. In other words, $\\bP(B \\mid A) = \\bP(B)$. That's what we mean by independence: the\nprobability of $B$ doesn't change if you condition on $A$.\n\nOur square of probabilities can be generated by multiplying together the probability of $A$ and the probability of $B$:\n\n![](\"http://i.imgur.com/pjwcoTn.png\")
\n\nThis picture demonstrates another way to define what it means for $A$ and $B$ to be independent:\n\n$$\\bP(A, B) = \\bP(A)\\bP(B)\\ .$$\n\n\n\nIn terms of factoring a joint distribution\n--\n\nLet's contrast independence with non-independence. Here's a picture of two ordinary, non-independent events $A$ and $B$:\n\n![](\"http://i.imgur.com/6ZHSR0l.png\")
\n\n(If the meaning of this picture isn't clear, take a look at [https://arbital.com/p/496](https://arbital.com/p/496).)\n\nWe have the red blocks for $\\bP(A)$ and the blue blocks for $\\bP(\\neg A)$ lined up in columns. This means we've [factored](https://arbital.com/p/factoring_probability) our\nprobability distribution using $A$ as the first factor: \n\n$$\\bP(A,B) = \\bP(A) \\bP(B \\mid A)\\ .$$\n\nWe could just as well have factored by $B$ first: $\\bP(A,B) = \\bP(B) \\bP( A \\mid B)\\ .$ Then we'd draw a picture like this:\n\n\n![](\"http://i.imgur.com/O0RNzxw.png\")
\n\n\n\n\nNow, here again is the picture of [two independent events](https://arbital.com/p/4cf) $A$ and $B$:\n\n\n\n\n\nIn this picture, there's red and blue lined-up columns for $\\bP(A)$ and $\\bP(\\neg A)$, and there's *also* dark and light lined-up rows for $\\bP(B)$ and\n$\\bP(\\neg B)$. It looks like we somehow [factored](https://arbital.com/p/factoring_probability) our probability distribution $\\bP$ using both $A$ and \n$B$ as the first factor. \n\nIn fact, this is exactly what happened: since $A$ and $B$ are [independent](https://arbital.com/p/4cf), we have that $\\bP(B \\mid A) = \\bP(B)$. So the diagram\nabove is actually factored according to $A$ first: $\\bP(A,B) = \\bP(A) \\bP(B \\mid A)$. It's just that $\\bP(B \\mid A)= \\bP(B) = \\bP(B \\mid \\neg A)$, since $B$\nis independent from $A$. So we don't need to have different ratios of dark to light (a.k.a. conditional probabilities of $B$) in the left and right columns:\n\n![](\"http://i.imgur.com/Nfiuz3d.png\")
\n\nIn this visualization, we can see what happens to the probability of $B$ when you condition on $A$ or on $\\neg A$: it doesn't change at all. The ratio of\n\\[area where $B$ happens\\](https://arbital.com/p/the) to \\[whole area\\](https://arbital.com/p/the), is the same as the ratio $\\bP(B \\mid A)$ where we only look at the area where $A$ happens, which is the\nsame as the ratio $\\bP(B \\mid \\neg A)$ where we only look at the area where $\\neg A$ happens. The fact that the probability of $B$ doesn't change when we\ncondition on $A$ is exactly what we mean when we say that $A$ and $B$ are independent.\n\nThe square diagram above is *also* factored according to $B$ first, using $\\bP(A,B) = \\bP(B) \\bP(A \\mid B)$. The red / blue ratios are the same in both rows\nbecause $\\bP(A \\mid B) = \\bP(A) = \\bP(A \\mid \\neg B)$, since $A$ and $B$ are independent:\n\n![](\"http://i.imgur.com/DfDljOL.png\")
\n\nWe couldn't do any of this stuff if the columns and rows didn't both line up. (Which is good, because then we'd have proved the false statement that any two\nevents are independent!)\n\nIn terms of multiplying marginal probabilities\n---\n\nAnother way to say that $A$ and $B$ are independent variables %note:We're using the [equivalence](https://arbital.com/p/event_variable_equivalence) between [https://arbital.com/p/event_probability\nevents](https://arbital.com/p/event_probability\nevents) and [binary variables](https://arbital.com/p/binary_random_variable).% is that for any truth values $t_A,t_B \\in \\{\\true, \\false\\},$\n\n$$\\bP(A = t_A, B= t_B) = \\bP(A = t_A)\\bP(B = t_B)\\ .$$\n\n\n\nSo the [joint probabilities](https://arbital.com/p/1rh) for $A$ and $B$ are computed by separately getting the probability of $A$ and the probability of $B$, and then\nmultiplying the two probabilities together. For example, say we want to compute the probability $\\bP(A, \\neg B) = \\bP(A = \\true, B = \\false)$. We start with\nthe [marginal probability](https://arbital.com/p/marginal_probability) of $A$:\n\n![](\"http://i.imgur.com/ZnxqSMo.png\")
\n\nand the probability of $\\neg B$:\n\n![](\"http://i.imgur.com/txRlJyE.png\")
\n\nand then we multiply them:\n\n![](\"http://i.imgur.com/GOOnTuF.png\")
\n\n\nWe can get all the joint probabilities this way. So we can visualize the whole joint distribution as the thing that you get when you multiply two independent\nprobability distributions together. We just overlay the two distributions: \n\n![](\"http://i.imgur.com/X4FSciB.png\")
\n\nTo be a little more mathematically elegant, we'd use the [topological product of two spaces](https://arbital.com/p/topological_product) shown earlier to draw the joint distribution\nas a product of the distributions of $A$ and $B$: \n\n", "date_published": "2016-06-16T14:55:54Z", "authors": ["Eric Rogstad", "Jaime Sevilla Molina", "Tsvi BT"], "summaries": [], "tags": [], "alias": "4cl"}
{"id": "5d3f34fec2e732a3b29415719a899b53", "title": "Transposition (as an element of a symmetric group)", "url": "https://arbital.com/p/transposition_in_symmetric_group", "source": "arbital", "source_type": "text", "text": "In a [https://arbital.com/p/-497](https://arbital.com/p/-497), a transposition is a permutation which has the effect of swapping two elements while leaving everything else unchanged.\nMore formally, it is a permutation of [order](https://arbital.com/p/order_of_a_group_element) $2$ which fixes all but two elements.\n\n%%%knows-requisite([https://arbital.com/p/4cg](https://arbital.com/p/4cg)):\nA transposition is precisely an element with cycle type $2$.\n%%%\n\n# Example\n\nIn $S_5$, the permutation $(12)$ is a transposition: it swaps $1$ and $2$ while leaving all three of the elements $3,4,5$ unchanged.\nHowever, the permutation $(124)$ is not a transposition, because it has order $3$, not order $2$.", "date_published": "2016-06-15T07:50:47Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Definition"], "alias": "4cn"}
{"id": "c3aeeb562751eba078eed6a16e977000", "title": "Every member of a symmetric group on finitely many elements is a product of transpositions", "url": "https://arbital.com/p/symmetric_group_is_generated_by_transpositions", "source": "arbital", "source_type": "text", "text": "Given a permutation $\\sigma$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, there is a finite sequence $\\tau_1, \\dots, \\tau_k$ of [transpositions](https://arbital.com/p/4cn) such that $\\sigma = \\tau_k \\tau_{k-1} \\dots \\tau_1$.\nEquivalently, symmetric groups are generated by their transpositions.\n\nNote that the transpositions might \"overlap\".\nFor example, $(123)$ is equal to $(23)(13)$, where the element $3$ appears in two of the transpositions.\n\nNote also that the sequence of transpositions is by no means uniquely determined by $\\sigma$.\n\n# Proof\n\nIt is enough to show that a [cycle](https://arbital.com/p/49f) is expressible as a sequence of transpositions.\nOnce we have this result, we may simply replace the successive cycles in $\\sigma$'s disjoint cycle notation by the corresponding sequences of transpositions, to obtain a longer sequence of transpositions which multiplies out to give $\\sigma$.\n\nIt is easy to verify that the cycle $(a_1 a_2 \\dots a_r)$ is equal to $(a_{r-1} a_r) (a_{r-2} a_r) \\dots (a_2 a_r) (a_1 a_r)$.\nIndeed, that product of transpositions certainly does not move anything that isn't some $a_i$; while if we ask it to evaluate $a_i$, then the $(a_1 a_r)$ does nothing to it, $(a_2 a_r)$ does nothing to it, and so on up to $(a_{i-1} a_r)$.\nThen $(a_i a_r)$ sends it to $a_r$; then $(a_{i+1} a_r)$ sends the resulting $a_r$ to $a_{i+1}$; then all subsequent transpositions $(a_{i+2} a_r), \\dots, (a_{r-1} a_r)$ do nothing to the resulting $a_{i+1}$.\nSo the output when given $a_i$ is $a_{i+1}$.\n\n# Why is this useful?\n\nIt can make arguments simpler: if we can show that some property holds for transpositions and that it is closed under products, then it must hold for the entire symmetric group.", "date_published": "2016-06-15T08:03:48Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4cp"}
{"id": "f36ad5d7b3332f0c53ac21097ffa673a", "title": "Order of a group element", "url": "https://arbital.com/p/order_of_a_group_element", "source": "arbital", "source_type": "text", "text": "Given an element $g$ of group $(G, +)$ (which henceforth we abbreviate simply as $G$), the order of $g$ is the number of times we must add $g$ to itself to obtain the identity element $e$.\n\n%%%knows-requisite([https://arbital.com/p/3gg](https://arbital.com/p/3gg)):\nEquivalently, it is the order of the group $\\langle g \\rangle$ generated by $g$: that is, the order of $\\{ e, g, g^2, \\dots, g^{-1}, g^{-2}, \\dots \\}$ under the inherited group operation $+$.\n%%%\n\nConventionally, the identity element itself has order $1$.\n\n# Examples\n\n%%%knows-requisite([https://arbital.com/p/497](https://arbital.com/p/497)):\nIn the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$, the order of an element is the [https://arbital.com/p/-least_common_multiple](https://arbital.com/p/-least_common_multiple) of its [cycle type](https://arbital.com/p/4cg).\n%%%\n%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):\nIn the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_6$, the order of the generator is $6$.\nIf we view $C_6$ as being the integers [modulo](https://arbital.com/p/modular_arithmetic) $6$ under addition, then the element $0$ has order $1$; the elements $1$ and $5$ have order $6$; the elements $2$ and $4$ have order $3$; and the element $3$ has order $2$.\n%%%\n\nIn the group $\\mathbb{Z}$ of [integers](https://arbital.com/p/48l) under addition, every element except $0$ has infinite order. $0$ itself has order $1$, being the identity.", "date_published": "2016-06-15T08:14:47Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Formal definition", "Needs clickbait"], "alias": "4cq"}
{"id": "e23ed3406771bf79a24c5df6f5b322a2", "title": "Strong Church Turing thesis", "url": "https://arbital.com/p/strong_Church_Turing_thesis", "source": "arbital", "source_type": "text", "text": "> Every realistic model of computation is [polynomial time reducible](https://arbital.com/p/) to [probabilistic Turing machines](https://arbital.com/p/)\n\nWhich amounts to saying that every computable process in the universe can be efficiently simulated by a probabilistic Turing machine.\n\nThe definition of *realistic model* appeals to intuition rather than a precise definition of realistic. A rule of thumb is that a computation model is realistic if it could be used to accurately model some physical process. For example, there is a clear relation between the model of [register machines](https://arbital.com/p/) and the inner workings of personal computers. On the other hand, a computational model which can access an [NP oracle](https://arbital.com/p/) does not have a physical counterpart.\n\nAs it happened with the standard [https://arbital.com/p/4bw](https://arbital.com/p/4bw), this is an inductive rather than a mathematically defined statement. However, unlike the standard CT thesis, it is [widely believed to be wrong](https://arbital.com/p/), primarily because [quantum computation](https://arbital.com/p/) stands as a highly likely counterexample to the thesis.\n\nWe remark that non-deterministic computation is not a candidate counterexample to the strong CT thesis, since it is not realistic. That said, we can find a relation between the two concepts: if [$P=NP$](https://arbital.com/p/4bd) and [$BQP\\subset NP$](https://arbital.com/p/) then $BQP = P$, so quantum computation can be efficiently simulated, and we lose the strongest contestant for a counterexample to Strong CT.\n\n#Consequences of falsehood\n\nThe falsehood of the Strong CT thesis opens up the possibility of the existence of physical processes that, while computable, cannot be modeled in a reasonable time with a classical computer. One consequence of this, that would also mean that if quantum computing is possible the speed .up it provides is non-trivial.\n\nEpistemic processes which assign priors using a penalty for computational time complexity are misguided if the Strong CT thesis is false. See for example [Levin search](https://arbital.com/p/).\n\n [probably not worth raising? human brains have crazy-low decoherence times.](https://arbital.com/p/comment:)", "date_published": "2016-06-16T10:46:01Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Jaime Sevilla Molina", "Patrick Stevens"], "summaries": [], "tags": ["Start", "Needs summary"], "alias": "4ct"}
{"id": "f4742cf8ff797d734ab2fb95c2e6527f", "title": "Category (mathematics)", "url": "https://arbital.com/p/category_mathematics", "source": "arbital", "source_type": "text", "text": "A **category** consists of a collection of objects with morphisms between them. A morphism $f$ goes from one object, say $X$, to another, say $Y$, and is drawn as an arrow from $X$ to $Y$. Note that $X$ may equal $Y$ (in which case $f$ is referred to as an [https://arbital.com/p/-endomorphism](https://arbital.com/p/-endomorphism)). The object $X$ is called the source or [domain](https://arbital.com/p/3js) of $f$ and $Y$ is called the target or [codomain](https://arbital.com/p/3lg) of $f$, though note that $f$ itself need not be a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) and $X$ and $Y$ need not be sets. This is written as $f: X \\rightarrow Y$.\n\nThese morphisms must satisfy three conditions:\n\n 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \\rightarrow Y$ and $g: Y \\rightarrow Z$, there exists a morphism $X \\rightarrow Z$, written as $g \\circ f$ or simply $gf$. \n 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \\rightarrow Y$, $g: Y \\rightarrow Z$ and $h:Z \\rightarrow W$ composition is associative, i.e., $h(gf) = (hg)f$.\n 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \\rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \\rightarrow X$ and $g:X \\rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.", "date_published": "2016-06-18T05:53:09Z", "authors": ["Eric Bruylant", "Mark Chimes", "Patrick Stevens"], "summaries": [], "tags": ["Work in progress"], "alias": "4cx"}
{"id": "b156c0cb6663dd716ab7680f4f4afe7a", "title": "Dihedral group", "url": "https://arbital.com/p/dihedral_group", "source": "arbital", "source_type": "text", "text": "The dihedral group $D_{2n}$ is the group of symmetries of the $n$-vertex [https://arbital.com/p/-regular_polygon](https://arbital.com/p/-regular_polygon).\n\n# Presentation\nThe dihedral groups have very simple [presentations](https://arbital.com/p/group_presentation): $$D_{2n} \\cong \\langle a, b \\mid a^n, b^2, b a b^{-1} = a^{-1} \\rangle$$\nThe element $a$ represents a rotation, and the element $b$ represents a reflection in any fixed axis.\n\n\n# Properties\n\n- The dihedral groups $D_{2n}$ are all non-abelian for $n > 2$. ([Proof.](https://arbital.com/p/4d0))\n- The dihedral group $D_{2n}$ is a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, generated by the elements $a = (123 \\dots n)$ and $b = (2, n)(3, n-1) \\dots (\\frac{n}{2}+1, \\frac{n}{2}+3)$ if $n$ is even, $b = (2, n)(3, n-1)\\dots(\\frac{n-1}{2}, \\frac{n+1}{2})$ if $n$ is odd.\n\n# Examples\n\n## $D_6$, the group of symmetries of the triangle\n\n\n\n\n# Infinite dihedral group\n\nThe infinite dihedral group has presentation $\\langle a, b \\mid b^2, b a b^{-1} = a^{-1} \\rangle$.\nIt is the \"infinite-sided\" version of the finite $D_{2n}$.\n\nWe may view the infinite dihedral group as being the subgroup of the group of [homeomorphisms](https://arbital.com/p/homeomorphism) of $\\mathbb{R}^2$ generated by a reflection in the line $x=0$ and a translation to the right by one unit.\nThe translation is playing the role of a rotation in the finite $D_{2n}$.", "date_published": "2016-06-16T18:38:00Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Definition", "Stub"], "alias": "4cy"}
{"id": "63a4bf593bdd3a3c52ac566ff6f6d5c8", "title": "Dihedral groups are non-abelian", "url": "https://arbital.com/p/dihedral_groups_are_non_abelian", "source": "arbital", "source_type": "text", "text": "Let $n \\geq 3$. Then the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) on $n$ vertices, $D_{2n}$, is not [abelian](https://arbital.com/p/3h2).\n\n# Proof\n\nThe most natural dihedral group [presentation](https://arbital.com/p/group_presentation) is $\\langle a, b \\mid a^n, b^2, bab^{-1} = a^{-1} \\rangle$.\nIn particular, $ba = a^{-1} b = a^{-2} a b$, so $ab = ba$ if and only if $a^2$ is the identity.\nBut $a$ is the rotation which has order $n > 2$, so $ab$ cannot be equal to $ba$.", "date_published": "2016-06-15T13:06:13Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Proof"], "alias": "4d0"}
{"id": "6e46d9faf7086f3d55b7d4d64dd3f124", "title": "Needs image", "url": "https://arbital.com/p/needs_image_meta_tag", "source": "arbital", "source_type": "text", "text": "Meta tag for pages which would benefit from having images.\n\nIf a specific image has been described, use the [https://arbital.com/p/5v6](https://arbital.com/p/5v6) tag.", "date_published": "2016-08-12T21:30:49Z", "authors": ["M Yass", "Alexei Andreev", "Stephanie Zolayvar", "Patrick Stevens", "Eric Bruylant", "Mark Chimes"], "summaries": [], "tags": [], "alias": "4d3"}
{"id": "f7d65b638334391f688f153ab0d8d6fd", "title": "Morphism", "url": "https://arbital.com/p/morphism", "source": "arbital", "source_type": "text", "text": "A morphism is the abstract representation of a [relation](https://arbital.com/p/3nt) between [mathematical objects](https://arbital.com/p/-mathematical_object).\n\nUsually, it is used to refer to [functions](https://arbital.com/p/3jy) mapping element of one set to another, but it may represent a more general notion of a relation in [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7).\n\n##Isomorphisms##\n\nTo understand a morphism, it is easier to first understand the concept of an [https://arbital.com/p/-4f4](https://arbital.com/p/-4f4). Two mathematical structures (say two [groups](https://arbital.com/p/-3gd)) are called **isomorphic** if they are indistinguishable using the information of the language and theory under consideration. \n\nImagine you are the Count von Count. You care only about counting things. You don't care what it is you count, you just care how many there are. You decide that you want to collect objects you count into boxes, and you consider two boxes equal if there are the same number of elements in both boxes. How do you know if two boxes have the same number of elements? You pair them up and see if there are any left over in either box. If there aren't any left over, then the boxes are \"bijective\" and the way that you paired them up is a [bijection](https://arbital.com/p/499). A bijection is a simple form of an isomorphism and the boxes are said to be isomorphic.\n\nFor example, the theory of groups only talks about the way that elements are combined via group operation (and whether they are the [https://arbital.com/p/-identity](https://arbital.com/p/-identity) or [inverses](https://arbital.com/p/-4sn), but that information is already given by the information of how elements are combined under the group operation (hereafter called multiplication). The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if their multiplication acts the same way.", "date_published": "2016-08-06T13:04:12Z", "authors": ["Patrick Stevens", "Eric Bruylant", "Mark Chimes", "Jaime Sevilla Molina", "Team Arbital"], "summaries": [], "tags": ["Category theory", "Work in progress"], "alias": "4d8"}
{"id": "ff33d7c6c480b4c4259e3fbbe09002e9", "title": "Isomorphism", "url": "https://arbital.com/p/isomorphism", "source": "arbital", "source_type": "text", "text": "A pair of mathematical structures are **isomorphic** to each other if they are \"essentially the same\", even if they aren't necessarily equal. \n\nAn **isomorphism** is a [https://arbital.com/p/-4d8](https://arbital.com/p/-4d8) between isomorphic structures which translates one to the other in a way that preserves all the relevant structure. An important property of an isomorphism is that it can be 'undone' by its [inverse](https://arbital.com/p/-4sn) isomorphism. \n\nAn isomorphism from an object to itself is called an **[automorphism](https://arbital.com/p/automorphism)**. They can be thought of as symmetries: different ways in which an object can be mapped onto itself without changing it.\n\n##Equality and Identity##\nThe simplest isomorphism is equality: if two things are equal then they are actually the same thing (and so not actually *two* things at all). Anything is obviously indistinguishable from itself under whatever measure you might use (it has any property in common with itself) and so regardless of the theory or language, anything is isomorphic to itself. This is represented by the [identity](https://arbital.com/p/-identity_function) (iso)morphism.\n\n%%%knows-requisite([https://arbital.com/p/3gd](https://arbital.com/p/3gd)):\n##[Group Isomorphisms](https://arbital.com/p/49x)##\nFor a more technical example, the theory of groups only talks about the way that elements are combined via group operation. The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if you can pair up the elements such that their group operations act the same way.\n%%%\n\n##Isomorphisms in Category Theory##\nIn [category theory](https://arbital.com/p/-4c7), an isomorphism is a morphism which has a two-sided [https://arbital.com/p/-4sn](https://arbital.com/p/-4sn). That is to say, $f:A \\to B$ is an isomorphism if there is a morphism $g: B \\to A$ where $f$ and $g$ cancel each other out.\n\nFormally, this means that both composites $fg$ and $gf$ are equal to identity morphisms (morphisms which 'do nothing' or declare an object equal to itself). That is, $gf = \\mathrm {id}_A$ and $fg = \\mathrm {id}_B$.", "date_published": "2016-10-20T22:07:16Z", "authors": ["Daniel Satanove", "Eric Rogstad", "Dylan Hendrickson", "Patrick Stevens", "Eric Bruylant", "Mark Chimes"], "summaries": [], "tags": ["Group isomorphism", "Category theory", "Morphism", "Needs exercises", "C-Class"], "alias": "4f4"}
{"id": "56eca286cfd8534a39f79405dfd70641", "title": "Square visualization of probabilities on two events: (example) Diseasitis", "url": "https://arbital.com/p/4fh", "source": "arbital", "source_type": "text", "text": "$$\n\\newcommand{\\bP}{\\mathbb{P}}\n$$\n\nsummary: $$ \\newcommand{\\bP}{\\mathbb{P}} $$\n\nWe can visualize the [https://arbital.com/p/22s](https://arbital.com/p/22s):\n\n![](\"http://i.imgur.com/ssokgJ2.png\")
\n\n\nThen we can see at a glance that the probability $\\bP(D \\mid B)$ of the patient having Diseasitis given that the tongue depressor is black, despite some intuitions, isn't all that big: \n\n$$\\bP(D \\mid B) = \\frac{\\bP(B, D)}{ \\bP(B, D) + \\bP(B, \\neg D)} = \\frac{3}{7} $$\n![](\"http://i.imgur.com/NJOydOD.png\")
\n\n\n\nFrom the [https://arbital.com/p/22s](https://arbital.com/p/22s): \n\n> You are screening a set of patients for a disease, which we'll call Diseasitis. Based on prior epidemiology, you expect that around 20% of the patients in the screening population will in fact have Diseasitis. You are testing for the presence of the disease using a tongue depressor containing a chemical strip. Among patients with Diseasitis, 90% turn the tongue depressor black. However, 30% of the patients without Diseasitis will also turn the tongue depressor black. Among all the patients with black tongue depressors, how many have Diseasitis? \n\nIt seems like, since Diseasitis very strongly predicts that the patient has a black tongue depressor, it should be the case that the [https://arbital.com/p/-1rj](https://arbital.com/p/-1rj) $\\bP( \\text{Diseasitis} \\mid \\text{black tongue depressor})$ is big. But actually, it turns out that a patient with a black tongue depressor is more likely than not to be completely Diseasitis-free.\n\nCan we see this fact at a glance? Below, we'll use the [https://arbital.com/p/-496](https://arbital.com/p/-496) to draw pictures and use our visual intuition.\n\n\nTo introduce some notation: our [prior probability](https://arbital.com/p/1rm) $\\bP(D)$ that the patient has Diseasitis is $0.2$. We think that if the patient is sick $(D)$, then it's 90% likely that the tongue depressor will turn black $(B)$: we assign conditional probability $\\bP(B \\mid D) = 0.9$. We assign conditional probability $\\bP(B \\mid \\neg D) = 0.3$ that the tongue depressor will be black even if the patient isn't sick. We want to know $\\bP(D \\mid B)$, the [posterior probability](https://arbital.com/p/1rp) that the patient has Diseasitis given that we've seen a black tongue depressor.\n\nIf we wanted to, we could solve this problem precisely using [https://arbital.com/p/1lz](https://arbital.com/p/1lz): \n\n$$\n\\begin{align}\n\\bP(D \\mid B) &= \\frac{\\bP(B \\mid D) \\bP(D)}{\\bP(B)}\\\\\n&= \\frac{0.9 \\times 0.2}{ \\bP(B, D) + \\bP(B, \\neg D)}\\\\\n&= \\frac{0.18}{ \\bP(D)\\bP(B \\mid D) + \\bP(\\neg D)\\bP(B \\mid \\neg D)}\\\\\n&= \\frac{0.18}{ 0.18 + 0.24}\\\\\n&= \\frac{0.18}{ 0.42} = \\frac{3}{7} \\approx 0.43\\ .\n\\end{align}\n$$\n\nSo even if we've seen a black tongue depressor, the patient is more likely to be healthy than not: $\\bP(D \\mid B) < \\bP(\\neg D \\mid B) \\approx 0.57$. \n\nNow, this calculation might be enlightening if you are a real expert at [https://arbital.com/p/1lz](https://arbital.com/p/1lz). A better calculation would probably be the [odds ratio form of Bayes's rule](https://arbital.com/p/1x9).\n\nBut either way, maybe there's still an intuition saying that, come on, if the tongue depressor is such a strong indicator of Diseasitis that $\\bP(B \\mid D) = 0.9$, it must be that $\\bP(D \\mid B) =big$.\n\nLet's use the [square visualization of probabilities](https://arbital.com/p/496) to make it really visibly obvious that $\\bP(D \\mid B) < \\bP(\\neg D \\mid B)$, and to figure out why $\\bP(B \\mid D) = big$ doesn't imply $\\bP(D \\mid B) =big$. \n\nWe start with the probability of $\\bP(D)$ (so we're [factoring](https://arbital.com/p/factoring_probability) our probabilities by $D$ first): \n\n![](\"http://i.imgur.com/oFctjqn.png\")
\n\nNow let's break up the red column, where $D$ is true and the patient has Diseasitis, into a block for the probability $\\bP(B \\mid D)$ that $B$ is also true, and a block for the probability $\\bP(\\neg B \\mid D)$ that $B$ is false.\n\n>Among patients with Diseasitis, 90% turn the tongue depressor black.\n\nThat is, in 90% of the outcomes where $D$ happens, $B$ also happens. So $0.9$ of the red column will be dark ($B$), and $0.1$ will be light:\n\n![](\"http://i.imgur.com/ra6maoF.png\")
\n\n\n>However, 30% of the patients without Diseasitis will also turn the tongue depressor black.\n\nSo we break up the blue $\\neg D$ column by $\\bP(B \\mid \\neg D) = 0.3$ and $\\bP(\\neg B \\mid \\neg D) = 0.7$: \n\n\n\n\nNow we would like to know the probability $\\bP(D \\mid B)$ of Diseasitis once we've observed that the tongue depressor is black. Let's break up our diagram by whether or not $B$ happens: \n\n![](\"http://i.imgur.com/yKv723w.png\")
\n\nConditioning on $B$ is like only looking at the part of our distribution where $B$ happens. So the probability $\\bP(D \\mid B)$ of $D$ conditioned on $B$ is the proportion of that area where $D$ also happens: \n\n![](\"http://i.imgur.com/imiJdSH.png\")
\n\nHere we can see why $\\bP(D \\mid B)$ isn't all that big. It's true that $\\bP(B,D)$ is big relative to $\\bP(\\neg B,D)$, since we know that $\\bP(B \\mid D)$ is big (patients with Diseasitis almost always have black tongue depressors): \n\n![](\"http://i.imgur.com/kJvtM1X.png\")
\n\nBut this ratio doesn't really matter if we want to know $\\bP(D \\mid B)$, the probability that a patient with a black tongue depressor has Diseasitis. What matters is that we also assign a reasonably high probability $\\bP(B, \\neg D)$ to the patient having a black tongue depressor but nevertheless *not* suffering from Diseasitis:\n\n\n\nSo even when we see a black tongue depressor, there's still a pretty high chance the patient is healthy anyway, and our [https://arbital.com/p/-1rp](https://arbital.com/p/-1rp) $\\bP(D\\mid B)$ is not that high. Recall our square of probabilities: \n\n\n\nWhen asked about $\\bP(D\\mid B)$, we think of the really high probability $\\bP(B\\mid D) = 0.9$:\n\n![](\"http://i.imgur.com/Hy2FyYP.png\")
\n\nReally, we should look at the part of our probability mass where $B$ happens, and see that a sizeable portion goes to places where $\\neg D$ happens, and the patient is healthy:\n\n![](\"http://i.imgur.com/Kk4IgfK.png\")
\n\nSide note\n---\n\nThe square visualization is very similar to [frequency diagrams](https://arbital.com/p/1x1), except we can just think in terms of probability mass rather than specifically frequency. Also, see that page for [waterfall diagrams](https://arbital.com/p/1x1), another way to visualize updating probabilities.", "date_published": "2016-06-18T08:41:24Z", "authors": ["Tsvi BT", "Team Arbital"], "summaries": [], "tags": [], "alias": "4fh"}
{"id": "b300f2a0a31331956b5e4d1208fbc36c", "title": "Computer Programming Familiarity", "url": "https://arbital.com/p/programming_familiarity", "source": "arbital", "source_type": "text", "text": "This page exists so that math explanations can be customized on the basis of whether or not the readers is familiar with coding. If you know how to code, then click the three dot icon and check the \"this requisite\" box.", "date_published": "2016-06-16T17:22:00Z", "authors": ["Kevin Clancy"], "summaries": ["This page exists so that math explanations can be customized on the basis of whether or not the readers is familiar with coding."], "tags": ["Stub"], "alias": "4fv"}
{"id": "7d79287f83043d50b80118e1f90e88cc", "title": "Group conjugate", "url": "https://arbital.com/p/group_conjugate", "source": "arbital", "source_type": "text", "text": "Two elements $x, y$ of a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ are *conjugate* if there is some $h \\in G$ such that $hxh^{-1} = y$.\n\n# Conjugacy as \"changing the worldview\"\n\nConjugating by $h$ is equivalent to \"viewing the world through $h$'s eyes\".\nThis is most easily demonstrated in the [https://arbital.com/p/-497](https://arbital.com/p/-497), where it [is a fact](https://arbital.com/p/4bh) that if $$\\sigma = (a_{11} a_{12} \\dots a_{1 n_1})(a_{21} \\dots a_{2 n_2}) \\dots (a_{k 1} a_{k 2} \\dots a_{k n_k})$$\nand $\\tau \\in S_n$, then $$\\tau \\sigma \\tau^{-1} = (\\tau(a_{11}) \\tau(a_{12}) \\dots \\tau(a_{1 n_1}))(\\tau(a_{21}) \\dots \\tau(a_{2 n_2})) \\dots (\\tau(a_{k 1}) \\tau(a_{k 2}) \\dots \\tau(a_{k n_k}))$$\n\nThat is, conjugating by $\\tau$ has \"caused us to view $\\sigma$ from the point of view of $\\tau$\".\n\nSimilarly, in the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) $D_{2n}$ on $n$ vertices, conjugation of the rotation by a reflection yields the inverse of the rotation: it is \"the rotation, but viewed as acting on the reflected polygon\".\nEquivalently, if the polygon is sitting on a glass table, conjugating the rotation by a reflection makes the rotation act \"as if we had moved our head under the table to look upwards first\".\n\nIn general, if $G$ is a group which [acts](https://arbital.com/p/3t9) as (some of) the symmetries of a certain object $X$ %%note:Which [we can always view as being the case](https://arbital.com/p/49b).%% then conjugation of $g \\in G$ by $h \\in G$ produces a symmetry $hgh^{-1}$ which acts in the same way as $g$ does, but on a copy of $X$ which has already been permuted by $h$.\n\n# Closure under conjugation\n\nIf a subgroup $H$ of $G$ is closed under conjugation by elements of $G$, then $H$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6).\nThe concept of a normal subgroup is extremely important in group theory.\n\n%%%knows-requisite([https://arbital.com/p/3t9](https://arbital.com/p/3t9)):\n# Conjugation action\n\nConjugation forms a [action](https://arbital.com/p/3t9).\nFormally, let $G$ act on itself: $\\rho: G \\times G \\to G$, with $\\rho(g, k) = g k g^{-1}$.\nIt is an exercise to show that this is indeed an action.\n%%hidden(Show solution):\nWe need to show that the identity acts trivially, and that products may be broken up to act individually.\n\n- $\\rho(gh, k) = (gh)k(gh)^{-1} = ghkh^{-1}g^{-1} = g \\rho(h, k) g^{-1} = \\rho(g, \\rho(h, k))$;\n- $\\rho(e, k) = eke^{-1} = k$.\n%%\n\nThe [stabiliser](https://arbital.com/p/group_stabiliser) of this action, $\\mathrm{Stab}_G(g)$ for some fixed $g \\in G$, is the set of all elements such that $kgk^{-1} = g$: that is, such that $kg = gk$.\nEquivalently, it is the [centraliser](https://arbital.com/p/group_centraliser) of $g$ in $G$: it is the subgroup of all elements which commute with $G$.\n\nThe [orbit](https://arbital.com/p/group_orbit) of the action, $\\mathrm{Orb}_G(g)$ for some fixed $g \\in G$, is the [https://arbital.com/p/-4bj](https://arbital.com/p/-4bj) of $g$ in $G$.\nBy the [https://arbital.com/p/-4l8](https://arbital.com/p/-4l8), this immediately gives that the size of a conjugacy class divides the [order](https://arbital.com/p/3gg) of the parent group.\n%%%", "date_published": "2016-06-20T07:05:54Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Math 2"], "alias": "4gk"}
{"id": "0c8e1b23e178d18f2ef1ea613910688e", "title": "There is only one logarithm", "url": "https://arbital.com/p/only_one_log", "source": "arbital", "source_type": "text", "text": "When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others? On this page, we'll answer that question, among others.\n\nIn [https://arbital.com/p/4bz](https://arbital.com/p/4bz), we saw that any $f$ with [domain](https://arbital.com/p/3js) [$\\mathbb R^+$](https://arbital.com/p/positive_reals) that satisfies the equation $f(x \\cdot y) = f(x) + f(y)$ for all $x$ and $y$ in its domain is either trivial (i.e., it sends all inputs to 0), or it is [isomorphic](https://arbital.com/p/4f4) to [$\\log_b$](https://arbital.com/p/3nd) for some base $b$. Thus, if we want a function's output to change by a constant that depends on $y$ every time its input changes by a factor of $y$, we only have one meaningful degree of freedom, and that's the choice of $b \\neq 1$ such that $f(b) = 1$. Once we choose which value of $b$ $f$ sends to 1, the entire behavior of the function is fully defined.\n\nHow much freedom does this choice give us? Almost none! To see this, let's consider the difference between choosing base $b$ as opposed to an alternative base $c.$ Say we have an input $x \\in \\mathbb R^+$ — what's the difference between $\\log_b(x)$ and $\\log_c(x)?$\n\nWell, $x = c^y$ for some $y$, because $x$ is positive. Thus, $\\log_c(x)$ $=$ $\\log_c(c^y)$ $=$ $y.$ By contrast, $\\log_b(x)$ $=$ $\\log_b(c^y)$ $=$ $y \\log_b(c).$ ([Refresher](https://arbital.com/p/4bz).) Thus, $\\log_c$ and $\\log_b$ disagree on $x$ only by a constant factor — namely, $\\log_b(c).$ And this is true for _any_ $x$ — you can get $\\log_c(x)$ by calculating $\\log_b(x)$ and dividing by $\\log_b(c):$ $$\\log_c(x) = \\frac{\\log_b(x)}{\\log_b(c)}.$$\n\nThis is a remarkable equation, in the sense that it's worth remarking on. No matter what base $b$ we choose for $f$, and no matter what input $x$ we put into $f$, if we want to figure out what we would have gotten if we chose base $c$ instead, all we need to do is calculate $f(c)$ and divide $f(x)$ by $f(c).$ In other words, the different logarithm functions actually aren't very different at all — each one has all the same information as all the others, and you can recover the behavior of $\\log_c$ using $\\log_b$ and a simple calculation!\n\n(By a symmetric argument, you can show that $\\log_b(x) = \\frac{\\log_c(x)}{\\log_c(b)},$ which implies that $\\log_b(c) = \\frac{1}{\\log_c(b)}$, a fact we already knew from from studying [https://arbital.com/p/-44l](https://arbital.com/p/-44l).)\n\nAbove, we asked:\n\n> When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others?\n\nFrom the fact that the [length of a written number grows logarithmically with the magnitude of the number](https://arbital.com/p/416) and the above equation, we can see that, no matter how large a number is, its base 10 representation differs in length from its base 12 representation only by a factor of $\\log_{10}(12) \\approx 1.08$. Similarly, the [binary](https://arbital.com/p/binary_notation) representation of a number is always about $\\log_2(10) \\approx 3.32$ times longer than its decimal representation. Because there is only one logarithm function (up to a multiplicative constant), which number base you use to represent numbers only affects the size of the representation by a multiplicative constant.\n\nSimilarly, if you ever want to convert between logarithmic measurements in different bases, you only need to perform a single multiplication (or division). For example, if someone calculated how many hours it took for the bacteria colony to triple, and you want to know how long it took to double, all you need to do is multiply by $\\log_3(2) \\approx 0.63.$ There is essentially only one logarithm function; the base merely defines the unit of measure. Given measurements taken in one base, you can easily convert to another.\n\nWhich base of the logarithm is best to use? Well, by the arguments above, it doesn't really matter which base you pick — it's easy enough to convert between bases if you need to. However, in computer science, [https://arbital.com/p/-3qq](https://arbital.com/p/-3qq), and many parts of physics, it's common to choose base 2. Why? Two reasons. First, because if you're designing a physical system to store data, a system that only has two stable states is the simplest available. Second, because if you need to pick an arbitrary amount that a thing has grown to call \"one growth,\" then doubling (i.e., \"it's now once again as large as it was when it started\") is a pretty natural choice. (For example, many people find the 'half life' of a radioactive element to be a more intuitive notion than its 'fifth life,' even though both are equally valid measures of radioactivity.)\n\nIn other parts of physics and engineering, the log base 10 is more common, because it has a natural relationship to the way we represent numbers (using a base 10 representation). That is, $\\log_{10}$ is one of the easiest logarithms to calculate in your head (because you can just [count the number of digits in a number's representation](https://arbital.com/p/416)), so it's often used by engineers.\n\nAmong mathematicians, the most common base of the logarithm is [$e$](https://arbital.com/p/e) $\\approx 2.718,$ a [https://arbital.com/p/-trancendental_number](https://arbital.com/p/-trancendental_number) that cannot be fully written down. Why this strange base? That's an [advanced topic](https://arbital.com/p/base_e_is_natural) not covered in this tutorial. In brief, though, if you look at the [derivative](https://arbital.com/p/47d) of the logarithm (i.e., the way the function changes if you perturb the input), if the input was $x$ and you perturb it a tiny amount, the output changes by about a factor of $\\frac{1}{x}$ times a constant $y$ that depends on the base of the logarithm, and $e$ is the base that causes that constant to be $1.$ For this reason and a handful of others, there is a compelling sense in which $\\log_e$ is the \"true form\" of the logarithm. (So much so that it has its own symbol, $\\ln,$ which is called the \"natural logarithm.\")\n\nRegardless of which base you choose, converting between bases is easy — and as equation (1) shows, you only need to have one logarithm function in your toolkit in order to understand any other logarithm. Because if you have $\\log_b$ and you want to know $\\log_c(x)$, you can just calculate $\\log_b(x)$ and divide it by $\\log_b(c)$, and then you're done. In other words, there is really only one logarithm function, and it doesn't matter which one you pick.\n\n[A visualization of the graph of the logarithm where you can change the base, but it changes only the labels on the axes while the curve itself remains completely static.](https://arbital.com/p/fixme:)", "date_published": "2016-06-20T20:50:28Z", "authors": ["Eric Rogstad", "Nate Soares", "Patrick Stevens"], "summaries": [], "tags": ["Needs summary", "Work in progress"], "alias": "4gm"}
{"id": "3883718ecbeebc0ae09fa3f0ee137722", "title": "The log lattice", "url": "https://arbital.com/p/log_lattice", "source": "arbital", "source_type": "text", "text": "[https://arbital.com/p/45q](https://arbital.com/p/45q) and other pages give physical interpretations of what logarithms are really doing. Now it's time to understand the raw numerics. Logarithm functions often output irrational, [https://arbital.com/p/-transcendental](https://arbital.com/p/-transcendental) numbers. For example, $\\log_2(3)$ starts with\n\n1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360\n\nWhat do these long strings of digits mean? Why these numbers, in particular? We already have the tools to answer those questions, all that remains is to put them together into a visualization. In [https://arbital.com/p/44l](https://arbital.com/p/44l), we saw why it is that $\\log_2(3)$ must be this number: The cost of a [3-digit](https://arbital.com/p/4sj) in terms of 2-digits is more than 1 and less than 2; and the cost of ten 3-digits = one $3^{10}$-digit in terms of 2-digits is more than 15 and less than 16, and the cost of a hundred 3-digits = one $3^{100}$-digit in terms of 2-digits is more than 158 and less than 159, and so on. The giant number above is telling us about the _entire sequence_ of $3^{n}$-digit costs, for every $n$. If we double it, we get the cost of a 9-digit in terms of 2-digits. If we triple it, we get the cost of a 27-digit in terms of 2-digits.\n\nIn fact, this number also interacts correctly with all the other outputs of $\\log_2.$ $\\log_2(2)=1,$ and if we add 1 to the number above, we get the cost of a 6-digit in terms of 2-digits.\n\nThat is, the outputs of the log base 2 form a gigantic lattice, with an often-irrational output corresponding to each input. The gigantic lattice is set up just so, such that whenever you start at $x$ on the left and go to $x \\cdot y,$ the output you get on the right is the output to that corresponds to $x$ plus the output that corresponds to $y$.\n\n\n\n_The values of the logarithm base 2 on integers from 1 to 10._\n\nThe log lattice is this giant lattice assigning a specific number to each number, such that multiplying on the left corresponds to adding on the right.\n\n\n\n_An illustration of some of the connections between the logs base 2 of powers of 3._\n\nIt's an intricate lattice that preservers a huge amount of structure — all the structure of the numbers, in fact, given that $\\log$ is invertible. $\\log_2(3)$ is a number that is simultaneously $1$ less than $\\log_2(6),$ and half of $\\log_2(9)$, and a tenth of $\\log_2(3^{10}),$ and $\\log_2(3^9)$ less than $\\log_2(3^{10}).$ The value of $\\log_2(3)$ has to satisfy a _massive_ number of constraints, in order to be precisely the number such that multiplication on the left corresponds to addition on the right. It's no surprise, then, that [it has no concise decimal expansion](https://arbital.com/p/4n8).\n\n\n\n_Notice the connections between the logs base 2 of the powers of 3. log 2 of 3 is a tenth of log 2 of 310 is a tenth of the log 2 of 3100, which is twice the log 2 of 350 which is twice the log 2 of 325. And, of course, the log 2 of 3101 is about 1.58 units larger than the log 2 of 3100._\n\nIn fact, the constraints on this log lattice are so tight that there's only one way to do it, up to a multiplicative constant. You can multiply everything on the right side by a constant, and that's the only thing you can do without disturbing the structure. For example, here's part of the log lattice viewed in base 3:\n\n\n\nIt's exactly the same as the log lattice viewed in base 2, except that everything on the right has been divided by about 1.58 (i.e., multiplied by about 0.63).\n\nWhat are logarithms doing? Well, fundamentally, there is a way to transform numbers such that what was once multiplication is now addition. If you were going to do multiplication to some numbers, you can transform them into this lattice, and then do addition, and then transform them back, and you'll get the same result. There's only one way to preserve all that structure, although if you want to view the structure, you need to (arbitrarily) choose some number on the left to call \"1\" on the right. Given that choice $b$ of translation, the log base $b$ taps into that intricate lattice.\n\n[https://arbital.com/p/visualization](https://arbital.com/p/visualization)", "date_published": "2016-07-30T01:12:49Z", "authors": ["Eric Rogstad", "Nate Soares", "Malcolm McCrimmon", "Alexei Andreev"], "summaries": [], "tags": ["B-Class"], "alias": "4gp"}
{"id": "21d15afffb06ba8aa76dd0f4bd0ac2d7", "title": "Formal definition", "url": "https://arbital.com/p/formal_definition_meta_tag", "source": "arbital", "source_type": "text", "text": "Meta tag for pages which give formal or brief jargon-heavy technical definitions of a concept. Formal definitions should be secondary [lenses](https://arbital.com/p/17b) when explanations are available.", "date_published": "2016-07-13T22:30:18Z", "authors": ["Eric Bruylant", "Jessica Taylor", "Patrick Stevens", "Alexei Andreev"], "summaries": [], "tags": ["Stub"], "alias": "4gs"}
{"id": "f619ea58b5c666dbf9d4525ac5665e55", "title": "Life in logspace", "url": "https://arbital.com/p/logspace", "source": "arbital", "source_type": "text", "text": "[https://arbital.com/p/4gp](https://arbital.com/p/4gp) hints at the reason that engineers, scientists, and AI researchers find logarithms so useful.\n\nAt this point, you may have a pretty good intuition for what logarithms are doing, and you may be saying to yourself\n\n> Ok, I understand logarithms, and I see how they could be a pretty useful concept sometimes. Especially if, like, someone happens to have an out-of-control bacteria colony lying around. I can also see how they're useful for measuring representations of things (like the length of a number when you write it down, or the cost of communicating a message). If I squint, I can _sort of_ see why it is that the brain encodes many percepts logarithmically (such that the tone that I perceive is the $\\log_2$ of the frequency of vibration in the air), as it's not too surprising that evolution occasionally hit upon this natural tool for representing things. I guess logs are pretty neat.\n\nThen you start seeing logarithms in other strange places — say, you learn that [AlphaGo](https://en.wikipedia.org/wiki/AlphaGo) primarily manipulated [logarithms of the probability](https://arbital.com/p/log_probability) that the human would take a particular action instead of manipulating the probabilities directly. Then maybe you learn that scientists, engineers, and navigators of yore used to pay large sums of money for giant, pre-computed tables of logarithm outputs.\n\nAt that point, you might say\n\n> Hey, wait, why do these logs keep showing up all over the place? Why are they _that_ ubiquitous? What's going on here?\n\nThe reason here is that _addition is easier than multiplication,_ and this is true both for navigators of yore (who had to do lots of calculations to keep their ships on course), and for modern AI algorithms.\n\nLet's say you have a whole bunch of numbers, and you know you're going to have to perform lots and lots of operations on them that involve taking some numbers you already have and multiplying them together. There are two ways you could do this: One is that you could bite the bullet and multiply them all, which gets pretty tiresome (if you're doing it by hand) and/or pretty expensive (if you're training a gigantic neural network on millions of games of go). Alternatively, you could put all your numbers through a log function (a one-time cost), and then perform all your operations using addition (much cheaper!), and then transform them back using [https://arbital.com/p/-exponentiation](https://arbital.com/p/-exponentiation) when you're done (another one-time cost).\n\nEmpirically, the second method tends to be faster, cheaper, and more convenient (at the cost of some precision, given that most log outputs are [transcendental](https://arbital.com/p/transcendental_number)).\n\nThis is the last insight about logarithms contained in this tutorial, and it is a piece of practical advice: If you have to perform a large chain of multiplications, you can often do it cheaper by first transforming your problem into \"log space\", where multiplication acts like addition (and exponentiation acts like multiplication). Then you can perform a bunch of additions instead, and transform the answer back into normal space at the end.", "date_published": "2016-07-29T22:26:10Z", "authors": ["Malcolm McCrimmon", "Nate Soares", "Eric Rogstad", "Alexei Andreev"], "summaries": [], "tags": ["B-Class"], "alias": "4h0"}
{"id": "071623e979361c5b32b0debe4e8ed312", "title": "The End (of the basic log tutorial)", "url": "https://arbital.com/p/log_tutorial_end", "source": "arbital", "source_type": "text", "text": "That concludes our introductory tutorial on logarithms! You have made it to the end.\n\nThroughout this tutorial, we saw that the logarithm base $b$ of $x$ calculates the number of $b$-factors in $x.$ Hopefully, this claim now means more to you than it once did. We've seen a number of different ways of interpreting what logarithms are doing, including:\n\n- $\\log_b(x) = y$ means [\"it takes about $y$ digits to write $x$ in base $b$.\"](https://arbital.com/p/416)\n- $\\log_b(x) = y$ means [\"it takes about $y$ $b$-digits to emulate an $x$-digit.\"](https://arbital.com/p/44l)\n- $\\log_b(x) = y$ means [\"if the space of possible messages to send goes up by a factor of $x$, then the cost, in $b$-digits, goes up by a factor of $y$](https://arbital.com/p/45q)\n- And, simply, $\\log_b(x) = y$ means that if you start with 1 and grow it by factors of $b$, then after $y$ iterations of this your result will be $x.$\n\nFor example, $\\log_2(100)$ counts the number of doublings that constitute a factor-of-100 increase. (The answer is more than 6 doublings, but slightly less than 7 doublings).\n\nWe've also seen that any function $f$ whose output grows by a constant (that depends on $y$) every time its input grows by a factor of $y$ is [very likely a logarithm function](https://arbital.com/p/4bz), and that, in essence, [https://arbital.com/p/-4gm](https://arbital.com/p/-4gm) function.\n\nWe've glanced at the [underlying structure](https://arbital.com/p/4gp) that all logarithm functions tap into, and we've briefly discussed [what makes working with logarithms so dang useful](https://arbital.com/p/4h0).\n\nThere are also a huge number of questions about, applications for, and extensions of the logarithm that we _didn't_ explore. Those include, but are not limited to:\n\n- Why is $e$ the natural base of the logarithm?\n- What is up with the link between logarithms, exponentials, and roots?\n- What is the derivative of $\\log_b(x)$ and why is it proportional to $\\frac{1}{x}$?\n- How can logarithms be efficiently calculated?\n- What happens when we extend logarithms to complex numbers, and why is the result a [https://arbital.com/p/-multifunction](https://arbital.com/p/-multifunction)?\n\nAnswering these questions will require an advanced tutorial on logarithms. Such a thing does not exist yet, but you can help make it happen.", "date_published": "2016-09-20T23:27:16Z", "authors": ["Nate Soares", "Alexei Andreev"], "summaries": [], "tags": ["B-Class"], "alias": "4h2"}
{"id": "48031479529c5f413eb60081680e5050", "title": "Normal subgroup", "url": "https://arbital.com/p/normal_subgroup", "source": "arbital", "source_type": "text", "text": "A *normal subgroup* $N$ of group $G$ is one which is closed under [conjugation](https://arbital.com/p/4gk): for all $h \\in G$, it is the case that $\\{ h n h^{-1} : n \\in N \\} = N$.\nIn shorter form, $hNh^{-1} = N$.\n\nSince [conjugacy is equivalent to \"changing the worldview\"](https://arbital.com/p/4gk), a normal subgroup is one which \"looks the same from the point of view of every element of $G$\".\n\nA subgroup of $G$ is normal if and only if it is the [kernel](https://arbital.com/p/49y) of some [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $G$ to some group $H$. ([Proof.](https://arbital.com/p/4h7))\n\n%%%knows-requisite([https://arbital.com/p/category_theory_equaliser](https://arbital.com/p/category_theory_equaliser)):\nFrom a category-theoretic point of view, the kernel of $f$ is an equaliser of an arrow $f$ with the zero arrow; it is therefore universal such that composition with $f$ yields zero.\n%%%", "date_published": "2016-06-18T13:43:43Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Definition", "Stub"], "alias": "4h6"}
{"id": "307c3e97643d0bcd09743442236b0c5e", "title": "Subgroup is normal if and only if it is the kernel of a homomorphism", "url": "https://arbital.com/p/subgroup_normal_iff_kernel_of_homomorphism", "source": "arbital", "source_type": "text", "text": "Let $N$ be a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$.\nThen $N$ is [normal](https://arbital.com/p/4h6) in $G$ if and only if there is a group $H$ and a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $\\phi:G \\to H$ such that the [kernel](https://arbital.com/p/49y) of $\\phi$ is $N$.\n\n# Proof\n\n## \"Normal\" implies \"is a kernel\"\nLet $N$ be normal, so it is closed under [conjugation](https://arbital.com/p/4gk).\nThen we may define the [https://arbital.com/p/-quotient_group](https://arbital.com/p/-quotient_group) $G/N$, whose elements are the [left cosets](https://arbital.com/p/group_coset) of $N$ in $G$, and where the operation is that $gN + hN = (g+h)N$.\nThis group is well-defined ([proof](https://arbital.com/p/quotient_by_subgroup_is_well_defined_iff_normal)).\n\nNow there is a homomorphism $\\phi: G \\to G/N$ given by $g \\mapsto gN$.\nThis is indeed a homomorphism, by definition of the group operation $gN + hN = (g+h)N$.\n\nThe kernel of this homomorphism is precisely $\\{ g : gN = N \\}$; that is simply $N$:\n\n- Certainly $N \\subseteq \\{ g : gN = N \\}$ (because $nN = N$ for all $n$, since $N$ is closed as a subgroup of $G$);\n- We have $\\{ g : gN = N \\} \\subseteq N$ because if $gN = N$ then in particular $g e \\in N$ (where $e$ is the group identity) so $g \\in N$.\n\n## \"Is a kernel\" implies \"normal\"\nLet $\\phi: G \\to H$ have kernel $N$, so $\\phi(n) = e$ if and only if $n \\in N$.\nWe claim that $N$ is closed under conjugation by members of $G$.\n\nIndeed, $\\phi(h n h^{-1}) = \\phi(h) \\phi(n) \\phi(h^{-1}) = \\phi(h) \\phi(h^{-1})$ since $\\phi(n) = e$.\nBut that is $\\phi(h h^{-1}) = \\phi(e)$, so $hnh^{-1} \\in N$.\n\nThat is, if $n \\in N$ then $hnh^{-1} \\in N$, so $N$ is normal.", "date_published": "2016-06-17T08:33:06Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4h7"}
{"id": "3717223c08c4e6987d727c04dc1735c8", "title": "Quotient by subgroup is well defined if and only if subgroup is normal", "url": "https://arbital.com/p/quotient_by_subgroup_is_well_defined_iff_normal", "source": "arbital", "source_type": "text", "text": "Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $N$ a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $G$.\nThen we may define the *quotient group* $G/N$ to be the set of [left cosets](https://arbital.com/p/4j4) $gN$ of $N$ in $G$, with the group operation that $gN + hN = (gh)N$.\nThis is well-defined if and only if $N$ is normal.\n\n# Proof\n\n## $N$ normal implies $G/N$ well-defined\n\nRecall that $G/N$ is well-defined if \"it doesn't matter which way we represent a coset\": whichever coset representatives we use, we get the same answer.\n\nSuppose $N$ is a normal subgroup of $G$.\nWe need to show that given two representatives $g_1 N = g_2 N$ of a coset, and given representatives $h_1 N = h_2 N$ of another coset, that $(g_1 h_1) N = (g_2 h_2)N$.\n\nSo given an element of $g_1 h_1 N$, we need to show it is in $g_2 h_2 N$, and vice versa.\n\nLet $g_1 h_1 n \\in g_1 h_1 N$; we need to show that $h_2^{-1} g_2^{-1} g_1 h_1 n \\in N$, or equivalently that $h_2^{-1} g_2^{-1} g_1 h_1 \\in N$.\n\nBut $g_2^{-1} g_1 \\in N$ because $g_1 N = g_2 N$; let $g_2^{-1} g_1 = m$.\nSimilarly $h_2^{-1} h_1 \\in N$ because $h_1 N = h_2 N$; let $h_2^{-1} h_1 = p$.\n\nThen we need to show that $h_2^{-1} m h_1 \\in N$, or equivalently that $p h_1^{-1} m h_1 \\in N$.\n\nSince $N$ is closed under conjugation and $m \\in N$, we must have that $h_1^{-1} m h_1 \\in N$;\nand since $p \\in N$ and $N$ is closed under multiplication, we must have $p h_1^{-1} m h_1 \\in N$ as required.\n\n## $G/N$ well-defined implies $N$ normal\n\nFix $h \\in G$, and consider $hnh^{-1} N + hN$.\nSince the quotient is well-defined, this is $(hnh^{-1}h) N$, which is $hnN$ or $hN$ (since $nN = N$, because $N$ is a subgroup of $G$ and hence is closed under the group operation).\nBut that means $hnh^{-1}N$ is the identity element of the quotient group, since when we added it to $hN$ we obtained $hN$ itself.\n\nThat is, $hnh^{-1}N = N$.\nTherefore $hnh^{-1} \\in N$.\n\nSince this reasoning works for any $h \\in G$, it follows that $N$ is closed under conjugation by elements of $G$, and hence is normal.", "date_published": "2016-06-20T06:58:24Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4h9"}
{"id": "88f82ae09ffaaaa88fe95afddc3b584e", "title": "Alternating group", "url": "https://arbital.com/p/alternating_group", "source": "arbital", "source_type": "text", "text": "The *alternating group* $A_n$ is defined as a certain subgroup of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$: namely, the collection of all elements which can be made by multiplying together an even number of [transpositions](https://arbital.com/p/4cn).\nThis is a well-defined notion ([proof](https://arbital.com/p/4hg)).\n\n%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):\n$A_n$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $S_n$; it is the [quotient](https://arbital.com/p/quotient_group) of $S_n$ by the [sign homomorphism](https://arbital.com/p/4hk).\n%%%\n\n# Examples\n\n- A [cycle](https://arbital.com/p/49f) of even length is an odd permutation in the sense that it can only be made by multiplying an odd number of transpositions.\nFor example, $(132)$ is equal to $(13)(23)$.\n- A cycle of odd length is an even permutation, in that it can only be made by multiplying an even number of transpositions.\nFor example, $(1354)$ is equal to $(54)(34)(14)$.\n\n- The alternating group $A_4$ consists precisely of twelve elements: the identity, $(12)(34)$, $(13)(24)$, $(14)(23)$, $(123)$, $(124)$, $(134)$, $(234)$, $(132)$, $(143)$, $(142)$, $(243)$.\n\n# Properties\n\n%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):\nThe alternating group $A_n$ is of [index](https://arbital.com/p/index_of_a_subgroup) $2$ in $S_n$.\nTherefore $A_n$ is [normal](https://arbital.com/p/4h6) in $S_n$ ([proof](https://arbital.com/p/4hl)).\nAlternatively we may give the homomorphism explicitly of which $A_n$ is the [kernel](https://arbital.com/p/49y): it is the [sign homomorphism](https://arbital.com/p/4hk).\n%%%\n\n- $A_n$ is generated by its $3$-cycles. ([Proof.](https://arbital.com/p/4hs))\n- $A_n$ is [simple](https://arbital.com/p/4jc). ([Proof.](https://arbital.com/p/4jb))\n- The [conjugacy classes](https://arbital.com/p/4bj) of $A_n$ are [easily characterised](https://arbital.com/p/alternating_group_five_conjugacy_classes).", "date_published": "2016-06-18T10:15:05Z", "authors": ["Patrick Stevens", "Team Arbital"], "summaries": [], "tags": ["Definition"], "alias": "4hf"}
{"id": "159775f0d8377136ab31b2d02faa7e51", "title": "The collection of even-signed permutations is a group", "url": "https://arbital.com/p/even_signed_permutations_form_a_group", "source": "arbital", "source_type": "text", "text": "The collection of elements of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ which are made by multiplying together an even number of permutations forms a subgroup of $S_n$.\n\nThis proves that the [https://arbital.com/p/-alternating_group](https://arbital.com/p/-alternating_group) $A_n$ is well-defined, if it is given as \"the subgroup of $S_n$ containing precisely that which is made by multiplying together an even number of transpositions\".\n\n# Proof\n\nFirstly we must check that \"I can only be made by multiplying together an even number of transpositions\" is a well-defined notion; this [is in fact true](https://arbital.com/p/4hh).\n\nWe must check the group axioms.\n\n- Identity: the identity is simply the product of no transpositions, and $0$ is even.\n- Associativity is inherited from $S_n$.\n- Closure: if we multiply together an even number of transpositions, and then a further even number of transpositions, we obtain an even number of transpositions.\n- Inverses: if $\\sigma$ is made of an even number of transpositions, say $\\tau_1 \\tau_2 \\dots \\tau_m$, then its inverse is $\\tau_m \\tau_{m-1} \\dots \\tau_1$, since a transposition is its own inverse.", "date_published": "2016-06-17T11:43:40Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4hg"}
{"id": "e0c9a9c0dc94332f8e4ed8d92b847488", "title": "The sign of a permutation is well-defined", "url": "https://arbital.com/p/sign_of_permutation_is_well_defined", "source": "arbital", "source_type": "text", "text": "The [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ contains elements which are made up from [transpositions](https://arbital.com/p/4cn) ([proof](https://arbital.com/p/4cp)).\nIt is a fact that if $\\sigma \\in S_n$ can be made by multiplying together an even number of transpositions, then it cannot be made by multiplying an odd number of transpositions, and vice versa.\n\n%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):\nEquivalently, there is a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $S_n$ to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_2 = \\{0,1\\}$, taking the value $0$ on those permutations which are made from an even number of permutations and $1$ on those which are made from an odd number.\n%%%\n\n# Proof\n\nLet $c(\\sigma)$ be the number of cycles in the [disjoint cycle decomposition](https://arbital.com/p/49f) of $\\sigma \\in S_n$, including singletons.\nFor example, $c$ applied to the identity yields $n$, while $c((12)) = n-1$ because $(12)$ is shorthand for $(12)(3)(4)\\dots(n-1)(n)$.\nWe claim that multiplying $\\sigma$ by a transposition either increases $c(\\sigma)$ by $1$, or decreases it by $1$.\n\nIndeed, let $\\tau = (kl)$.\nEither $k, l$ lie in the same cycle in $\\sigma$, or they lie in different ones.\n\n- If they lie in the same cycle, then $$\\sigma = \\alpha (k a_1 a_2 \\dots a_r l a_s \\dots a_t) \\beta$$ where $\\alpha, \\beta$ are disjoint from the central cycle (and [hence commute](https://arbital.com/p/49g) with $(kl)$).\nThen $\\sigma (kl) = \\alpha (k a_s \\dots a_t)(l a_1 \\dots a_r) \\beta$, so we have broken one cycle into two.\n- If they lie in different cycles, then $$\\sigma = \\alpha (k a_1 a_2 \\dots a_r)(l b_1 \\dots b_s) \\beta$$ where again $\\alpha, \\beta$ are disjoint from $(kl)$.\nThen $\\sigma (kl) = \\alpha (k b_1 b_2 \\dots b_s l a_1 \\dots a_r) \\beta$, so we have joined two cycles into one.\n\nTherefore $c$ takes even values if there are evenly many transpositions in $\\sigma$, and odd values if there are odd-many transpositions in $\\sigma$.\n\nMore formally, let $\\sigma = \\alpha_1 \\dots \\alpha_a = \\beta_1 \\dots \\beta_b$, where $\\alpha_i, \\beta_j$ are transpositions.\n%%%knows-requisite([https://arbital.com/p/modular_arithmetic](https://arbital.com/p/modular_arithmetic)):\n(The following paragraph is more succinctly expressed as: \"$c(\\sigma) \\equiv n+a \\pmod{2}$ and also $\\equiv n+b \\pmod{2}$, so $a \\equiv b \\pmod{2}$.\")\n%%%\nThen $c(\\sigma)$ flips odd-to-even or even-to-odd for each integer $1, 2, \\dots, a$; it also flips odd-to-even or even-to-odd for each integer $1, 2, \\dots, b$.\nTherefore $a$ and $b$ must be of the same [parity](https://arbital.com/p/even_odd_parity).", "date_published": "2016-06-28T12:14:29Z", "authors": ["Patrick Stevens", "AM AM"], "summaries": [], "tags": [], "alias": "4hh"}
{"id": "7e3a5ca536d4af7014f6db5032c8f3b6", "title": "Isomorphism: Intro (Math 0)", "url": "https://arbital.com/p/Isomorphism_intro_math_0", "source": "arbital", "source_type": "text", "text": "If two things are essentially the same from a certain perspective, and they only differ in unimportant details, then they are **isomorphic**.\n\n##Comparing Amounts##\nConsider the Count von Count. He cares only about counting things. He doesn't care what they are, just how many there are. He decides that he wants to collect items into plastic crates, and he considers two crates equal if both contain the same number of items. \n\n\n\nNow Elmo comes to visit, and he wants to impress the Count, but Elmo is not great at counting. Without counting them explicitly, how can Elmo tell if two crates contain the same number of items?\n\nWell, he can take one item out of each crate and put the pair to one side.\n\n\n\n He continues pairing items up in this way and when one crate runs out he checks if there are any left over in the other crate. If there aren't any left over, then he knows there were the same number of items in both crates. \n\n\n\nSince the Count von Count only cares about counting things, the two crates are basically equivalent, and might as well be the same crate to him. Whenever two objects are the same from a certain perspective, we say that they are **isomorphic**.\n\nIn this example, the way in which the crates were the same is that each item in one crate could be paired with an item in the other.\n\n\n\n\n\nThis wouldn't have been possible if the crates had different numbers of items in them.\n\n\n\n\n\n\n\nWhenever you can match each item in one collection with exactly one item in another collection, we say that the collections are **[bijective](https://arbital.com/p/-499)** and the way you paired them is a **[bijection](https://arbital.com/p/-499)**. A bijection is a specific kind of isomorphism.\n\n\n\nNote that there might be many different bijections between two bijective things.\n\n\n\nIn fact, all that counting involves is pairing up the things you want to count, either with your fingers or with the concepts of 'numbers' in your head. If there are as many objects in one crate as there are numbers from one to seven, and there are as many objects in another crate as numbers from one to seven, then both crates contain the same number of objects.\n\n##Comparing Maps##\nNow imagine that you have a map of the London Underground. Such a map is not to scale, nor does it even show how the tracks bend or which station is in which direction compared to another. They only record which stations are connected. \n\n\n\nYour Spanish friend is coming to visit and you want to get them a version of the map in Spanish. But on the Spanish maps, the shape of the tracks is different and you can't read Spanish. What's more, not all the maps are of the London Underground! What do you do? Well, given a Spanish map and your English map, you can try to match up the stations (through trial and error) and if the stations are all *connected to each other in the same ways* on both maps then you know they are both of the same train system. \n\nMore precisely, consider the following smaller (fictional) example. There are stations *Trafalgal*, *Marybone*, *Oxbridge*, *Eastweston* and *Charlesburrough* on the English map. *Marybone* is connected to all the other stations, and *Charlesburrough* is connected to *Eastweston*. (There are no other connections)\n\n\n\n%note: We don't want to worry about whether stations are connected to themselves. You can just assume no station is ever connected to itself.%\n\n%note: If one station is connected to another, then the second station is also connected to the first. So since *Marybone* is connected to *Eastweston* then *Eastweston* is connected to *Marybone*. If it seems silly to even mention this fact, then don't worry to much. It's just that we might just have easily decided that there's a one-way train running from *Marybone* to *Eastweston* but not in the other direction.% \n\nAssume the Spanish map has the following stations: *Patata*, *Huesto*, *Carbon*, *Esteoeste*, and *Puente de Buey*. Assume also that *Huesto* is connected to every other station, and that *Carbon* is connected to *Esteoeste*. (Again, there are no other connections).\n\n\n\nThen the two maps are essentially the same for your purposes. They are **isomorphic** (as [graphs](https://arbital.com/p/-graph), in fact), and the way that you matched the stations on the one map with those on the other is an **isomorphism**.\n\n## Isomorphisms##\nImagine that you have a *London Underground Official Spanish-to-English Train Station Dictionary* that tells you how to translate the names of the train stations. So, for example, you can use it to convert *Patata* to *Trafalgal*. Then this dictionary is an **isomorphism** from the Spanish map to the English one. \n\nIn particular, the dictionary translates *Huesto* to *Marybone*, *Patata* to *Tralfalgal*, *Puento de Buey* to *Oxbridge*, *Esteoeste* to *Eastweston*, and *Carbon* to *Charlesburrough*.\n\nYou could also get a *London Underground Official* **English-to-Spanish** *Train Station Dictionary*. Then, if you were to use this dictionary to translate *Tralfalgal*, you'd get back *Patata*. Hence your first original translation of that station from English to Spanish has been undone.\n\nIn particular, the dictionary translates *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon*.\n\n\n\n\nIn fact, if you take the English map and translate all of the stations into Spanish using the one dictionary and then translate back, you'd get back to where you started. Similarly, if you translated the Spanish map from Spanish into English with the one dictionary and back to Spanish with the other you'd get the Spanish map back. Hence both of these dictionaries are complete [inverses](https://arbital.com/p/-inverse) of each other.\n\n\nIn fact, in [category theory](https://arbital.com/p/-4c7), this is exactly the definition of an isomorphism: if you have some translation ([https://arbital.com/p/-4d8](https://arbital.com/p/-4d8)) such that you can find a backwards translation (morphism in the opposite direction), and using the one translation after the other is for all intents and purposed the same as not having translated anything at all (i.e., no important information is lost in translation), then the original translation is an **isomorphism**. (In fact, both of them are isomorphisms).\n\nWhat if you had a different pair of dictionaries. \n\nIn particular, what if the Spanish-to-English dictionary translates *Huesto* to *Marybone*, *Puento de Buey* to *Tralfalgal*, *Patata* to *Oxbridge*, *Carbon* to *Eastweston*, and *Esteoeste* to *Charlesburrough*?\n\nThen if we translate from Spanish into English, and then translate back with the original English-to-Spanish dictionary, then *Patata* is first translated to *Oxbridge*, but then it is translated back into *Puente de Buey*. So it does not reverse the translation. Hence this is not an isomorphism.\n\nHowever, what if the English-to-Spanish Dictionary translates *Marybone* to *Huesto*, *Oxbridge* to *Patata*. *Eastweston* to *Puento de Buey*, *Tralfalgal* to *Esteoeste*, and *Charlesburrough* to *Carbon*? Then the translations are reverses of each other. Hence this is another isomorphism. There may be many isomorphisms between two isomorphic maps.\n\n\n\n##Non-Isomorphic Maps##\nImagine you had the English map from above. As a reminder the stations were:\n*Marybone*, *Eastweston*, *Charlesburrough*, *Tralfalgal*, and *Oxbridge*.\n\nIf, now, you find a Spanish map with only four stations on it, it can't possibly be isomorphic to your English map; there would be some station appearing on the English map which isn't named on the Spanish one. \n\n\n\nSimilarly, if the Spanish map has six stations, then they aren't isomorphic either since there is an extra station on the Spanish map not appearing on the English one.\n\n\n\nWhat if there are five stations on the Spanish map. Is it then definitely isomorphic to the English one?\n\nRecall that:\n*Marybone* is connected to every other station, and *Eastweston* is connected to *Charlesburrough*.\n\nBut what if now instead on the Spanish map, *Huesto* is still connected to everything, but nothing else is connected to anything else. \n\n\n\nThen the translation taking *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon* is not an isomorphism. *Eastweston* is connected to *Charlesburrough* on the English map, but the corresponding stations on the Spanish map, *Esteoeste* and *Carbon*, are not connected to each other. Hence this translation is not an isomorphism, since under these translations, the maps represent different things.\n\nBut even though this way of pairing up the stations isn't an isomorphism, maybe there is another way of pairing them up which is? But no, even this is doomed to failure because the *number of connections* in both cases is different. For example, *Eastweston* is connected to two stations, but no station on the Spanish map is connected to two other stations. Hence there cannot be any translation that works. No isomorphism exists between the two maps.\n\nIf no isomorphism exists between two structures, then they are **non-isomorphic**.\n\nNotice, in fact, that the two maps do not have the same total number of connections. There are five connections on the English map, but only four on the Spanish map. Hence they cannot be isomorphic. \n\nWhat if the Spanish map has the following connections? *Patata* to everything except *Carbon*, and *Carbon* to *Huesto*and *Esteoeste*. Then there still cannot be an isomorphism, since, again, *Marybone* is connected to four other stations, but nothing on the Spanish map is connected to four stations. In this case, both maps have five connections. Hence even if both maps have the same total number of connections, they may still be non-isomorphic.\n\n\n\n\n##Comparing Weights##\nNot all isomorphisms need be mappings between structures. Consider if you work at the post-office and must weigh packages. You do not care about the size and shape of the packages, only their weight. Then you consider two packages isomorphic if their weights are equal. \n\nImagine, then, that you have two packages, say one containing a book %note: *The Official London Underground History of Train Stations*% and the other is a plastic crate %note: \"To the Count, with love\"%. \n\nYou also have a half-broken pair of brass scales: they have a pair of pans on which items can be placed. \n\n\n\n\n\nHowever, they can only tip to the left or remain flat.\n\n\n\n\nIf the item on the left is heavier than the one on the right, then the scales tilt left.\n\n\n\nOtherwise, if they are of equal weight or the item on the left is lighter than the one on the right, the scales remain level.\n\n\n\n\n\n\nPlace the book on the left pan of the scale, and the crate on the right. If the scales balance then either the book is lighter than the crate or it is the same weight as the crate. Now swap them. If they remain level, then either the crate is lighter than the book or it is the same weight as the book. Since the book cannot be lighter than the crate whilst the crate is simultaneously lighter than the book, they must be the same weight. Hence they are isomorphic.\n\nThis very act of balancing the scales is an isomorphism. It has an inverse: just swap the two packages around! Start with the book on the left pan and the crate on the right. Then place the crate on the left pan and the book on the right. The fact that the scales balance both times tells you (the obvious fact) that the book weighs the same as itself. Since you already know this, doing this actually tells you as much about the book's weight compared to itself as doing nothing at all.\n\nIf this last part seems silly or confusing, don't worry too much about it. It's just to illustrate how the idea of an isomorphism is intricately tied with having an inverse.\n\n##An Isomorphism Joke##\nA man walks into a bar. He is surprised at how the patrons are acting. One of them says a number, like “forty-two”, and the rest break into laughter. He asks the bartender what’s going on. The bartender explains that they all come here so often that they’ve memorized all of each other’s jokes, and instead of telling them explicitly, they just give each a number, say the number, and laugh appropriately. The man is intrigued, so he shouts “Two thousand!”. He is shocked to find everyone laughs uproariously, the loudest he's heard that evening. Perplexed, he turns to the bartender and says “They laughed so much more at mine than at any of the others.\" \"Well of course,\" the bartender answers matter-of-factly, \"they've never heard that one before!”", "date_published": "2016-07-13T19:32:29Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Mark Chimes", "Team Arbital"], "summaries": [], "tags": ["Math 0", "Bijective function", "B-Class"], "alias": "4hj"}
{"id": "29d521d5b4ec1bae4cb014c60592c179", "title": "Sign homomorphism (from the symmetric group)", "url": "https://arbital.com/p/sign_homomorphism_symmetric_group", "source": "arbital", "source_type": "text", "text": "The sign [homomorphism](https://arbital.com/p/47t) is given by sending a permutation $\\sigma$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ to $0$ if we can make $\\sigma$ by multiplying together an even number of [transpositions](https://arbital.com/p/4cn), and to $1$ otherwise.\n\n%%%knows-requisite([https://arbital.com/p/modular_arithmetic](https://arbital.com/p/modular_arithmetic)):\nEquivalently, it is given by sending $\\sigma$ to the number of transpositions making it up, [modulo](https://arbital.com/p/modular_arithmetic) $2$.\n%%%\n\nThe sign homomorphism [is well-defined](https://arbital.com/p/4hh).\n\n%%%knows-requisite([https://arbital.com/p/quotient_group](https://arbital.com/p/quotient_group)):\nThe [https://arbital.com/p/-alternating_group](https://arbital.com/p/-alternating_group) is obtained by taking the [quotient](https://arbital.com/p/quotient_group) of the symmetric group by the sign homomorphism.\n%%%", "date_published": "2016-06-17T12:13:38Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Definition"], "alias": "4hk"}
{"id": "6c4b49c1b8fd6dcf6bd93e6e571973c5", "title": "Alternating group is generated by its three-cycles", "url": "https://arbital.com/p/alternating_group_generated_by_three_cycles", "source": "arbital", "source_type": "text", "text": "The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$ is generated by its $3$-[cycles](https://arbital.com/p/49f).\nThat is, every element of $A_n$ can be made by multiplying together $3$-cycles only.\n\n# Proof\n\nThe product of two [transpositions](https://arbital.com/p/4cn) is a product of $3$-cycles: \n\n- $(ij)(kl) = (ijk)(jkl)$\n- $(ij)(jk) = (ijk)$\n- $(ij)(ij) = e$.\n\nTherefore any permutation which is a product of evenly-many transpositions (that is, all of $A_n$) is a product of $3$-cycles, because we can group up successive pairs of transpositions.\n\nConversely, every $3$-cycle is in $A_n$ because $(ijk) = (ij)(jk)$.", "date_published": "2016-06-17T13:23:26Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4hs"}
{"id": "b711a3a66acd43d12512e3c908642c1a", "title": "Group coset", "url": "https://arbital.com/p/group_coset", "source": "arbital", "source_type": "text", "text": "Given a subgroup $H$ of [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$, the *left cosets* of $H$ in $G$ are sets of the form $\\{ gh : h \\in H \\}$, for some $g \\in G$.\nThis is written $gH$ as a shorthand.\n\nSimilarly, the *right cosets* are the sets of the form $Hg = \\{ hg: h \\in H \\}$.\n\n# Examples\n%%%knows-requisite([https://arbital.com/p/497](https://arbital.com/p/497)):\n## Symmetric group\n\nIn $S_3$, the [https://arbital.com/p/-497](https://arbital.com/p/-497) on three elements, we can list the elements as $\\{ e, (123), (132), (12), (13), (23) \\}$, using [cycle notation](https://arbital.com/p/49f).\nDefine $A_3$ (which happens to have a name: the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf)) to be the subgroup with elements $\\{ e, (123), (132) \\}$.\n\nThen the coset $(12) A_3$ has elements $\\{ (12), (12)(123), (12)(132) \\}$, which is simplified to $\\{ (12), (23), (13) \\}$.\n\nThe coset $(123)A_3$ is simply $A_3$, because $A_3$ is a subgroup so is closed under the group operation. $(123)$ is already in $A_3$.\n%%%\n\n\n\n# Properties\n\n- The left cosets of $H$ in $G$ [partition](https://arbital.com/p/set_partition) $G$. ([Proof.](https://arbital.com/p/4j5))\n- For any pair of left cosets of $H$, there is a [bijection](https://arbital.com/p/499) between them; that is, all the cosets are all the same size. ([Proof.](https://arbital.com/p/4j8))\n\n# Why are we interested in cosets?\n\nUnder certain conditions (namely that the subgroup $H$ must be [normal](https://arbital.com/p/4h6)), we may define the [https://arbital.com/p/-quotient_group](https://arbital.com/p/-quotient_group), a very important concept; see the page on [\"left cosets partition the parent group\"](https://arbital.com/p/4j5) for a glance at why this is useful.\n\n\nAdditionally, there is a key theorem whose usual proof considers cosets ([Lagrange's theorem](https://arbital.com/p/lagrange_theorem_on_subgroup_size)) which strongly restricts the possible sizes of subgroups of $G$, and which itself is enough to classify all the groups of [order](https://arbital.com/p/3gg) $p$ for $p$ [prime](https://arbital.com/p/prime_number).\nLagrange's theorem also has very common applications in [https://arbital.com/p/-number_theory](https://arbital.com/p/-number_theory), in the form of the [Fermat-Euler theorem](https://arbital.com/p/fermat_euler_theorem).", "date_published": "2016-06-17T15:58:15Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Formal definition", "Needs clickbait"], "alias": "4j4"}
{"id": "9c027a495b263865572d7eca86062f80", "title": "Left cosets partition the parent group", "url": "https://arbital.com/p/left_cosets_partition_parent_group", "source": "arbital", "source_type": "text", "text": "Given a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ and a subgroup $H$, the [left cosets](https://arbital.com/p/4j4) of $H$ in $G$ [partition](https://arbital.com/p/set_partition) $G$, in the sense that every element of $g$ is in precisely one coset.\n\n# Proof\nFirstly, every element is in a coset: since $g \\in gH$ for any $g$.\nSo we must show that no element is in more than one coset.\n\nSuppose $c$ is in both $aH$ and $bH$.\nThen we claim that $aH = cH = bH$, so in fact the two cosets $aH$ and $bH$ were the same.\nIndeed, $c \\in aH$, so there is $k \\in H$ such that $c = ak$.\nTherefore $cH = \\{ ch : h \\in H \\} = \\{ akh : h \\in H \\}$.\n\nExercise: $\\{ akh : h \\in H \\} = \\{ ar : r \\in H \\}$.\n%%hidden(Show solution):\nSuppose $akh$ is in the left-hand side.\nThen it is in the right-hand side immediately: letting $r=kh$.\n\nConversely, suppose $ar$ is in the right-hand side.\nThen we may write $r = k k^{-1} r$, so $a k k^{-1} r$ is in the right-hand side; but then $k^{-1} r$ is in $H$ so this is exactly an object which lies in the left-hand side.\n%%\n\nBut that is just $aH$.\n\nBy repeating the reasoning with $a$ and $b$ interchanged, we have $cH = bH$; this completes the proof.\n\n# Why is this interesting?\n\nThe fact that the left cosets partition the group means that we can, in some sense, \"compress\" the group $G$ with respect to $H$.\nIf we are only interested in $G$ \"up to\" $H$, we can deal with the partition rather than the individual elements, throwing away the information we're not interested in.\n\nThis concept is most importantly used in defining the [https://arbital.com/p/-4tq](https://arbital.com/p/-4tq).\nTo do this, the subgroup must be [normal](https://arbital.com/p/4h6) ([proof](https://arbital.com/p/4h9)).\nIn this case, the collection of cosets itself inherits a group structure from the parent group $G$, and the structure of the quotient group can often tell us a lot about the parent group.", "date_published": "2016-06-28T07:29:03Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4j5"}
{"id": "47c9435ffeb999cf2c5f4eb6cc4f1c95", "title": "Church-Turing thesis: Evidence for the Church-Turing thesis", "url": "https://arbital.com/p/evidence_for_CT_thesis", "source": "arbital", "source_type": "text", "text": "As the Church-Turing thesis is not a [proper mathematical sentence](https://arbital.com/p/) we cannot prove it. However, we can collect [https://arbital.com/p/-3s](https://arbital.com/p/-3s) to increase our confidence in its correctness.\n\n#The inductive argument\nEvery computational model we have seen so far is [reducible](https://arbital.com/p/) to Turing's model.\n\nIndeed, the thesis was originally formulated independently by Church and Turing in reference to two different computational models ([https://arbital.com/p/Turing_machines](https://arbital.com/p/Turing_machines) and [https://arbital.com/p/Lambda_Calculus](https://arbital.com/p/Lambda_Calculus) respectively). When they were shown to be [equivalent](https://arbital.com/p/equivilance_relation) it was massive evidence in favor of both of them.\n\nA non-exhaustive list of models which can be shown to be reducible to Turing machines are:\n\n* [https://arbital.com/p/Lambda_calculus](https://arbital.com/p/Lambda_calculus)\n* [https://arbital.com/p/Quantum_computation](https://arbital.com/p/Quantum_computation)\n* [Non-deterministic_Turing_machines](https://arbital.com/p/Nondeterministic_Turing_machines)\n* [https://arbital.com/p/Register_machines](https://arbital.com/p/Register_machines)\n* The set of [https://arbital.com/p/-recursive_functions](https://arbital.com/p/-recursive_functions)\n\n#Lack of counterexamples\nPerhaps the strongest argument we have for the CT thesis is that there is not a widely accepted candidate to a counterexample of the thesis. This is unlike the [https://arbital.com/p/-4ct](https://arbital.com/p/-4ct), where quantum computation stands as a likely counterexample.\n\nOne may wonder whether the computational models which use a source of [randomness](https://arbital.com/p/) (such as quantum computation or [https://arbital.com/p/-probabilistic_Turing_machines](https://arbital.com/p/-probabilistic_Turing_machines)) are a proper counterexample to the thesis: after all, Turing machines are fully [https://arbital.com/p/-deterministic](https://arbital.com/p/-deterministic), so they cannot simulate randomness.\n\nTo properly explain this issue we have to recall what it means for a quantum computer or a probabilistic Turing machine to compute something: we say that such a device computes a function $f$ if for every input $x$ the probability of the device outputting $f(x)$ is greater or equal to some arbitrary constant greater than $1/2$. Thus we can compute $f$ in a classical machine, for there exists always the possibility of simulating every possible outcome of the randomness to deterministically compute the probability distribution of the output, and output as a result the possible outcome with greater probability.\n\nThus, randomness is reducible to Turing machines, and the CT thesis holds.", "date_published": "2016-06-20T17:59:25Z", "authors": ["Eric Bruylant", "Jaime Sevilla Molina"], "summaries": [], "tags": [], "alias": "4j7"}
{"id": "400da54ae4510760f5a636e14bd71598", "title": "Left cosets are all in bijection", "url": "https://arbital.com/p/left_cosets_biject", "source": "arbital", "source_type": "text", "text": "Let $H$ be a subgroup of $G$. \nThen for any two [left cosets](https://arbital.com/p/4j4) of $H$ in $G$, there is a [https://arbital.com/p/-499](https://arbital.com/p/-499) between the two cosets.\n\n# Proof\n\nLet $aH, bH$ be two cosets.\nDefine the function $f: aH \\to bH$ by $x \\mapsto b a^{-1} x$.\n\nThis has the correct [codomain](https://arbital.com/p/3lg): if $x \\in aH$ (so $x = ah$, say), then $ba^{-1} a x = bx$ so $f(x) \\in bH$.\n\nThe function is [injective](https://arbital.com/p/4b7): if $b a^{-1} x = b a^{-1} y$ then (pre-multiplying both sides by $a b^{-1}$) we obtain $x = y$.\n\nThe function is [surjective](https://arbital.com/p/4bg): given $b h \\in b H $, we want to find $x \\in aH$ such that $f(x) = bh$.\nLet $x = a h$ to obtain $f(x) = b a^{-1} a h = b h$, as required.", "date_published": "2016-06-17T16:00:59Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4j8"}
{"id": "62247dccd97eb7668f9dc4d057d189cf", "title": "The alternating groups on more than four letters are simple", "url": "https://arbital.com/p/alternating_group_is_simple", "source": "arbital", "source_type": "text", "text": "Let $n > 4$ be a [https://arbital.com/p/-45h](https://arbital.com/p/-45h). Then the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$ on $n$ elements is [simple](https://arbital.com/p/simple_group).\n\n# Proof\n\nWe go by [induction](https://arbital.com/p/mathematical_induction) on $n$.\nThe base case of $n=5$ [we treat separately](https://arbital.com/p/4jf).\n\nFor the inductive step: let $n \\geq 6$, and suppose $H$ is a nontrivial [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $A_n$.\nUltimately we aim to show that $H$ contains the $3$-[cycle](https://arbital.com/p/49f) $(123)$; since $H$ [is a union of conjugacy classes](https://arbital.com/p/subgroup_normal_iff_union_of_conjugacy_classes), and since the $3$-cycles [form a conjugacy class](https://arbital.com/p/splitting_conjugacy_classes_in_alternating_group) in $A_n$, this means every $3$-cycle is in $H$ and [hence](https://arbital.com/p/4hs) $H = A_n$.\n\n## Lemma: $H$ contains a member of $A_{n-1}$\nTo start, we will show that at least $H$ contains *something* from $A_{n-1}$ (which we view as all the elements of $A_n$ which don't change the letter $n$). %%note:Recall that $A_n$ [acts](https://arbital.com/p/3t9) naturally on the set of \"letters\" $\\{1,2,\\dots, n \\}$ by permutation.%%\n(This approach has to be useful for an induction proof, because we need some way of introducing the simplicity of $A_{n-1}$.)\nThat is, we will show that there is some $\\sigma \\in H$ with $\\sigma \\not = e$, such that $\\sigma(n) = n$.\n\nLet $\\sigma \\in H$, where $\\sigma$ is not the identity.\n$\\sigma$ certainly sends $n$ somewhere; say $\\sigma(n) = i$, where $i \\not = n$ (since if it is, we're done immediately).\n\nThen if we can find some $\\sigma' \\in H$, not equal to $\\sigma$, such that $\\sigma'(n) = i$, we are done: $\\sigma^{-1} \\sigma'(n) = n$.\n\n$\\sigma$ must move something other than $i$ (that is, there must be $j \\not = i$ such that $\\sigma(j) \\not = j$), because if it did not, it would be the transposition $(n i)$, which is not in $A_n$ because it is an odd number of transpositions.\nHence $\\sigma(j) \\not = j$ and $j \\not = i$; also $j \\not = n$ because if it were, \n\nNow, since $n \\geq 6$, we can pick $x, y$ distinct from $n, i, j, \\sigma(j)$.\nThen set $\\sigma' = (jxy) \\sigma (jxy)^{-1}$, which must lie in $H$ because $H$ is closed under [conjugation](https://arbital.com/p/4gk).\n\nThen $\\sigma'(n) = i$; and $\\sigma' \\not = \\sigma$, because $\\sigma'(j) = \\sigma(y)$ which is not equal to $\\sigma(j)$ (since $y \\not = j$).\nHence $\\sigma'$ and $\\sigma$ have different effects on $j$ so they are not equal.\n\n## Lemma: $H$ contains all of $A_{n-1}$\nNow that we have shown $H$ contains some member of $A_{n-1}$.\nBut $H \\cap A_{n-1}$ is normal in $A_n$, because $H$ is normal in $A_n$ and it is [_intersect _subgroup _is _normal the intersection of a subgroup with a normal subgroup](https://arbital.com/p/4h6).\nTherefore by the inductive hypothesis, $H \\cap A_{n-1}$ is either the trivial group or is $A_{n-1}$ itself.\n\nBut $H \\cap A_{n-1}$ is certainly not trivial, because our previous lemma gave us a non-identity element in it; so $H$ must actually contain $A_{n-1}$.\n\n## Conclusion\n\nFinally, $H$ contains $A_{n-1}$ so it contains $(123)$ in particular; so we are done by the initial discussion.\n\n# Behaviour for $n \\leq 4$\n- $A_1$ is the trivial group so is vacuously not simple.\n- $A_2$ is also the trivial group. \n- $A_3$ is isomorphic to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_3$ on three generators, so it is simple: it has no nontrivial proper subgroups, let alone normal ones.\n- $A_4$ has the following normal subgroup (the [Klein four-group](https://arbital.com/p/klein_four_group)): $\\{ e, (12)(34), (13)(24), (14)(23) \\}$. Therefore $A_4$ is not simple.", "date_published": "2016-06-17T17:19:24Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Stub"], "alias": "4jb"}
{"id": "6b3025e57bbfcf556b8f9dec4acbb8bd", "title": "Simple group", "url": "https://arbital.com/p/simple_group", "source": "arbital", "source_type": "text", "text": "A simple group is a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) with no [normal subgroups](https://arbital.com/p/4h6).", "date_published": "2016-06-17T16:06:49Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Definition", "Stub"], "alias": "4jc"}
{"id": "c2a6aa2a6edda4cead06ce5df0b58b3c", "title": "The alternating group on five elements is simple", "url": "https://arbital.com/p/alternating_group_five_is_simple", "source": "arbital", "source_type": "text", "text": "The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements is [simple](https://arbital.com/p/4jc).\n\n# Proof\n\nRecall that $A_5$ has [order](https://arbital.com/p/3gg) $60$, so [Lagrange's theorem](https://arbital.com/p/4jn) states that any subgroup of $A_5$ has order dividing $60$.\n\nSuppose $H$ is a normal subgroup of $A_5$, which is not the trivial subgroup $\\{ e \\}$.\nIf $H$ has order divisible by $3$, then by [Cauchy's theorem](https://arbital.com/p/4l6) there is a $3$-[cycle](https://arbital.com/p/49f) in $H$ (because the $3$-cycles are the only elements with order $3$ in $A_5$).\nBecause $H$ [is a union of conjugacy classes](https://arbital.com/p/4jw), and because the $3$-cycles [form a conjugacy class in $A_n$ for $n > 4$](https://arbital.com/p/4kv), $H$ would therefore contain *every* $3$-cycle; but then [it would be the entire alternating group](https://arbital.com/p/4hs).\n\nIf instead $H$ has order divisible by $2$, then there is a double transposition such as $(12)(34)$ in $H$, since these are the only elements of order $2$ in $A_5$.\nBut then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12)(34)$ and $(15)(34)$, so it contains $(15)(34)(12)(34) = (125)$.\nHence as before $H$ contains every $3$-cycle so is the entire alternating group.\n\nSo $H$ must have order exactly $5$, by [Lagrange's theorem](https://arbital.com/p/4jn); so it contains an element of order $5$ since [prime order groups are cyclic](https://arbital.com/p/4jh).\n\nThe only such elements of $A_n$ are $5$-cycles; but the conjugacy class of a $5$-cycle is of size $12$, which is too big to fit in $H$ which has size $5$.", "date_published": "2016-06-28T06:23:28Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4jf"}
{"id": "39903915e64ca91d97dfb7ceaac5e34a", "title": "Prime order groups are cyclic", "url": "https://arbital.com/p/prime_order_group_is_cyclic", "source": "arbital", "source_type": "text", "text": "Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) whose [order](https://arbital.com/p/3gg) is equal to $p$, a [https://arbital.com/p/-4mf](https://arbital.com/p/-4mf).\nThen $G$ is [isomorphic](https://arbital.com/p/49x) to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_p$ of order $p$.\n\n# Proof\nPick any non-identity element $g$ of the group.\n\nBy [Lagrange's theorem](https://arbital.com/p/4jn), the subgroup generated by $g$ has size $1$ or $p$ (since $p$ was prime).\nBut it can't be $1$ because the only subgroup of size $1$ is the trivial subgroup.\n\nHence the subgroup must be the entire group.", "date_published": "2016-06-20T06:47:24Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4jh"}
{"id": "d4ebd0c5e0785748f256157c445ae952", "title": "Lagrange theorem on subgroup size", "url": "https://arbital.com/p/lagrange_theorem_on_subgroup_size", "source": "arbital", "source_type": "text", "text": "Lagrange's Theorem states that if $G$ is a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $H$ a subgroup, then the [order](https://arbital.com/p/3gg) $|H|$ of $H$ divides the order $|G|$ of $G$.\nIt generalises to infinite groups: the statement then becomes that the [left cosets](https://arbital.com/p/4j4) form a [partition](https://arbital.com/p/set_partition), and for any pair of cosets, there is a [bijection](https://arbital.com/p/499) between them.\n\n# Proof\n\nIn full generality, the cosets [form a partition](https://arbital.com/p/4j5) and [are all in bijection](https://arbital.com/p/4j8).\n\nTo specialise this to the finite case, we have divided the $|G|$ elements of $G$ into buckets of size $|H|$ (namely, the cosets), so $|G|/|H|$ must in particular be an integer.", "date_published": "2016-06-17T19:24:51Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4jn"}
{"id": "0b6b67d54cd2b7e3819ba2c870fea6b8", "title": "The alternating group on five elements is simple: Simpler proof", "url": "https://arbital.com/p/simplicity_of_alternating_group_five_simpler_proof", "source": "arbital", "source_type": "text", "text": "If there is a non-trivial [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) $H$ of the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements, then [it is a union of conjugacy classes](https://arbital.com/p/4jw).\nAdditionally, by [Lagrange's theorem](https://arbital.com/p/4jn), the [order](https://arbital.com/p/3gg) of a subgroup must divide the order of the group, so the total size of $H$ must divide $60$.\n\nWe can list the [conjugacy classes of $A_5$](https://arbital.com/p/alternating_group_five_conjugacy_classes); they are of size $1, 20, 15, 12, 12$ respectively.\n\nBy a brute-force check, no sum of these containing $1$ can possibly divide $60$ (which is the size of $A_5$) unless it is $1$ or $60$.\n\n# The brute-force check\n\nWe first list the [divisors](https://arbital.com/p/number_theory_divisor) of $60$: they are $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$$\n\nSince the subgroup $H$ must contain the identity, it must contain the conjugacy class of size $1$.\nIf it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $13$ (since the smallest other class is of size $12$); so it is allowed to be one of $15$, $20$, $30$, or $60$.\nSince $H$ is also assumed to be a *proper* subgroup, it cannot be $A_5$ itself, so in fact $60$ is also banned.\n\n## The class of size $20$\nIf $H$ then contains the conjugacy class of size $20$, then $H$ can only be of size $30$ because we have already included $21$ elements.\nBut there is no way to add just $9$ elements using conjugacy classes of size bigger than or equal to $12$.\n\nSo $H$ cannot contain the class of size $20$.\n\n## The class of size $15$\n\nIn this case, $H$ is allowed to be of size $20$ or $30$, and we have already found $16$ elements of it. So there are either $4$ or $14$ elements left to find; but we are only allowed to add classes of size exactly $12$, so this can't be done either.\n\n## The classes of size $12$\n\nWhat remains is two classes of size $12$, from which we can make $1+12 = 13$ or $1+12+12 = 25$.\nNeither of these divides $60$, so these are not legal options either.\n\nThis exhausts the search, and completes the proof.", "date_published": "2016-06-17T19:38:57Z", "authors": ["Team Arbital", "Patrick Stevens"], "summaries": [], "tags": [], "alias": "4jt"}
{"id": "c7ff8f308bb6a4e840e5c5801f5d4d70", "title": "Subgroup is normal if and only if it is a union of conjugacy classes", "url": "https://arbital.com/p/subgroup_normal_iff_union_of_conjugacy_classes", "source": "arbital", "source_type": "text", "text": "Let $H$ be a subgroup of the [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$.\nThen $H$ is [normal](https://arbital.com/p/4h6) in $G$ if and only if it can be expressed as a [union](https://arbital.com/p/set_union) of [conjugacy classes](https://arbital.com/p/4bj).\n\n# Proof\n\n$H$ is normal in $G$ if and only if $gHg^{-1} = H$ for all $g \\in G$; equivalently, if and only if $ghg^{-1} \\in H$ for all $h \\in H$ and $g \\in G$.\n\nBut if we fix $h \\in H$, then the statement that $ghg^{-1} \\in H$ for all $g \\in G$ is equivalent to insisting that the conjugacy class of $h$ is contained in $H$.\nTherefore $H$ is normal in $G$ if and only if, for all $h \\in H$, the conjugacy class of $h$ lies in $H$.\n\nIf $H$ is normal, then it is clearly a union of conjugacy classes (namely $\\cup_{h \\in H} C_h$, where $C_h$ is the conjugacy class of $h$).\n\nConversely, if $H$ is not normal, then there is some $h \\in H$ such that the conjugacy class of $h$ is not wholly in $H$; so $H$ is not a union of conjugacy classes because it contains $h$ but not the entire conjugacy class of $h$.\n(Here we have used that the [https://arbital.com/p/-conjugacy_classes_partition_the_group](https://arbital.com/p/-conjugacy_classes_partition_the_group).)\n\n# Interpretation\n\nA normal subgroup is one which is fixed under conjugation; the most natural (and, indeed, the smallest) objects which are fixed under conjugation are conjugacy classes; so this criterion tells us that to obtain a *subgroup* which is fixed under conjugation, it is necessary and sufficient to assemble these objects (the conjugacy classes), which are themselves the smallest objects which are fixed under conjugation, into a group.", "date_published": "2016-06-17T19:37:34Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4jw"}
{"id": "9c027442545253c9d0e985eaf53a04dc", "title": "Conjugacy classes of the alternating group on five elements", "url": "https://arbital.com/p/alternating_group_five_conjugacy_classes", "source": "arbital", "source_type": "text", "text": "This page lists the [conjugacy classes](https://arbital.com/p/4bj) of the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements.\nSee a [different lens](https://arbital.com/p/4l0) for a derivation of this result using less theory.\n\n$A_5$ has size $5!/2 = 60$, where the exclamation mark denotes the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function.\nWe will assume access to [the conjugacy class table of $S_5$](https://arbital.com/p/4bk) the [https://arbital.com/p/-497](https://arbital.com/p/-497) on five elements; $A_5$ is a [quotient](https://arbital.com/p/quotient_group) of $S_5$ by the [sign homomorphism](https://arbital.com/p/4hk).\n\nWe have that a conjugacy class splits if and only if its [cycle type](https://arbital.com/p/49f) is all odd, all distinct. ([Proof.](https://arbital.com/p/4kv))\nThis makes the classification of conjugacy classes very easy.\n\n# The table\n\nWe must remove all the lines of [$S_5$'s table](https://arbital.com/p/4bk) which correspond to odd permutations (that is, those which are the product of odd-many [transpositions](https://arbital.com/p/4cn)). Indeed, those lines are classes which are not even in $A_5$.\n\nWe are left with cycle types $(5)$, $(3, 1, 1)$, $(2, 2, 1)$, $(1,1,1,1,1)$.\nOnly the $(5)$ cycle type can split into two, by the splitting condition.\nIt splits into the class containing $(12345)$ and the class which is $(12345)$ conjugated by odd permutations in $S_5$.\nA representative for that latter class is $(12)(12345)(12)^{-1} = (21345)$.\n\n$$\\begin{array}{|c|c|c|c|}\n\\hline\n\\text{Representative}& \\text{Size of class} & \\text{Cycle type} & \\text{Order of element} \\\\ \\hline\n(12345) & 12 & 5 & 5 \\\\ \\hline\n(21345) & 12 & 5 & 5 \\\\ \\hline\n(123) & 20 & 3,1,1 & 3 \\\\ \\hline\n(12)(34) & 15 & 2,2,1 & 2 \\\\ \\hline\ne & 1 & 1,1,1,1,1 & 1 \\\\ \\hline\n\\end{array}$$", "date_published": "2016-06-18T13:41:33Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4kr"}
{"id": "d15fbc97003683db1283bda74765530a", "title": "Splitting conjugacy classes in alternating group", "url": "https://arbital.com/p/splitting_conjugacy_classes_in_alternating_group", "source": "arbital", "source_type": "text", "text": "Recall that in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, the notion of \"[https://arbital.com/p/-4bj](https://arbital.com/p/-4bj)\" coincides with that of \"has the same [cycle type](https://arbital.com/p/4cg)\" ([proof](https://arbital.com/p/4bh)).\nIt turns out to be the case that when we descend to the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$, the conjugacy classes nearly all remain the same; the exception is those classes in $S_n$ which have cycle type all odd and all distinct.\nThose classes split into exactly two classes in $A_n$.\n\n# Proof\n\n## Conjugate implies same cycle type\n\nThis direction of the proof is identical to [the same direction](https://arbital.com/p/4bh) in the proof of the corresponding result on symmetric groups.\n\n## Converse: splitting condition\n\nRecall before we start that an even-length cycle can only be written as the product of an *odd* number of transpositions, and vice versa.\n\nThe question is: is every $\\tau \\in A_n$ with the same cycle type as $\\sigma \\in A_n$ conjugate (in $A_n)$ to $\\sigma$?\nIf so, then the conjugacy class in $A_n$ of $\\sigma$ is just the same as that of $\\sigma$ in $S_n$; if not, then the conjugacy class in $S_n$ must break into two pieces in $A_n$, namely $\\{ \\rho \\sigma \\rho^{-1} : \\rho \\ \\text{even} \\}$ and $\\{ \\rho \\sigma \\rho^{-1} : \\rho \\ \\text{odd} \\}$.\n(Certainly these are conjugacy classes: they only contain even permutations so they are subsets of $A_n$, while everything in either class is conjugate in $A_n$ because the definition only depends on the [parity](https://arbital.com/p/even_odd_parity) of $\\rho$.)\n\nLet $\\sigma = c_1 \\dots c_k$, and $\\tau = c_1' \\dots c_k'$ of the same cycle type: $c_i = (a_{i1} \\dots a_{i r_i})$, $c_i' = (b_{i1} \\dots b_{i r_i})$.\n\nDefine $\\rho$ to be the permutation which takes $a_{ij}$ to $b_{ij}$, but otherwise does not move any other letter.\nThen $\\rho \\sigma \\rho^{-1} = \\tau$, so if $\\rho$ lies in $A_n$ then we are done: $\\sigma$ and $\\tau$ are conjugate in $A_n$.\n\nThat is, we may assume without loss of generality that $\\rho$ is odd.\n\n### If any of the cycles is even in length\n\nSuppose without loss of generality that $r_1$, the length of the first cycle, is even.\nThen we can rewrite $c_1' = (b_{11} \\dots b_{1r_1})$ as $(b_{12} b_{13} \\dots b_{1 r_1} b_{11})$, which is the same cycle expressed slightly differently.\n\nNow $c_1'$ is even in length, so it is an odd permutation (being a product of an odd number of transpositions, $(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \\dots (b_{13} b_{11})(b_{12} b_{11})$).\nHence $\\rho c_1'$ is an even permutation.\n\nBut conjugating $\\tau$ by $\\rho c_1'$ yields $\\sigma$:\n$$\\sigma = \\rho \\tau \\rho^{-1} = \\rho c_1' (c_1'^{-1} \\tau c_1') c_1'^{-1} \\rho^{-1}$$\nwhich is the result of conjugating $c_1'^{-1} \\tau c_1' = \\tau$ by $\\rho c_1'$.\n\n(It is the case that $c_1'^{-1} \\tau c_1' = \\tau$, because $c_1'$ commutes with all of $\\tau$ except with the first cycle $c_1'$, so the expression is $c_1'^{-1} c_1' c_1' c_2' \\dots c_k'$, where we have pulled the final $c_1'$ through to the beginning of the expression $\\tau = c_1' c_2' \\dots c_k'$.)\n\n%%hidden(Example):\nSuppose $\\sigma = (12)(3456), \\tau = (23)(1467)$.\n\nWe have that $\\tau$ is taken to $\\sigma$ by conjugating with $\\rho = (67)(56)(31)(23)(12)$, which is an odd permutation so isn't in $A_n$.\nBut we can rewrite $\\tau = (32)(1467)$, and the new permutation we'll conjugate with is $\\rho c_1' = (67)(56)(31)(23)(12)(32)$, where we have appended $c_1' = (23) = (32)$.\nIt is the case that $\\rho$ is an even permutation and hence is in $A_n$, because it is the result of multiplying the odd permutation $\\rho$ by the odd permutation $c_1'$.\n\nNow the conjugation is the composition of two conjugations: first by $(32)$, to yield $(32)\\tau(32)^{-1} = (23)(1467)$ (which is $\\tau$ still!), and then by $\\rho$.\nBut we constructed $\\rho$ so as to take $\\tau$ to $\\sigma$ on conjugation, so this works just as we needed.\n%%\n\n## If all the cycles are of odd length, but some length is repeated\nWithout loss of generality, suppose $r_1 = r_2$ (and label them both $r$), so the first two cycles are of the same length: say $\\sigma$'s version is $(a_1 a_2 \\dots a_r)(c_1 c_2 \\dots c_r)$, and $\\tau$'s version is $(b_1 b_2 \\dots b_r)(d_1 d_2 \\dots d_r)$.\n\nThen define $\\rho' = \\rho (b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$.\nSince $r$ is odd and $\\rho$ is an odd permutation, $\\rho'$ is an even permutation.\n\nNow conjugating by $\\rho'$ is the same as first conjugating by $(b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$ and then by $\\rho$.\n\nBut conjugating by $(b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$ takes $\\tau$ to $\\tau$, because it has the effect of replacing all the $b_i$ by $d_i$ and all the $d_i$ by $b_i$, so it simply swaps the two cycles round.\n\nHence the conjugation of $\\tau$ by $\\rho'$ yields $\\sigma$, and $\\rho'$ is in $A_n$. \n\n%%hidden(Example):\nSuppose $\\sigma = (123)(456), \\tau = (154)(237)$.\n\nThen conjugation of $\\tau$ by $\\rho = (67)(35)(42)(34)(25)$, an odd permutation, yields $\\sigma$.\n\nNow, if we first perform the conjugation $(12)(53)(47)$, we take $\\tau$ to itself, and then performing $\\rho$ yields $\\sigma$.\nThe combination of $\\rho$ and $(12)(53)(47)$ is an even permutation, so it does line in $A_n$.\n%%\n\n## If all the cycles are of odd length, and they are all distinct\n\nIn this case, we are required to check that the conjugacy class *does* split.\nRemember, we started out by supposing $\\sigma$ and $\\tau$ have the same cycle type, and they are conjugate in $S_n$ by an odd permutation (so they are not obviously conjugate in $A_n$); we need to show that indeed they are *not* conjugate in $A_n$.\n\nIndeed, the only ways to rewrite $\\tau$ into $\\sigma$ (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in $\\sigma$. There is no choice about which $\\tau$-cycle we take to which $\\sigma$-cycle, because all the cycle lengths are distinct.\nBut the permutation $\\rho$ which takes the $\\tau$-cycles to the $\\sigma$-cycles is odd, so is not in $A_n$.\n\nMoreover, since each cycle is odd, we can't get past the problem by just cycling round the cycle (for instance, by taking the cycle $(123)$ to the cycle $(231)$), because that involves conjugating by the cycle itself: an even permutation, since the cycle length is odd.\nComposing with the even permutation can't take $\\rho$ from being odd to being even.\n\nTherefore $\\tau$ and $\\sigma$ genuinely are not conjugate in $A_n$, so the conjugacy class splits.\n\n%%hidden(Example):\nSuppose $\\sigma = (12345)(678), \\tau = (12345)(687)$ in $A_8$.\n\nThen conjugation of $\\tau$ by $\\rho = (87)$, an odd permutation, yields $\\sigma$.\nCan we do this with an *even* permutation instead?\n\nConjugating $\\tau$ by anything at all must keep the cycle type the same, so the thing we conjugate by must take $(12345)$ to $(12345)$ and $(687)$ to $(678)$.\n\nThe only ways of $(12345)$ to $(12345)$ are by conjugating by some power of $(12345)$ itself; that is even.\nThe only ways of taking $(687)$ to $(678)$ are by conjugating by $(87)$, or by $(87)$ and then some power of $(678)$; all of these are odd.\n\nTherefore the only possible ways of taking $\\tau$ to $\\sigma$ involve conjugating by an odd permutation $(678)^m(87)$, possibly alongside some powers of an even permutation $(12345)$; therefore to get from $\\tau$ to $\\sigma$ requires an odd permutation, so they are in fact not conjugate in $A_8$.\n%%\n\n# Example\n\nIn $A_7$, the cycle types are $(7)$, $(5, 1, 1)$, $(4,2,1)$, $(3,2,2)$, $(3,3,1)$, $(3,1,1,1,1)$, $(1,1,1,1,1,1,1)$, and $(2,2,1,1,1)$.\nThe only class which splits is the $7$-cycles, of cycle type $(7)$; it splits into a pair of half-sized classes with representatives $(1234567)$ and $(12)(1234567)(12)^{-1} = (2134567)$.", "date_published": "2016-06-28T07:44:43Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4kv"}
{"id": "bf97588cc3afc769f94588b2745eebd9", "title": "Decision problem", "url": "https://arbital.com/p/decision_problem", "source": "arbital", "source_type": "text", "text": "A **decision problem** is a subset $D$ of a set of instances $A$, where generally $A$ is the set of finite [https://arbital.com/p/-bitstrings](https://arbital.com/p/-bitstrings) $\\{0,1\\}^*$.\n\nIn plain English, a decision problem is composed by a number of members of a collection, which generally share a common property. \n\nIn plainer English, a decision problem is a question of the form \"does bitstring $w$ have property $p$, yes or no?\" \n\nEvery question that fits the set form (is this string in the set?) can be re-expressed as \"does this bitstring have the property of being in the set?\" Everything that fits the property form can be re-expressed in set form as well, because a property implicitly specifies a set of things that have the property. Thus, the two forms are equivalent.\n\nIt is called a decision problem because we are trying to *decide* whether a given bitstring belongs to the set or not.\n\nA **solution** of the problem would consist in a procedure which given an instance $w$ of $A$ decides correctly in finite time whether $w$ belongs to $D$ or not. That is, a solution consists in a way of identifying the common trait shared by the members of $D$ which distinguish them from the rest of instances in $A$.\n\nWe say that a problem is [https://arbital.com/p/-decidable](https://arbital.com/p/-decidable) if there exists a solution which works for every instance. Trivially, finite decision problems are decidable, since we can have a look-up list with every element which we would consult every time we are given an instance to classify. If the instance corresponds to an element in the list, we accept it, and otherwise reject it. \n\nWe say that a decision problem is [https://arbital.com/p/-semidecidable](https://arbital.com/p/-semidecidable) if there is a procedure which, in finite time, identifies members of $D$, but fails to [https://arbital.com/p/-halt](https://arbital.com/p/-halt) and reject instances which do not belong to $D$.\n\nDecision problems can be classified according to their difficulty of solving in [complexity classes](https://arbital.com/p/), and are a central object of study in [https://arbital.com/p/-49w](https://arbital.com/p/-49w).\n\n#Examples\n##Graph connectedness\nSuppose we encode every possible finite graph, up to isomorphism, in the collection of finite bitstrings. Then we could define:\n$$\nCONNECTED = \\{s\\in\\{0,1\\}^*:\\text{$s$ represents a connected graph}\\}\n$$\nWhich would be a [decidable decision problem](https://arbital.com/p/graph_connectedness).\n\n##Tautology identification\nLet $TAUTOLOGY$ be the collection of formulas of [https://arbital.com/p/-first_order_logic](https://arbital.com/p/-first_order_logic) which are true for every [interpretation](https://arbital.com/p/mathematical_interpretation). Then [$TAUTOLOGY$ is a semidecidable decision problem](https://arbital.com/p/), as if a formula is [https://arbital.com/p/-valid](https://arbital.com/p/-valid) we can search for a proof, but otherwise, we do not have an effective procedure to decide that a formula is not valid.\n\n##Primality\nLet:\n$$\nPRIMES = \\{ x\\in \\mathbb{N}:\\text{$x$ is prime}\\}\n$$\nThen $PRIMES$ is a decidable decision problem.\n\nWe can alternately define\n$$\nPRIMES = \\{s\\in\\{0,1\\}^*:\\text{$s$ represent a prime number in base $2$}\\}\n$$\nThis illustrates that we can specify a particular decision problem in different ways.", "date_published": "2016-07-04T16:24:40Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Jaime Sevilla Molina", "Alex Appel"], "summaries": ["A decision problem is a question of the form \"does [https://arbital.com/p/-bitstring](https://arbital.com/p/-bitstring) $w$ have [https://arbital.com/p/-property](https://arbital.com/p/-property) $p$, yes or no?\""], "tags": ["Start"], "alias": "4kz"}
{"id": "3876354db66e9cef9f4118cf9599e2ff", "title": "Conjugacy classes of the alternating group on five elements: Simpler proof", "url": "https://arbital.com/p/conjugacy_classes_alternating_five_simpler", "source": "arbital", "source_type": "text", "text": "The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements has [conjugacy classes](https://arbital.com/p/4bj) very similar to those of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$, but where one of the classes has split in two.\n\nNote that $A_5$ has $60$ elements, since it is precisely half of the elements of $S_5$ which has $5! = 120$ elements (where the exclamation mark is the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function).\n\n# Table\n\n$$\\begin{array}{|c|c|c|c|}\n\\hline\n\\text{Representative}& \\text{Size of class} & \\text{Cycle type} & \\text{Order of element} \\\\ \\hline\n(12345) & 12 & 5 & 5 \\\\ \\hline\n(21345) & 12 & 5 & 5 \\\\ \\hline\n(123) & 20 & 3,1,1 & 3 \\\\ \\hline\n(12)(34) & 15 & 2,2,1 & 2 \\\\ \\hline\ne & 1 & 1,1,1,1,1 & 1 \\\\ \\hline\n\\end{array}$$\n\n# Working\n\nFirstly, the identity is in a class of its own, because $\\tau e \\tau^{-1} = \\tau \\tau^{-1} = e$ for every $\\tau$. \n\nNow, by the same reasoning as in [the proof](https://arbital.com/p/4bh) that conjugate elements must have the same cycle type in $S_n$, that result also holds in $A_n$.\n\nHence we just need to see whether any of the cycle types comprise more than one conjugacy class.\n\nRecall that the available cycle types are $(5)$, $(3,1,1)$, $(2,2,1)$, $(1,1,1,1,1)$ (the last of which is the identity and we have already considered it).\n\n## Double-transpositions\n\nAll the double-transpositions are conjugate (so the $(2,2,1)$ cycle type does not split):\n\n- $(ab)(cd)$ is conjugate to $(ab)(ce)$ if we conjugate by $(ab)(de)$; symmetrically this covers all the cases where one of the two transpositions remains the same.\n- $(ab)(cd)$ is conjugate to $(ac)(bd)$ by $(cba)$; this covers the case that $e$ is not introduced.\n- $(ab)(cd)$ is conjugate to $(ac)(be)$ by $(bc)(de)$; this covers the remaining cases that $e$ is introduced and neither of the two transpositions remains the same.\n\n## Three-cycles\n\nAll the three-cycles are conjugate (so the $(3,1,1)$ cycle type does not split): \n\n- $(abc)$ is conjugate to $(acb)$ by $(bc)(de)$, so three-cycles are conjugate to their permutations.\n- $(abc)$ is conjugate to $(abd)$ by $(cde)$; this covers the case of introducing a single new element to the cycle.\n- $(abc)$ is conjugate to $(ade)$ by $(bd)(ce)$; this covers the case of introducing two new elements to the cycle.\n\n## Five-cycles\n\nThis class does split: I claim that $(12345)$ and $(21345)$ are not conjugate.\n(Once we have this, then the class must split into two chunks, since $\\{ \\rho (12345) \\rho^{-1}: \\rho \\ \\text{even} \\}$ is closed under conjugation in $A_5$, and $\\{ \\rho (12345) \\rho^{-1}: \\rho \\ \\text{odd} \\}$ is closed under conjugation in $A_5$.\nThe first is the conjugacy class of $(12345)$ in $A_5$; the second is the conjugacy class of $(21345) = (12)(12345)(12)^{-1}$.\nThe only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)\n\nRecall that $\\tau (12345) \\tau^{-1} = (\\tau(1), \\tau(2), \\tau(3), \\tau(4), \\tau(5))$, so we would need a permutation $\\tau$ such that $\\tau$ sends $1$ to $2$, $2$ to $1$, $3$ to $3$, $4$ to $4$, and $5$ to $5$.\nThe only such permutation is $(12)$, the transposition, but that is not actually a member of $A_5$.\n\nHence in fact $(12345)$ and $(21345)$ are not conjugate.", "date_published": "2016-06-18T13:07:31Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": [], "alias": "4l0"}
{"id": "78f1d53b6fdba907879776b75bdb9e5e", "title": "Poset: Exercises", "url": "https://arbital.com/p/poset_exercises", "source": "arbital", "source_type": "text", "text": "Try these exercises to test your poset knowledge.\n\n# Corporate Ladder\n\nImagine a company with five distinct roles: CEO, marketing manager, marketer, IT manager, and IT worker. At this company, if the CEO gives an order, everyone else must follow it. If an order comes from an IT manager, only IT workers are obligated to follow it. Similarly, if an order comes from a marketing manager, only marketers are obligated to follow it. Nobody is obligated to follow orders from marketers or IT workers.\n\nDo the workers at this company form a poset under the \"obligated to follow orders from\" relation?\n\n%%%hidden(Show solution):\nWhile not technically a poset due to its lack of reflexivity, it is pretty close. It is actually a strict ordering, whose underlying partial order could be obtained by making the reasonable assumption that each worker will follow her own orders.\n%%%\n\n# Bag Inclusion\n\nWe can define a notion of [power sets](https://arbital.com/p/power_set) for [bags](https://arbital.com/p/3jk) as follows. Let $X$ be a set, then we use $\\mathcal{M}(X)$ to denote the set of all bags containing elements of $X$. Let $A \\in \\mathcal{M}(X)$. The multiplicity function of $A$, $1_A : X \\rightarrow \\mathbb N$ maps each element of $X$ to the number of times that element occurs in $A$. We can use multiplicity functions to define a inclusion relation $\\subseteq$ for bags. For $A, B \\in \\mathcal M(X)$, we write $A \\subseteq B$ whenever for all $x \\in X$, $1_A(x) \\leq 1_B(x)$.\n\nDoes $\\mathcal{M}(X)$ form a poset under the bag inclusion relation $\\subseteq$? If so, prove it. Otherwise, show that it does not satisfy one of the three poset properties. \n\n# Duality\n\nGive the dual of the following proposition. \n\nFor all posets $P$ and all $p, q \\in P$, $q \\prec p$ implies that $\\{ r \\in P~|~r \\leq p \\}$ is a superset of $\\{ r \\in P~|~r \\leq q\\}$.\n\n%%hidden(Show solution):\nFor all posets $P$ and all $p, q \\in P$, $q \\succ p$ implies that $\\{ r \\in P~|~r \\geq p \\}$ is a superset of $\\{ r \\in P~|~r \\geq q\\}$ (where $q \\succ p$ means $p \\prec q$).\n%%\n\n# Hasse diagrams\n\nLet $X = \\{ x, y, z \\}$. Draw a Hasse diagram for the poset $\\langle \\mathcal P(X), \\subseteq \\rangle$ of the power set of $X$ ordered by inclusion.\n\n%%hidden(Show solution):\n\n\n%%%comment:\ndot source:\n\ndigraph G {\n node [= 0.1, height = 0.1](https://arbital.com/p/width)\n edge [= \"none\"](https://arbital.com/p/arrowhead)\n e [= \"{}\"](https://arbital.com/p/label)\n x [= \"{x}\"](https://arbital.com/p/label)\n y [= \"{y}\"](https://arbital.com/p/label)\n z [= \"{z}\"](https://arbital.com/p/label)\n xy [= \"{x,y}\"](https://arbital.com/p/label)\n xz [= \"{x,z}\"](https://arbital.com/p/label)\n yz [= \"{y,z}\"](https://arbital.com/p/label)\n xyz [= \"{x,y,z}\"](https://arbital.com/p/label)\n\n rankdir = BT;\n e -> x\n e -> y\n e -> z\n x -> xy\n x -> xz\n y -> xy\n y -> yz\n z -> xz\n z -> yz\n xy -> xyz\n xz -> xyz\n yz -> xyz\n}\n%%%\n\n%%\n\n#Hasse diagrams (encore)\n\nIs it possible to draw a Hasse diagram for any poset?\n\n%%hidden(Show solution):\nNote that our description of Hasse diagrams made use of the covers relation $\\prec$. The covers relation, however, is not adequate to describe the structure of many posets. Consider the poset $\\langle \\mathbb R, \\leq \\rangle$ of the real numbers ordered by the standard comparison $\\leq$. We have $0 < 1$, but how would we convey that with a Hasse diagram? The problem is that $0$ has no covers, even though it is not a maximal element in $\\mathbb R$. In fact, for any $x \\in \\mathbb R$ such that $x > 0$, we can find a $y \\in \\mathbb R$ such that $0 < y < x$. This \"infinite density\" of $\\mathbb R$ makes it impossible to depict using a Hasse diagram.\n%%", "date_published": "2016-08-22T16:24:37Z", "authors": ["Eric Bruylant", "Kevin Clancy", "Mark Chimes", "Chris Pasek"], "summaries": [], "tags": ["Exercise "], "alias": "4l1"}
{"id": "e630cfe771380d4d0306e6686d5f1a96", "title": "Cauchy's theorem on subgroup existence", "url": "https://arbital.com/p/cauchy_theorem_on_subgroup_existence", "source": "arbital", "source_type": "text", "text": "Cauchy's theorem states that if $G$ is a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $p$ is a [prime](https://arbital.com/p/4mf) dividing $|G|$ the [order](https://arbital.com/p/3gg) of $G$, then $G$ has a subgroup of order $p$. Such a subgroup is necessarily [cyclic](https://arbital.com/p/47y) ([proof](https://arbital.com/p/4jh)).\n\n# Proof\n\nThe proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.\n\nLet $$X = \\{ (x_1, x_2, \\dots, x_p) : x_1 x_2 \\dots x_p = e \\}$$ the collection of $p$-[https://arbital.com/p/-tuple](https://arbital.com/p/-tuple)s of elements of the group such that the group operation applied to the tuple yields the identity.\nObserve that $X$ is not empty, because it contains the tuple $(e, e, \\dots, e)$.\n\nNow, the cyclic group $C_p$ of order $p$ [acts](https://arbital.com/p/3t9) on $X$ as follows: $$(h, (x_1, \\dots, x_p)) \\mapsto (x_2, x_3, \\dots, x_p, x_1)$$ where $h$ is the generator of $C_p$.\nSo a general element $h^i$ acts on $X$ by sending $(x_1, \\dots, x_p)$ to $(x_{i+1}, x_{i+2} , \\dots, x_p, x_1, \\dots, x_i)$.\n\nThis is indeed a group action (exercise).\n\n%%hidden(Show solution):\n\n- It certainly outputs elements of $X$, because if $x_1 x_2 \\dots x_p = e$, then $$x_{i+1} x_{i+2} \\dots x_p x_1 \\dots x_i = (x_1 \\dots x_i)^{-1} (x_1 \\dots x_p) (x_1 \\dots x_i) = (x_1 \\dots x_i)^{-1} e (x_1 \\dots x_i) = e$$\n- The identity acts trivially on the set: since rotating a tuple round by $0$ places is the same as not permuting it at all.\n- $(h^i h^j)(x_1, x_2, \\dots, x_p) = h^i(h^j(x_1, x_2, \\dots, x_p))$ because the left-hand side has performed $h^{i+j}$ which rotates by $i+j$ places, while the right-hand side has rotated by first $j$ and then $i$ places and hence $i+j$ in total.\n%%\n\nNow, fix $\\bar{x} = (x_1, \\dots, x_p) \\in X$.\n\nBy the [Orbit-Stabiliser theorem](https://arbital.com/p/4l8), the [orbit](https://arbital.com/p/4v8) $\\mathrm{Orb}_{C_p}(\\bar{x})$ of $\\bar{x}$ divides $|C_p| = p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\\bar{x} \\in X$.\n\nNow, what is the size of the set $X$?\n%%hidden(Show solution):\nIt is $|G|^{p-1}$.\n\nIndeed, a single $p$-tuple in $X$ is specified precisely by its first $p$ elements; then the final element is constrained to be $x_p = (x_1 \\dots x_{p-1})^{-1}$.\n%%\n\nAlso, the orbits of $C_p$ acting on $X$ partition $X$ ([proof](https://arbital.com/p/4mg)).\nSince $p$ divides $|G|$, we must have $p$ dividing $|G|^{p-1} = |X|$.\nTherefore since $|\\mathrm{Orb}_{C_p}((e, e, \\dots, e))| = 1$, there must be at least $p-1$ other orbits of size $1$, because each orbit has size $p$ or $1$: if we had fewer than $p-1$ other orbits of size $1$, then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$, and all the remaining orbits would have to be of size $p$, contradicting that $p \\mid |X|$.\n\n\nHence there is indeed another orbit of size $1$; say it is the singleton $\\{ \\bar{x} \\}$ where $\\bar{x} = (x_1, \\dots, x_p)$.\n\nNow $C_p$ acts by cycling $\\bar{x}$ round, and we know that doing so does not change $\\bar{x}$, so it must be the case that all the $x_i$ are equal; hence $(x, x, \\dots, x) \\in X$ and so $x^p = e$ by definition of $X$.", "date_published": "2016-06-30T12:08:56Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Proof"], "alias": "4l6"}
{"id": "a7a88c91bfa649be194015684fc5a08b", "title": "Equaliser (category theory)", "url": "https://arbital.com/p/category_theory_equaliser", "source": "arbital", "source_type": "text", "text": "In [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7), an *equaliser* of a pair of arrows $f, g: A \\to B$ is an object $E$ and a universal arrow $e: E \\to A$ such that $ge = fe$.\nExplicitly, $ge = fe$, and for any object $X$ and arrow $x: X \\to A$ such that $fx = gx$, there is a unique factorisation $\\bar{x} : X \\to A$ such that $e \\bar{x} = x$.", "date_published": "2016-06-18T13:45:48Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Formal definition", "Stub"], "alias": "4l7"}
{"id": "00465a9e0766763f9891ec8d97e08e35", "title": "Orbit-stabiliser theorem", "url": "https://arbital.com/p/orbit_stabiliser_theorem", "source": "arbital", "source_type": "text", "text": "summary(Technical): Let $G$ be a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd), [acting](https://arbital.com/p/3t9) on a set $X$. Let $x \\in X$.\nWriting $\\mathrm{Stab}_G(x)$ for the [stabiliser](https://arbital.com/p/4mz) of $x$, and $\\mathrm{Orb}_G(x)$ for the [orbit](https://arbital.com/p/4v8) of $x$, we have $$|G| = |\\mathrm{Stab}_G(x)| \\times |\\mathrm{Orb}_G(x)|$$ where $| \\cdot |$ refers to the size of a set.\n\nLet $G$ be a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd), [acting](https://arbital.com/p/3t9) on a set $X$. Let $x \\in X$.\nWriting $\\mathrm{Stab}_G(x)$ for the [stabiliser](https://arbital.com/p/4mz) of $x$, and $\\mathrm{Orb}_G(x)$ for the [orbit](https://arbital.com/p/4v8) of $x$, we have $$|G| = |\\mathrm{Stab}_G(x)| \\times |\\mathrm{Orb}_G(x)|$$ where $| \\cdot |$ refers to the size of a set.\n\nThis statement generalises to infinite groups, where the same proof goes through to show that there is a [bijection](https://arbital.com/p/499) between the [left cosets](https://arbital.com/p/4j4) of the group $\\mathrm{Stab}_G(x)$ and the orbit $\\mathrm{Orb}_G(x)$.\n\n# Proof\n\nRecall that the [https://arbital.com/p/-4lt](https://arbital.com/p/-4lt) of the parent group.\n\nFirstly, it is enough to show that there is a bijection between the left cosets of the stabiliser, and the orbit.\nIndeed, then $$|\\mathrm{Orb}_G(x)| |\\mathrm{Stab}_G(x)| = |\\{ \\text{left cosets of} \\ \\mathrm{Stab}_G(x) \\}| |\\mathrm{Stab}_G(x)|$$\nbut the right-hand side is simply $|G|$ because an element of $G$ is specified exactly by specifying an element of the stabiliser and a coset.\n(This follows because the [cosets partition the group](https://arbital.com/p/4j5).)\n\n## Finding the bijection\n\nDefine $\\theta: \\mathrm{Orb}_G(x) \\to \\{ \\text{left cosets of} \\ \\mathrm{Stab}_G(x) \\}$, by $$g(x) \\mapsto g \\mathrm{Stab}_G(x)$$\n\nThis map is well-defined: note that any element of $\\mathrm{Orb}_G(x)$ is given by $g(x)$ for some $g \\in G$, so we need to show that if $g(x) = h(x)$, then $g \\mathrm{Stab}_G(x) = h \\mathrm{Stab}_G(x)$.\nThis follows: $h^{-1}g(x) = x$ so $h^{-1}g \\in \\mathrm{Stab}_G(x)$.\n\nThe map is [injective](https://arbital.com/p/4b7): if $g \\mathrm{Stab}_G(x) = h \\mathrm{Stab}_G(x)$ then we need $g(x)=h(x)$.\nBut this is true: $h^{-1} g \\in \\mathrm{Stab}_G(x)$ and so $h^{-1}g(x) = x$, from which $g(x) = h(x)$.\n\nThe map is [surjective](https://arbital.com/p/4bg): let $g \\mathrm{Stab}_G(x)$ be a left coset.\nThen $g(x) \\in \\mathrm{Orb}_G(x)$ by definition of the orbit, so $g(x)$ gets taken to $g \\mathrm{Stab}_G(x)$ as required.\n\nHence $\\theta$ is a well-defined bijection.", "date_published": "2016-07-01T02:46:57Z", "authors": ["Mark Chimes", "Patrick Stevens", "Alexei Andreev"], "summaries": ["A [group action](https://arbital.com/p/-3t9) of a group $G$ acting on a set $X$ describes how $G$ sends elements of $X$ to other elements of $X$. Given a specific element $x \\in X$, the [stabiliser](https://arbital.com/p/-4mz) is all those elements of the group which send $x$ back to itself, and the [orbit](https://arbital.com/p/-4v8) of $x$ is all the elements to which $x$ can get sent. \n\nThis theorem tells you that $G$ is divided into equal-sized pieces using $x$. Each piece \"looks like\" the stabilizer of $x$ (and is the same size), and the orbit of $x$ tells you how to \"move the piece around\" over $G$ in order to cover it. \n\nPut another way, each element $y$ in the orbit of $x$ is transformed \"in the same way\" by $G$ relative to $y$. \n\nThis theorem is closely related to [Lagrange's Theorem](https://arbital.com/p/-4jn)."], "tags": [], "alias": "4l8"}
{"id": "eccc782bfb359cef03409b4f9a717fce", "title": "Needs exercises", "url": "https://arbital.com/p/needs_exercises_meta_tag", "source": "arbital", "source_type": "text", "text": "One way the user can make sure they've understood the material is to do [exercises](https://arbital.com/p/4n2). They should usually be on their own [lens](https://arbital.com/p/17b), as additional \"homework problems\" that users can solve to get more practice. They can also be embedded in the explanation pages, so the user can solve them as they are learning, especially on [path](https://arbital.com/p/1rt) pages where the reader has asked for exercises.\n\nMany pages could be improved by adding exercises, but priority should be given to: \n\n* Pages which don't currently have exercises.\n* Pages which have a full explanation of a concept (rather than just [defienitions](https://arbital.com/p/4gs)).\n* Pages which are a requisite that many other concepts rely on.\n* Pages which are mostly self-contained, rather than overviews of a large and multifaceted concept (exercises should go with child pages explaining the subtopics).", "date_published": "2016-06-20T20:55:38Z", "authors": ["Eric Bruylant", "Mark Chimes", "Alexei Andreev"], "summaries": [], "tags": [], "alias": "4lg"}
{"id": "2a01921c6d100fc88892772cbc88d98b", "title": "Join and meet: Exercises", "url": "https://arbital.com/p/poset_join_exercises", "source": "arbital", "source_type": "text", "text": "Try these exercises to test your knowledge of joins and meets.\n\n\nTangled up \n--------------------\n\n\n\nDetermine whether or not the following joins and meets exist in the poset depicted by the above Hasse diagram. \n\n$c \\vee b$\n\n%%hidden(Show solution):\nThis join does *not* exist, because $i$ and $j$ are incomparable upper bounds of $\\{ c, b \\}$, and no smaller upper bounds of $\\{ c, b \\}$ exist.\n%%\n\n$g \\vee e$\n\n%%hidden(Show solution):\nThis join exists. $g \\vee e = j$.\n%%\n\n$j \\wedge f$ \n\n%%hidden(Show solution):\nThis meet does *not* exist, because $d$ and $b$ are incomparable lower bounds of $\\{ j, f \\}$, and no larger lower bounds of $\\{ j, f \\}$ exist.\n%%\n\n$\\bigvee \\{a,b,c,d,e,f,g,h,i,j,k,l\\}$\n\n%%hidden(Show solution):\nThis join exists. It is $l$.\n%%\n\nJoin fu\n------------\n\nLet $P$ be a poset, $S \\subseteq P$, and $p \\in P$. Prove that if both $\\bigvee S$ and $(\\bigvee S) \\vee p$ exist then $\\bigvee (S \\cup \\{p\\})$ exists as well, and $(\\bigvee S) \\vee p = \\bigvee (S \\cup \\{p\\})$.\n\n%%hidden(Show solution):\nFor any $X \\subset P$, let $X^U$ denote the set of upper bounds of $X$. The above proposition follows from the fact that $\\{\\bigvee S, p\\}^U = (S \\cup p)^U$, which is apparent from the following chain of bi-implications:\n\n$q \\in \\{\\bigvee S, p\\}^U \\iff$ \n\nfor all $s \\in S, q \\geq \\bigvee S \\geq s$, and $q \\geq p \\iff$\n\n$q \\in (S \\cup \\{p\\})^U$.\n\nIf two subsets of a poset have the same set of upper bounds, then either they both lack a least upper bound, or both have the same least upper bound.\n%%\n\nMeet fu\n------------\n\nLet $P$ be a poset, $S \\subseteq P$, and $p \\in P$. Prove that if both $\\bigwedge S$ and $(\\bigwedge S) \\wedge p$ exist then $\\bigwedge (S \\cup \\{p\\})$ exists as well, and $(\\bigwedge S) \\wedge p = \\bigwedge(S \\cup \\{p\\})$.\n\n%%hidden(Show solution):\nNote that the proposition we are trying to prove here is the dual of the one stated in join fu. Thanks to the duality principle, this theorem therefore comes for free with our solution to join fu.\n%%\n\nQuite a big join\n--------------------------\n\nIn the poset $\\langle \\mathbb N, | \\rangle$ discussed in [https://arbital.com/p/43s](https://arbital.com/p/43s), does $\\bigvee \\mathbb N$ exist? If so, what is it?", "date_published": "2016-07-29T19:24:27Z", "authors": ["Eric Bruylant", "Kevin Clancy"], "summaries": [], "tags": ["Exercise "], "alias": "4ll"}
{"id": "5284cc0e6d93659517c9cee7ccb3a3bd", "title": "Stabiliser is a subgroup", "url": "https://arbital.com/p/stabiliser_is_a_subgroup", "source": "arbital", "source_type": "text", "text": "Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) which [acts](https://arbital.com/p/3t9) on the [set](https://arbital.com/p/-3jz) $X$.\nThen for every $x \\in X$, the [stabiliser](https://arbital.com/p/4mz) $\\mathrm{Stab}_G(x)$ is a [subgroup](https://arbital.com/p/-subgroup) of $G$.\n\n# Proof\n\nWe must check the group axioms.\n\n- The [identity](https://arbital.com/p/-54p), $e$, is in the stabiliser because $e(x) = x$; this is part of the definition of a group action.\n- [Closure](https://arbital.com/p/-3gy) is satisfied: if $g(x) = x$ and $h(x) = x$, then $(gh)(x) = g(h(x))$ by definition of a group action, but that is $g(x) = x$.\n- [Associativity](https://arbital.com/p/-3h4) is inherited from the parent group.\n- [Inverses](https://arbital.com/p/-inverse_mathematics): if $g(x) = x$ then $g^{-1}(x) = g^{-1} g(x) = e(x) = x$.", "date_published": "2016-07-07T15:26:16Z", "authors": ["Dylan Hendrickson", "Mark Chimes", "Patrick Stevens"], "summaries": ["Given a [group](https://arbital.com/p/-3gd) $G$ [acting](https://arbital.com/p/-3t9) on a [set](https://arbital.com/p/-3jz) $X$, the [stabiliser](https://arbital.com/p/-4mz) of some [element](https://arbital.com/p/-element_mathematics) $x \\in X$ is a [subgroup](https://arbital.com/p/-subgroup) of $G$."], "tags": ["Proof"], "alias": "4lt"}
{"id": "d5c7e66e64dc32e12b1eae8d8dcdbe6e", "title": "Prime number", "url": "https://arbital.com/p/prime_number", "source": "arbital", "source_type": "text", "text": "A [https://arbital.com/p/-45h](https://arbital.com/p/-45h) $n > 1$ is *prime* if it has no [divisors](https://arbital.com/p/divisor_number_theory) other than itself and $1$.\nEquivalently, it has the property that if $n \\mid ab$ %%note:That is, $n$ divides the product $ab$%% then $n \\mid a$ or $n \\mid b$.\nConventionally, $1$ is considered to be neither prime nor [composite](https://arbital.com/p/composite_number) (i.e. non-prime).\n\n# Examples\n\n- The number $2$ is prime, because its divisors are $1$ and $2$; therefore it has no divisors other than itself and $1$.\n- The number $3$ is also prime, as are $5, 7, 11, 13, \\dots$.\n- The number $4$ is not prime; neither are $6, 8, 9, 10, 12, \\dots$.\n\n# Properties\n\n- There are infinitely many primes. ([Proof.](https://arbital.com/p/54r))\n- Every natural number may be written as a product of primes; moreover, this can only be done in one way (if we count \"the same product but with the order swapped\" as being the same: for example, $2 \\times 3 = 3 \\times 2$ is just one way of writing $6$). ([Proof.](https://arbital.com/p/fundamental_theorem_of_arithmetic))\n\n# How to find primes\n\nIf we want to create a list of all the primes below a given number, or the first $n$ primes for some fixed $n$, then an efficient way to do it is the [Sieve of Eratosthenes](https://arbital.com/p/sieve_of_eratosthenes).\n(There are other sieves available, but Eratosthenes is the simplest.)\n\nThere are many [tests](https://arbital.com/p/primality_testing) for primality and for compositeness.\n\n# More general concept\n\nThis definition of \"prime\" is, in a more general [ring-theoretic](https://arbital.com/p/3gq) setting, known instead as the property of [irreducibility](https://arbital.com/p/5m1).\nConfusingly, there is a slightly different notion in this ring-theoretic setting, which goes by the name of \"prime\"; this notion has [a separate page on Arbital](https://arbital.com/p/prime_element_ring_theory).\nIn the ring of integers, the two ideas of \"prime\" and \"irreducible\" actually coincide, but that is because the integers form a ring with several very convenient properties: in particular, being a [Euclidean domain](https://arbital.com/p/euclidean_domain), they are a [https://arbital.com/p/-principal_ideal_domain](https://arbital.com/p/-principal_ideal_domain) (PID), and [PIDs have unique factorisation](https://arbital.com/p/pid_implies_ufd).", "date_published": "2016-07-27T18:03:30Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Formal definition", "Stub"], "alias": "4mf"}
{"id": "0741dbb6a65310ab40d0a66ece3b149f", "title": "Group orbits partition", "url": "https://arbital.com/p/group_orbits_partition", "source": "arbital", "source_type": "text", "text": "Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd), [acting](https://arbital.com/p/3t9) on the set $X$.\nThen the [orbits](https://arbital.com/p/group_orbit) of $X$ under $G$ form a [partition](https://arbital.com/p/set_partition) of $X$.\n\n# Proof\n\nWe need to show that every element of $X$ is in an orbit, and that if $x \\in X$ lies in two orbits then they are the same orbit.\n\nCertainly $x \\in X$ lies in an orbit: it lies in the orbit $\\mathrm{Orb}_G(x)$, since $e(x) = x$ where $e$ is the identity of $G$.\n(This follows by the definition of an action.)\n\nSuppose $x$ lies in both $\\mathrm{Orb}_G(a)$ and $\\mathrm{Orb}_G(b)$, where $a, b \\in X$.\nThen $g(a) = h(b) = x$ for some $g, h \\in G$.\nThis tells us that $h^{-1}g(a) = b$, so in fact $\\mathrm{Orb}_G(a) = \\mathrm{Orb}_G(b)$; it is an exercise to prove this formally.\n\n%%hidden(Show solution):\nIndeed, if $r \\in \\mathrm{Orb}_G(b)$, then $r = k(b)$, say, some $k \\in G$.\nThen $r = k(h^{-1}g(a)) = kh^{-1}g(a)$, so $r \\in \\mathrm{Orb}_G(a)$.\n\nConversely, if $r \\in \\mathrm{Orb}_G(a)$, then $r = m(b)$, say, some $m \\in G$.\nThen $r = m(g^{-1}h(b)) = m g^{-1} h (b)$, so $r \\in \\mathrm{Orb}_G(b)$.\n%%", "date_published": "2016-06-20T06:55:28Z", "authors": ["Patrick Stevens"], "summaries": [], "tags": ["Proof"], "alias": "4mg"}
{"id": "83a5773dba39ca418cf27f76d38d4c96", "title": "Product (Category Theory)", "url": "https://arbital.com/p/product_category_theory", "source": "arbital", "source_type": "text", "text": "This simultaneously captures the concept of a product of [sets](https://arbital.com/p/-3jz), [posets](https://arbital.com/p/-3rb), [groups](https://arbital.com/p/-3gd), [topological spaces](https://arbital.com/p/-topological_space) etc. In addition, like any [universal construction](https://arbital.com/p/-universal_property), this characterization does not differentiate between [isomorphic](https://arbital.com/p/-4f4) versions of the product, thus allowing one to abstract away from an arbitrary, [specific construction](https://arbital.com/p/-specific_construction_category_theory).\n\n##Definition##\nGiven a pair of objects $X$ and $Y$ in a category $\\mathbb{C}$, the **product** of $X$ and $Y$ is an object $P$ *along with a pair of morphisms* $f: P \\rightarrow X$ and $g: P \\rightarrow Y$ satisfying the following [universal](https://arbital.com/p/-universal_property) condition:\n\nGiven any other object $W$ and morphisms $u: W \\rightarrow X$ and $v:W \\rightarrow Y$ there is a *unique* morphism $h: W \\rightarrow P$ such that $fh = u$ and $gh = v$.", "date_published": "2016-06-24T01:55:45Z", "authors": ["Mark Chimes", "Nate Soares", "Patrick Stevens", "Alexei Andreev"], "summaries": [], "tags": ["Stub"], "alias": "4mj"}
{"id": "95bf03188287ec8e7ed8f5c5c56384da", "title": "Empirical probabilities are not exactly 0 or 1", "url": "https://arbital.com/p/cromwells_rule", "source": "arbital", "source_type": "text", "text": "[Cromwell's Rule](https://en.wikipedia.org/wiki/Cromwell%27s_rule) in statistics argues that no empirical proposition should be assigned a subjective probability of *exactly* $0$ or $1$ - it is always *possible* to be mistaken. (Some argue that this rule should be generalized to logical facts as well.)\n\nA probability of exactly $0$ or $1$ corresponds to infinite [log odds](https://arbital.com/p/1zh), and would require infinitely [strong](https://arbital.com/p/22x) evidence to reach starting from any finite [prior](https://arbital.com/p/1rm). To put it another way, if you don't start out infinitely certain of a fact before making any observations (before you were born), you won't reach infinite certainty after any finite number of observations involving finite probabilities.\n\nAll sensible [universal priors](https://arbital.com/p/4mr) seem so far to have the property that they never assign probability exactly $0$ or $1$ to any predicted future observation, since their hypothesis space is always broad enough to include an imaginable state of affairs in which the future is different from the past.\n\nIf you did assign a probability of exactly $0$ or $1,$ you would be unable to [update](https://arbital.com/p/1ly) no matter how much contrary evidence you observed. [Prior odds](https://arbital.com/p/1rb) of 0 : 1 (or 1 : 0), times any finite [likelihood ratio](https://arbital.com/p/1rq), end up yielding 0 : 1 (or 1 : 0).\n\nAs Rafal Smigrodski put it:\n\n> \"I am not totally sure I have to be always unsure. Maybe I could be legitimately sure about something. But once I assign a probability of 1 to a proposition, I can never undo it. No matter what I see or learn, I have to reject everything that disagrees with the axiom. I don't like the idea of not being able to change my mind, ever.\"", "date_published": "2016-07-06T16:01:13Z", "authors": ["Eric Bruylant", "Eliezer Yudkowsky", "Alexei Andreev"], "summaries": ["[Cromwell's Rule](https://en.wikipedia.org/wiki/Cromwell%27s_rule) in statistics forbids us to assign probabilities of exactly $0$ or $1$ to any empirical proposition - that is, it is always possible to be mistaken.\n\n- Probabilities of exactly $0$ or $1$ correspond to infinite [log odds](https://arbital.com/p/1zh), meaning that no finite amount of evidence can ever suffice to reach them, or overturn them. Once you assign probability $0$ or $1,$ you can never change your mind.\n- Sensible [universal priors](https://arbital.com/p/4mr) never assign probability exactly $0$ or $1$ to any predicted future observation - their hypothesis space is always broad enough to include a scenario where the future is different from the past."], "tags": ["C-Class"], "alias": "4mq"}
{"id": "e9bdbaf95d38b29b98050bd9696272d2", "title": "Universal prior", "url": "https://arbital.com/p/universal_prior", "source": "arbital", "source_type": "text", "text": "A \"universal\" [prior](https://arbital.com/p/27p) is a [probability distribution](https://arbital.com/p/3tb) that assigns positive probability to *every thinkable hypothesis*, for some reasonable meaning of \"every thinkable hypothesis\". A central example would be [Solomonoff induction](https://arbital.com/p/11w), in which the observations are a sequence of bits, the universal prior is \"every possible computer program that generates bit sequences\", and every such computer program starts with a positive probability.", "date_published": "2016-06-21T18:41:52Z", "authors": ["Eliezer Yudkowsky"], "summaries": [], "tags": ["Start"], "alias": "4mr"}
{"id": "515e48bfca69c79540aa844eda3923ae", "title": "Stabiliser (of a group action)", "url": "https://arbital.com/p/group_stabiliser", "source": "arbital", "source_type": "text", "text": "Let the [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ [act](https://arbital.com/p/3t9) on the set $X$.\nThen for each element $x \\in X$, the *stabiliser* of $x$ under $G$ is $\\mathrm{Stab}_G(x) = \\{ g \\in G: g(x) = x \\}$.\nThat is, it is the collection of elements of $G$ which do not move $x$ under the action.\n\nThe stabiliser of $x$ is a subgroup of $G$, for any $x \\in X$. ([Proof.](https://arbital.com/p/4lt))\n\nA closely related notion is that of the [orbit](https://arbital.com/p/4v8) of $x$, and the very important [Orbit-Stabiliser theorem](https://arbital.com/p/4l8) linking the two.", "date_published": "2016-06-28T20:08:10Z", "authors": ["Mark Chimes", "Patrick Stevens", "Alexei Andreev"], "summaries": ["The stabilizer of an element $x$ under the action of a group $G$ is the set ([actually a subgroup](https://arbital.com/p/4lt)) of elements of $G$ which leave $x$ unchanged."], "tags": ["Needs summary", "Formal definition", "Stub"], "alias": "4mz"}
{"id": "b01471992bfbaad021e753b204e9cbc3", "title": "Exercise", "url": "https://arbital.com/p/exercise_meta_tag", "source": "arbital", "source_type": "text", "text": "One way readers can reinforce or make sure they've understood a concept is to do exercises. These pages give a collection of questions for this purpose.", "date_published": "2016-06-20T20:29:27Z", "authors": ["Eric Bruylant"], "summaries": [], "tags": [], "alias": "4n2"}
{"id": "516b42b4f7945fa7545e040eef2fcb90", "title": "Why is the decimal expansion of log2(3) infinite?", "url": "https://arbital.com/p/log2_of_3_never_ends", "source": "arbital", "source_type": "text", "text": "$\\log_2(3)$ starts with\n\n1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360\n\nand goes on indefinitely. Why is it 1.58... in particular? Well, it takes more than one but less than two [binary digits](https://arbital.com/p/binary_digit) to encode a [3-digit](https://arbital.com/p/4sj), so $\\log_2(3)$ must be between 1 and 2. ([Wait, what?](https://arbital.com/p/427)). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $10 \\cdot \\log_2(3)$ must be between 15 and 16, which means $1.5 < \\log_2(3) < 1.6.$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $1.58 < \\log_2(3) < 1.59.$ And so on. Because no power of 3 is ever equal to any power of 2, $10^n \\cdot \\log_2(3)$ will never quite be a whole number, no matter how large $n$ is.\n\nThus, $\\log_2(3)$ has no finite decimal expansion, because $3$ is not a [rational](https://arbital.com/p/4zq) [https://arbital.com/p/-power](https://arbital.com/p/-power) of $2$. Using this argument, we can see that $\\log_b(x)$ is an integer if (and only if) $x$ is a power of $b$, and that $\\log_b(x)$ only has a finite expansion if some power of $x$ is a power of $b.$", "date_published": "2016-07-04T13:55:58Z", "authors": ["Nate Soares"], "summaries": ["It takes more than one but less than two [binary digits](https://arbital.com/p/binary_digit) to encode a [3-digit](https://arbital.com/p/4sj), so $\\log_2(3)$ must be between 1 and 2. ([Wait, what?](https://arbital.com/p/427)). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $10 \\cdot \\log_2(3)$ must be between 15 and 16, which means $1.5 < \\log_2(3) < 1.6.$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $1.58 < \\log_2(3) < 1.59.$ And so on. Because no power of 3 is ever equal to any power of 2, $10^n \\cdot \\log_2(3)$ will never quite be a whole number, no matter how large $n$ is."], "tags": ["Start"], "alias": "4n8"}
{"id": "ff5ab615ee81918a3e032d3326567cc7", "title": "Guide", "url": "https://arbital.com/p/guide_meta_tag", "source": "arbital", "source_type": "text", "text": "Meta tag for the start page of a [multi-page guide](https://arbital.com/p/327).", "date_published": "2016-06-21T20:10:52Z", "authors": ["Eric Bruylant"], "summaries": [], "tags": [], "alias": "4pj"}
{"id": "c4c48d41e32c62c0ed3230c32fa18092", "title": "Needs brief summary", "url": "https://arbital.com/p/needs_brief_summary_meta_tag", "source": "arbital", "source_type": "text", "text": "Pages which have a long or in-depth main summary should also have a brief summary. These show up as an alternate tab on the [summary popover](https://arbital.com/p/1kl) when a user hovers over a [greenlink](https://arbital.com/p/17f). Add this tag to pages that need to have a brief summary.\n\nPages may also require a [general](https://arbital.com/p/433) or [technical](https://arbital.com/p/4pt) summary.", "date_published": "2016-06-22T15:11:40Z", "authors": ["Eric Bruylant", "Mark Chimes", "Nate Soares"], "summaries": [], "tags": ["Meta tags which request an edit to the page"], "alias": "4q2"}
{"id": "c0052e4c6a9bf9e7ba86fc6a44822331", "title": "n-digit", "url": "https://arbital.com/p/n_digit", "source": "arbital", "source_type": "text", "text": "An $n$-digit is a physical object that can be stably placed into any of $n$ distinguishable states. For example, a coin (which can be placed heads or tails) and a single bit of memory on a computer (which either has a high volt level or a low volt level) are both examples of 2-digits. A [https://arbital.com/p/-42d](https://arbital.com/p/-42d) is an example of a 10-digit. One die is an example of a 6-digit; two dice together are an example of a 36-digit (because they can be placed in 36 different ways).\n\nWhat does and doesn't count as an $n$-digit depends on context and convention: For example, if you want to communicate a message to me by placing a penny heads-side up and choosing whether to point Abraham Lincoln's face either north, south, east, or west, then, for the purposes of the two of us, that penny is a 4-digit rather than a 2-digit. The definition of \"stably placed\" is also a bit up-for-grabs: If you're writing a computer program and need to store a [256-message](https://arbital.com/p/3v9) in short-term memory, then a byte of RAM will do, but if you need to store the same 256-message for a long period of time, you may need to use a less temporary 256-digit (such as a hard drive).\n\nNote that it's possible to emulate $m$-digits using $n$-digits, in general. If $m < n$ then an $n$-digit is trivially an $m$-digit (i.e., you can use a digit wheel like a 7-digit in a pinch), and if $m > n$ then, given enough $n$-digits, you can make do. For example, 3 coins can be used to encode an 8-digit. See also [https://arbital.com/p/emulating_digits](https://arbital.com/p/emulating_digits).", "date_published": "2016-06-24T02:58:44Z", "authors": ["Nate Soares"], "summaries": [], "tags": ["Non-standard terminology"], "alias": "4sj"}
{"id": "aaa26ef8a8b00e033f7db0847e3dbb08", "title": "Emulating digits", "url": "https://arbital.com/p/emulating_digits", "source": "arbital", "source_type": "text", "text": "In general, given enough $n$-digits, you can emulate an $m$-digit, for any $m, n \\in$ [$\\mathbb N$](https://arbital.com/p/45h). If $m < n,$ you can emulate an $m$-digit using just one $n$-digit — in other words, you can use a [https://arbital.com/p/-42d](https://arbital.com/p/-42d) like a $7$-digit if you want to, by just ignoring three of the possible ways to set the digit wheel. If $m > n,$ things are a bit more difficult, but only slightly.\n\nBasically, with 2 $n$-digits, you can emulate a $n^2$-digit, as follows. Using your two $n$-digits, encode a number $(x, y)$ where $0 \\le x < n$ and $0 \\le y < n$. Interpret $(x, y)$ as $xn + y.$ You have now encoded a number between 0 (if $x = y = 0$) and $n^2 - 1$ (if $x = y = n-1$). Congratulations, you just used two $n$-digits to make an $n^2$ digit!\n\nYou can use the same strategy to emulate $n^3$-digits (interpret $(x, y, z)$ as $xn^2 + yn + z$), $n^4$-digits (you get the picture), and so on. Now, to emulate an $m$-digit, just pick an exponent $a$ such that $n^a > m,$ collect $a$ copies of an $n$-digit, and you're done.\n\nThis isn't necessarily the most efficient way to use $n$-digits to encode $m$-digits. For example, if $m$ is 1,000,001 and $n$ is 10, then you need seven 10-digits. Seven 10-digits are enough to emulate a 10-million-digit, whereas $m$ is a mere million-and-one-digit — paying for a 10-million-digit when all you needed was an $m$-digit seems a bit excessive. For some different methods you can use to recover your losses when encoding one type of digit using another type of digit, see [https://arbital.com/p/44l](https://arbital.com/p/44l) and [https://arbital.com/p/3ty](https://arbital.com/p/3ty). (These techniques are fairly useful in practice, given that modern computers encode everything using [bits](https://arbital.com/p/3p0), i.e. 2-digits, and so it's useful to know how to efficiently encode $m$-messages using bits when $m$ is pretty far from the nearest power of 2.)", "date_published": "2016-06-25T15:14:02Z", "authors": ["Eric Rogstad", "Nate Soares"], "summaries": [], "tags": [], "alias": "4sk"}
{"id": "3aefc32ba788404cc19d36361079f74d", "title": "Decimal notation", "url": "https://arbital.com/p/decimal_notation", "source": "arbital", "source_type": "text", "text": "Seventeen is the number that represents as many things as there are x marks at the end of this sentence: xxxxxxxxxxxxxxxxx. Writing out numbers by saying \"the number representing how many things there are in this pile:\" gets unwieldy when the pile gets large. Thus, we _represent_ numbers using the [numerals](https://arbital.com/p/numeral) 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Specifically, we write the number representing this many things: xxx as \"3\", and the number representing this many things: xxxxxxxxxxx as 11, and the number seventeen as \"17\". This is called \"decimal notation,\" because there are ten different symbols that we use. Numbers don't have to be written down in decimal notation, it's also possible to write them down in other notations such as [https://arbital.com/p/-binary_notation](https://arbital.com/p/-binary_notation). Some numbers can't even be written out in decimal notation (in full); consider, for example, the number $e$ which, in decimal notation, starts out with the digits 2.71828... and just keeps going.\n\n# How decimal notation works\n\nHow do you know that 17 is the number that represents the number of xs in this sequence: xxxxxxxxxxxxxxxxx? In practice, you know this because the rules of decimal notation were ingrained in you in a young child. But do you know those rules explicitly? Could you write out a series of rules for taking in some input symbols like '2', '4', and '6' and using those to figure out how many pebbles to add to a pile?\n\nThe answer, of course, is this many:\n\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\n\nBut how do we perform that conversion in general?\n\nIn short, the number 246 represents $(2 \\cdot 100) + (4 \\cdot 10) + (6 \\cdot 1),$ so as long as we know how to do addition and multiplication, and as long as we know what the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean, and as long as we know how to get to [powers](https://arbital.com/p/power) of 10 (1, 10, 100, 1000, ...), then we can explicitly understand decimal notation.\n\n(What do the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean? By convention, they represent as many things as are in the following ten sequences of xs: , x, xx, xxx, xxxx, xxxxx, xxxxxx, xxxxxxx, xxxxxxxx, and xxxxxxxxx, respectively.)\n\nThis explanation assumes that you're already quite familiar with decimal notation. Explaining decimal notation from scratch to someone who doesn't already know it (which was a task people _actually had to do_ back when half the world was using Roman numerals, a much less convenient system for representing numbers) is a fun task; to see what that looks like, refer to [https://arbital.com/p/+representing_numbers_from_scratch](https://arbital.com/p/+representing_numbers_from_scratch).\n\n# Other common notations\n\nThe above text made use of [https://arbital.com/p/-unary_notation](https://arbital.com/p/-unary_notation), which is a method of representing numbers by making a number of marks that correspond to the represented number. For example, in unary notation, 17 is written xxxxxxxxxxxxxxxxx (or ||||||||||||||||| or whatever, the actual marks don't matter). This is perhaps somewhat easier to understand, but writing large numbers like 93846793284756 gets rather ungainly.\n\nHistorical notations include [Roman numerals](https://arbital.com/p/https://en.wikipedia.org/wiki/Roman_numerals), which were a pretty bad way to represent numbers. (It took humanity quite some time to find _good_ tools for representing numbers; the decimal notation that's been ingrained in your head since early childhood is the result of many centuries worth of effort. It's much harder to invent good representations of numbers when you don't even have good tools for writing down and reasoning about numbers. Furthermore, the modern tools for representing numbers aren't necessarily ideal!)\n\nCommon notations in modern times (aside from decimal notation) include [https://arbital.com/p/-binary_notation](https://arbital.com/p/-binary_notation) (often used by computers), [https://arbital.com/p/-hexadecimal_notation](https://arbital.com/p/-hexadecimal_notation) (which is a useful format for humans reading binary notation). Binary notation and hexadecimal notation are very similar to decimal notation, with the difference that binary uses only two distinct symbols (instead of ten), and hexadecimal uses sixteen.", "date_published": "2016-07-04T05:32:35Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Nate Soares", "Michael Cohen"], "summaries": [], "tags": ["Needs brief summary"], "alias": "4sl"}
{"id": "4ac4ae84cf7114e3f73cdd44016be1cd", "title": "Inverse function", "url": "https://arbital.com/p/inverse_function", "source": "arbital", "source_type": "text", "text": "If a function $g$ is the inverse of a function $f$, then $g$ undoes $f$, and $f$ undoes $g$. In other words, $g(f(x)) = x$ and $f(g(y)) = y$. An inverse function takes as its [domain](https://arbital.com/p/3js) the [range](https://arbital.com/p/3lm) of the original function, and the [range](https://arbital.com/p/3lm) of the inverse function is the [domain](https://arbital.com/p/3js) of the original. To put that another way, if $f$ maps $A$ onto $B$, then $g$ maps $B$ back onto $A$. To indicate the inverse of a function $f$, we write $f^{-1}$.\n\n## Examples ##\n\n$$y=f(x) = x^3\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x=f^{-1}(y) = y^{1/3}$$\n$$y=f(x) = e^x\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x=f^{-1}(y) = ln(y)$$\n$$y=f(x) = x+4\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x=f^{-1}(y) = y-4$$", "date_published": "2016-07-07T15:47:27Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Eric Rogstad", "Michael Cohen"], "summaries": [], "tags": ["Formal definition"], "alias": "4sn"}
{"id": "67612619466b0a029fdee35fb770b67b", "title": "Quotient group", "url": "https://arbital.com/p/quotient_group", "source": "arbital", "source_type": "text", "text": "summary(brief): \nA **quotient group** $G/N$ of a [group](https://arbital.com/p/-3gd) $G$ by a [normal subgroup](https://arbital.com/p/-4h6) $N$ is obtained by dividing up the group into pieces ([equivalence classes](https://arbital.com/p/-equivalence_class)), and then treating everything in one class the same way (by treating each class as a single element). The quotient group has a group structure defined on it based on the original structure of $G$, that works 'basically the same as $G$ up to equivalence'.\n \n\nsummary(technical): \nGiven a [group](https://arbital.com/p/-3gd) $(G, \\bullet)$ and a [normal subgroup](https://arbital.com/p/-4h6) $N \\unlhd G$. The **quotient** of $G$ by $N$, written $G/N$, has as [underlying set](https://arbital.com/p/-3gz) the set of (left)-[cosets](https://arbital.com/p/-4j4) of $N$ in $G$ and as operation $\\circ$ which is defined as $aN \\circ bN = (a \\bullet b) N$, where $xN = \\{xn : n \\in N\\}$ for each $x \\in G$. \n\nThe operation $\\circ$ is [well defined](https://arbital.com/p/well_defined) in the sense that if other representatives $a'$ and $b'$ are chosen such that \n$a'N = aN$ and $b'N = bN$ then also $(a' \\bullet b')N = (a \\bullet b)N$. \n\nThere is a [canonical](https://arbital.com/p/-canonical) [homomorphism](https://arbital.com/p/-47t) (sometimes called the [projection](https://arbital.com/p/-quotient_projection)) $\\phi: G \\rightarrow G/N: a \\mapsto aN$.\n\nThis is a special case of a [quotient](https://arbital.com/p/-quotient_universal_algebra) from universal algebra.\n\n\nsummary(examples):\nGiven the [group](https://arbital.com/p/-3gd) of [integers](https://arbital.com/p/-48l) $\\mathbb{Z}$, and the [normal subgroup](https://arbital.com/p/-nomral_subgroup) $2 \\mathbb{Z}$ of all even numbers, we can form the group $\\mathbb{Z}/2\\mathbb{Z}$. This group has only two [elements](https://arbital.com/p/-element) and [only cares](https://arbital.com/p/-personification_in_mathematics) if a number is odd or even. It tells us that the sum of an odd and an even number is odd, that the sum of two even numbers is even, and the sum of two odd numbers is also even!\n\n\n\nsummary(motivation):\nLet's say we have a group. Maybe the group is kinda large and unwieldy, and we want to find an easier way to think about it. Or maybe we just want to focus on a certain aspect of the group. Some of the actions will change things in ways we just don't really care about, or don't mind ignoring for now. So let's create a group homomorphism that will map all these actions to the identity action in a new group. The image of this homomorphism will be a group much like the first, except that it will ignore all the effects that come from those actions that we're ignoring - just what we wanted! This new group is called the quotient group.\n\n\n#The basic idea\nLet's say we have a [group](https://arbital.com/p/3gd). Maybe the group is kinda large and unwieldy, and we want to find an easier way to think about it. Or maybe we just want to focus on a certain aspect of the group. Some of the actions will change things in ways we just don't really care about, or don't mind ignoring for now. So let's create a [group homomorphism](https://arbital.com/p/47t) that will map all these actions to the identity action in a new group. The image of this homomorphism will be a group much like the first, except that it will ignore all the effects that come from those actions that we're ignoring - just what we wanted! This new group is called the quotient group. \n\n#Definition\n\nWe start with our group $G$. The actions we want to ignore form a group $N$, which must be a [normal subgroup](https://arbital.com/p/4h6) of $G$. The quotient group is then called $G/N$, and has a canonical homomorphism $\\phi: G \\rightarrow G/N$ which maps $g \\in G$ to the [coset](https://arbital.com/p/4j4) $gN$. \n\n## The divisor group\nIn the definition, we require the divisor $N$, to be a normal subgroup of $G$. Why? Well first, let's see why requiring $N$ to be a group makes sense. Remember that $N$ has the actions whose effects we want to ignore. So it makes sense that it should contain the identity action, which has no effect. It also is reasonable that it would be closed under the group operation - doing two things we don't care about shouldn't change anything we care about. Together, these two properties imply it is a [subgroup](https://arbital.com/p/subgroup): $N \\le G$. \n\nA subgroup is great, but it isn't quite good enough by itself to work here. That's because we want the quotient group to preserve the overall structure of the group, i.e. it should preserve the group multiplication. In other words, there needs to be a [group homomorphism](https://arbital.com/p/47t) $\\phi$ from $G$ to $G/N$. Since $N$ is the subgroup of things we want to ignore, all its actions should get mapped to the identity action under this homomorphism. That means it's the [kernel](https://arbital.com/p/49y) of the homomorphism $\\phi$, which means it's a [normal subgroup](https://arbital.com/p/4h6): $N \\trianglelefteq G$. \n\n## Cosets\nWhat exactly are the elements of the new group? They are [equivalence classes](https://arbital.com/p/equivalence_class) of actions, the sets $gN = \\{gn : n \\in N\\}$ where $g \\in G$, also known as a [coset](https://arbital.com/p/4j4). The identity element is the set $N$ itself. Multiplication is defined by $g_1N \\cdot g_2N = (g_1g_2)N$. \n\n# Generalizes the idea of a quotient\nWhat gives a quotient group the right to call itself a quotient? If $G$ and $N$ both have finite order, then $|G/N| = |G|/|N|$, which can be proved by the fact that $G/N$ consists of the cosets of $N$ in $G$, and that these [cosets are the same size](https://arbital.com/p/4j8), and [partition](https://arbital.com/p/4j5) $G$. \n\n\n\n#Example\nSuppose you have a collection of objects, and you need to split them into two equal groups. So you are trying to determine under what circumstances changing the number of objects will affect this property. You notice that changing the size of the collection by certain numbers such as 0, 2, 4, 24, and -6 doesn't affect this property. \n\nThe set of different size changes can be modeled as the additive group of integers $\\mathbb Z$. The changes that don't affect this property also form a group: $2\\mathbb Z = \\{2n : n\\in \\mathbb Z\\}$. Exercise: verify that this is a normal subgroup of $\\mathbb Z$. \n\nThis subgroup gives us two cosets: $0 + 2\\mathbb Z$ and $1 + 2\\mathbb Z$ (remember that $+$ is the group operation in this example), which are the elements of our quotient group. We will give them their conventional names: $\\text{even}$ and $\\text{odd}$, and we can apply the coset multiplication rule to see that $\\text{even}+ \\text{even} = \\text{even}$, $\\text{even} + \\text{odd} = \\text{odd}$, and $\\text{odd} + \\text{odd} = \\text{even}$. \n\nInstead of thinking about specific numbers, and how they will change our ability to split our collection of objects into two equal groups, we now have reduced the problem to its essence. Only the parity matters, and it follows the simple rules of the quotient group we discovered. \n\n#See also \n\n - [Lagrange's theorem](https://arbital.com/p/4jn).\n - [The first isomorphism theorem](https://arbital.com/p/first_isomorphism_theorem).", "date_published": "2016-07-01T01:13:15Z", "authors": ["Eric Rogstad", "Mark Chimes", "Adele Lopez"], "summaries": ["Given a [group](https://arbital.com/p/-3gd) $G$ with operation $\\bullet$ and a special kind of [subgroup](https://arbital.com/p/-subgroup) $N \\leq G$ called the \"[normal subgroup](https://arbital.com/p/-4h6)\", there is a way of \"dividing\" $G$ by $N$, written $G/N$. Usually this is defined so that each [element](https://arbital.com/p/-element_mathematics) of $G/N$ is a [subset](https://arbital.com/p/-subset) of $G$. In fact, each of them is an [equivalence class](https://arbital.com/p/-equivalence_class) of elements in $G$, so called because we think of everything in one equivalence class as being equivalent. One of the equivalence classes is $N$, and each of the other equivalence classes can be seen as $N$ \"shifted around\" inside $G$. Because they are equivalence classes, each element of $G$ only occurs in exactly one of them.\n\nThis collection $G/N$ of equivalence classes has a [group structure](https://arbital.com/p/-group_structure) based on the structure on $G$. Write the [operation](https://arbital.com/p/-group_operation) as $\\circ$. Say $A$ and $B$ are two equivalence classes in $G/N$. Then $A \\circ B$ is defined by taking an element $a \\in A$ and $b \\in B$, multiplying them as $a \\bullet b$ and checking in which equivalence class this product ends up. \n\nThere are some things to check, namely that the equivalence class in which we end up is the same regardless of which elements in $A$ and $B$ we happen to pick. This is guaranteed by the fact that $N$ is normal in $G$.\n\nThere is also a [group homomorphism](https://arbital.com/p/-47t) $\\phi$ from $G$ to $G/N$ that \"collapses\" the elements by sending each element in $G$ to the equivalence class in which it lies. This is called collapsing because lots of elements in $G$ are sent to the same element in $G/N$, so we can see $G/N$ as a collapsed version of $G$."], "tags": [], "alias": "4tq"}
{"id": "2085488f2f3138bdd382189304492b3b", "title": "Exponential", "url": "https://arbital.com/p/exponential", "source": "arbital", "source_type": "text", "text": "An exponential is a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) that can be represented by some [https://arbital.com/p/-constant](https://arbital.com/p/-constant) taken to the [power](https://arbital.com/p/power_mathematics) of a [https://arbital.com/p/-variable](https://arbital.com/p/-variable). The name comes from the fact that the variable is the *exponent* of the expression.\n\n\n## Exponential Growth\n\nExponentials are most useful in describing growth patterns where the growth rate is [proportional](https://arbital.com/p/4w3) to the amount of the thing that's growing. They can be represented by the formula: $f(x) = c \\times a^x$, where $c$ is the starting value and $a$ is the growth factor.\n\nThe classic example of exponential growth is [https://arbital.com/p/-compound_interest](https://arbital.com/p/-compound_interest). If you have \\$100 in a bank account that gives you 2% interest every year, then every year your money is multiplied by $1.02$. This means you can represent your account balance as $f(x) = 100 \\times 1.02^x$, where $x$ is the number of years your money has been in the account.\n\nAnother example is a dividing cell. If one cell is placed into an infinite culture and splits once every hour, the number of cells in the culture after $x$ hours is $f(x) = 1 \\times 2^x$ (assuming none of the cells die).\n\n\n### Recursive definition\n\nWe mentioned earlier that in an exponential function, the growth rate is proportional to the amount of the thing that's growing. In the compound interest example, we can write each value in terms of the previous value: $f(x) = f(x-1) \\times 1.02$. Therefore, the amount of growth at every step can be represented as $\\Delta f(x) = f(x+1) - f(x) = 0.02 \\times f(x)$.\n\nThis makes exponential growth a *memoryless growth function*, as the growth rate depends only on current information. Compare this to [https://arbital.com/p/-simple_interest](https://arbital.com/p/-simple_interest), where the interest only grows at a percentage of the initial value. Then we would have to write: $f(x) = f(x-1) + 0.02 \\times f(0)$, and because $f(0)$ is constant while $f(x)$ continues to grow, we cannot express the growth rate in terms of only current information — we have to keep \"in memory\" the initial balance to calculate the growth rate.", "date_published": "2016-07-04T05:34:11Z", "authors": ["Eric Bruylant", "Nate Soares", "Joe Zeng"], "summaries": ["The exponent base $b$ of a number $x$, written $b^x,$ is what you get when you multiply 1 by $b$, $x$ times. For example, the exponent base 10 of 3 is $10^3$ and pronounced \"10 to the power 3\" is 1000, because $10 \\cdot 10 \\cdot 10 = 1000$. Similarly, $2^4=16,$ because $2 \\cdot 2 \\cdot 2 \\cdot 2 = 16.$\n\nThe input $x$ may be fractional. For example, $10^{1/2}$ is the number you get when you multiply 1 by 10 half a time, where \"multiplying by 10 half a time\" means multiplying by a number $n$ such that if you multiplied by $n$ twice, you would have multiplied by 10 once. In this case, $n \\approx 3.16,$ because then $n \\cdot n \\approx 10.$"], "tags": ["Start", "Needs lenses"], "alias": "4ts"}
{"id": "226ed4f2608c42dfba086266e12ad6a9", "title": "Most complex things are not very compressible", "url": "https://arbital.com/p/most_complexity_incompressible", "source": "arbital", "source_type": "text", "text": "Although the [halting problem](https://arbital.com/p/46h) means we can't *prove* it doesn't happen, it would nonetheless be *extremely surprising* if some 100-state Turing machine turned out to print the exact text of Shakespeare's *Romeo and Juliet.* Unless something was specifically generated by a simple algorithm, the Vast supermajority of data structures that *look* like they have high [algorithmic complexity](https://arbital.com/p/5v) actually *do* have high algorithmic complexity. Since there are at most $2^{101}$ programs that can be specified with at most 100 bits (in any particular language), we can't fit all the 1000-bit data structures into all the 100-bit programs. While *Romeo and Juliet* is certainly highly compressible, relative to most random bitstrings of the same length, it would be shocking for it to compress *all the way down* to a 100-state Turing machine. There just aren't enough 100-state Turing machines for one of their outputs to be *Romeo and Juliet*. Similarly, if you start with a 10 kilobyte text file, and 7zip compresses it down to 2 kilobytes, no amount of time spent trying to further compress the file using other compression programs will ever get that file down to 1 byte. For any given compressor there's at most 256 starting files that can ever be compressed down to 1 byte, and your 10-kilobyte text file almost certainly isn't one of them.\n\nThis takes on defensive importance with respect to refuting the probability-theoretic fallacy, \"Oh, sure, Occam's Razor seems to say that this proposition is complicated. But how can you be sure that this apparently complex proposition wouldn't turn out to be generated by some very simple mechanism?\" If we consider a [partition](https://arbital.com/p/1rd) of 10,000 possible propositions, collectively having a 0.01% probability on average, then all the arguments in the world for why various propositions might have unexpectedly high probability, must still add up to an average probability of 0.01%. It can't be the case that after considering that proposition 1 might have secretly high probability, and considering that proposition 2 might have secretly high probability, and so on, we end up assigning 5% probability to each proposition, because that would be a total probability of 500. If we assign prior probabilities using an algorithmic-complexity Occam prior as in [Solomonoff induction](https://arbital.com/p/11w), then the observation that \"most apparently complex things can't be further compressed into an amazingly simple Turing machine\", is the same observation as that \"most apparently Occam-penalized propositions can't turn out to be simpler than they look\" or \"most apparently subjectively improbable things can't turn out to have unseen clever arguments that would validly make them more subjectively probable\".", "date_published": "2016-06-27T00:06:10Z", "authors": ["Eliezer Yudkowsky"], "summaries": [], "tags": ["Start"], "alias": "4v5"}
{"id": "9a184632db24b13bb41959f02aadda5c", "title": "Group orbit", "url": "https://arbital.com/p/group_orbit", "source": "arbital", "source_type": "text", "text": "When we have a [group](https://arbital.com/p/3gd) [acting](https://arbital.com/p/3t9) on a set, we are often interested in how the group acts on a particular [element](https://arbital.com/p/group_element). One natural way to do this is to look at the set of all the elements that different group actions will take the starting element to. This is called the orbit. \n\n#Definition\nLet $X$ be a set, with element $x \\in X$, and let $G$ be a group acting on $X$. Then the orbit of $x$ is $Gx = \\{gx : g \\in G\\}$. \n\n#Properties\nThe set $X$ is partitioned by group orbits - each element defines its own orbit, and two orbits containing the same element must be the same, because every action has an inverse action. This gives the fact that [cosets partition a group](https://arbital.com/p/4j5) as the special case where $X$ is a subgroup of $G$. \n#See Also\n\n - [The orbit-stabiliser theorem](https://arbital.com/p/4l8).", "date_published": "2016-06-28T21:01:45Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Adele Lopez"], "summaries": [], "tags": ["Formal definition", "Needs clickbait"], "alias": "4v8"}
{"id": "f0b42b591f7187b0aa1a5c857ae7cff8", "title": "Lagrange theorem on subgroup size: Intuitive version", "url": "https://arbital.com/p/lagrange_theorem_on_subgroup_size_intuitive", "source": "arbital", "source_type": "text", "text": "Given a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$, it may have many [subgroups](https://arbital.com/p/576).\nSo far, we know almost nothing about those subgroups; it would be great if we had some way of restricting them.\n\nAn example of such a restriction, which we do already know, is that a subgroup $H$ of $G$ has to have [size](https://arbital.com/p/3gg) less than or equal to the size of $G$ itself.\nThis is because $H$ is contained in $G$, and if the set $X$ is contained in the set $Y$ then the size of $X$ is less than or equal to the size of $Y$.\n(This would have to be true for any reasonable definition of \"size\"; the [usual definition](https://arbital.com/p/4w5) certainly has this property.)\n\nLagrange's Theorem gives us a much more powerful restriction: not only is the size $|H|$ of $H$ less than or equal to $|G|$, but in fact $|H|$ divides $|G|$.\n\n%%hidden(Example: subgroups of the cyclic group on six elements):\n*A priori*, all we know about the subgroups of the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_6$ of order $6$ is that they are of order $1, 2, 3, 4, 5$ or $6$.\n\nLagrange's Theorem tells us that they can only be of order $1, 2, 3$ or $6$: there are no subgroups of order $4$ or $5$.\nLagrange tells us nothing about whether there *are* subgroups of size $1,2,3$ or $6$: only that if we are given a subgroup, then it is of one of those sizes.\n\nIn fact, as an aside, there are indeed subgroups of sizes $1,2,3,6$:\n\n- the subgroup containing only the identity is of order $1$\n- the \"improper\" subgroup $C_6$ is of order $6$\n- subgroups of size $2$ and $3$ are guaranteed by [Cauchy's theorem](https://arbital.com/p/4l6).\n\n%%\n\n# Proof\n\nIn order to show that $|H|$ divides $|G|$, we would be done if we could divide the elements of $G$ up into separate buckets of size $|H|$.\n\nThere is a fairly obvious place to start: we already have one bucket of size $|H|$, namely $H$ itself (which consists of some elements of $G$).\nCan we perhaps use this to create more buckets of size $|H|$?\n\nFor motivation: if we think of $H$ as being a collection of symmetries (which we can do, by [Cayley's Theorem](https://arbital.com/p/49b) which states that all groups may be viewed as collections of symmetries), then we can create more symmetries by \"tacking on elements of $G$\".\n\nFormally, let $g$ be an element of $G$, and consider $gH = \\{ g h : h \\in H \\}$.\n\nExercise: every element of $G$ does have one of these buckets $gH$ in which it lies.\n%%hidden(Show solution):\nThe element $g$ of $G$ is contained in the bucket $gH$, because the identity $e$ is contained in $H$ and so $ge$ is in $gH$; but $ge = g$.\n%%\n\nExercise: $gH$ is a set of size $|H|$. %%note:More formally put, [https://arbital.com/p/-4j8](https://arbital.com/p/-4j8).%%\n%%hidden(Show solution):\nIn order to show that $gH$ has size $|H|$, it is enough to match up the elements of $gH$ [bijectively](https://arbital.com/p/499) with the elements of $|H|$.\n\nWe can do this with the [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) $H \\to gH$ taking $h \\in H$ and producing $gh$.\nThis has an [inverse](https://arbital.com/p/4sn): the function $gH \\to H$ which is given by pre-multiplying by $g^{-1}$, so that $gx \\mapsto g^{-1} g x = x$.\n%%\n\nNow, are all these buckets separate? Do any of them overlap?\n\nExercise: if $x \\in rH$ and $x \\in sH$ then $rH = sH$. That is, if any two buckets intersect then they are the same bucket. %%note:More formally put, [https://arbital.com/p/4j5](https://arbital.com/p/4j5).%%\n%%hidden(Show solution):\nSuppose $x \\in rH$ and $x \\in sH$.\n\nThen $x = r h_1$ and $x = s h_2$, some $h_1, h_2 \\in H$.\n\nThat is, $r h_1 = s h_2$, so $s^{-1} r h_1 = h_2$.\nSo $s^{-1} r = h_2 h_1^{-1}$, so $s^{-1} r$ is in $H$ by closure of $H$.\n\nBy taking inverses, $r^{-1} s$ is in $H$.\n\nBut that means $\\{ s h : h \\in H \\}$ and $\\{ r h : h \\in H\\}$ are equal.\nIndeed, we show that each is contained in the other.\n\n- if $a$ is in the right-hand side, then $a = rh$ for some $h$. Then $s^{-1} a = s^{-1} r h$; but $s^{-1} r$ is in $H$, so $s^{-1} r h$ is in $H$, and so $s^{-1} a$ is in $H$.\nTherefore $a \\in s H$, so $a$ is in the left-hand side.\n- if $a$ is in the left-hand side, then $a = sh$ for some $h$. Then $r^{-1} a = r^{-1} s h$; but $r^{-1} s$ is in $H$, so $r^{-1} s h$ is in $H$, and so $r^{-1} a$ is in $H$.\nTherefore $a \\in rH$, so $a$ is in the right-hand side.\n%%\n\nWe have shown that the \"[cosets](https://arbital.com/p/4j4)\" $gH$ are all completely disjoint and are all the same size, and that every element lies in a bucket; this completes the proof.", "date_published": "2016-08-27T12:11:30Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": ["Lagrange's Theorem gives us a restriction on the size of a subgroup of a finite group: namely, subgroups have size dividing the parent group."], "tags": [], "alias": "4vz"}
{"id": "ef998743a16cd66299c16790b2831c7e", "title": "Proportion", "url": "https://arbital.com/p/proportion", "source": "arbital", "source_type": "text", "text": "A **proportion** is a representation of one value as a [https://arbital.com/p/-fraction](https://arbital.com/p/-fraction) or [https://arbital.com/p/-multiple](https://arbital.com/p/-multiple) of another value. It is used to provide a sense of relative magnitude.\n\nIf some value $a$ is proportional to another value $b$, it means that $a$ can be expressed as $c \\times b$, where $c$ is some constant, also known as the *constant of proportionality* (or sometimes just the *proportion*) of $a$ to $b$.\n\nFor example, no matter how large a domino is, it is always twice as long as it is wide. Then, the proportion of the length compared to the width can be said to be $2$, and we can write $l = 2w$ where $w$ is the width and $l$ is the length.\n\n\n## Notation\n\nTo write that $a$ is proportional to $b$, the notation $a \\propto b$ is used. For example, the surface area of an object of a specific shape is proportional to the square of its length, so we write $A \\propto L^2$ where $A$ is area and $L$ is length.\n\n\n## Percentages\n\nPercentages are a way of describing proportions relative to the number 100 to make them easier to understand. When people refer to a \"percentage basis\" for calculating something, they are specifically talking about this type of proportion.\n\nTo get a percentage from a proportion, simply multiply it by 100. In our domino example, we multiply $2$ by $100$ to find that the length is equal to $200\\%$ of the width.\n\nPercentages only work with dimensionless proportions — that is, where the constant of proportionality has no [units](https://arbital.com/p/unit). For example, if a car requires $6.7$ litres of gas to travel $100$ kilometres, it makes no sense to say that the fuel efficiency of the car (which is a constant of proportionality for fuel used versus distance travelled) is $6.7\\%$ litres per kilometre. However, when comparing the fuel efficiencies of two cars, if one car has an efficiency of $5.2$ L/100km and another has one of $6.8$ L/100km, it makes sense to say that the first car is $131\\%$ as efficient as the second.\n\n\n## Common examples of proportions\n\nThe [https://arbital.com/p/-circumference](https://arbital.com/p/-circumference) of a [https://arbital.com/p/-circle](https://arbital.com/p/-circle) is proportional to its [https://arbital.com/p/-diameter](https://arbital.com/p/-diameter). We write $C \\propto d$, and the constant of proportionality is $\\pi = 3.14159265\\ldots$.\n\nThe [https://arbital.com/p/-area](https://arbital.com/p/-area) of a circle is proportional to the square of its radius. We write $A \\propto r^2$, and the constant of proportionality is also $\\pi$.\n\nThe rate of growth of an [https://arbital.com/p/-4ts](https://arbital.com/p/-4ts) function is proportional to the current value of the function. We write $\\frac{df}{dt} \\propto f$, or $\\Delta f \\propto f$ if considering discrete values.\n\nThe gravitational potential energy of an object in a constant gravitational field is proportional to its height relative to some zero point. We write $E \\propto h$, and the constant of proportionality is the weight of the object, its mass multiplied by the gravitational acceleration in the field.\n\nThe amount of current flowing through a specific wire is proportional to the potential difference, or voltage, between the endpoints of the wire. We write $I \\propto V$, and the constant of proportionality is the resistance of that wire.\n\nIn an ideal gas, the pressure of the gas is proportional to its temperature, if the volume and number of particles in the gas is held constant. This is one property of the ideal gas law in physics. We write $P \\propto T$.", "date_published": "2016-07-04T17:39:25Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": [], "tags": ["Needs image", "C-Class"], "alias": "4w3"}
{"id": "bb01f09aadb748a086486b8b92fcdda7", "title": "Cardinality", "url": "https://arbital.com/p/cardinality", "source": "arbital", "source_type": "text", "text": "summary: If $A$ is a finite set then the cardinality of $A$, denoted $|A|$, is the number of elements $A$ contains. When $|A| = n$, we say that $A$ is a set of cardinality $n$. There exists a [bijection](https://arbital.com/p/499) from any finite set of cardinality $n$ to the set $\\{0, ..., (n-1)\\}$ containing the first $n$ natural numbers.\n\nTwo infinite sets have the same cardinality if there exists a bijection between them. Any set in bijective correspondence with [$\\mathbb N$](https://arbital.com/p/45h) is called __countably infinite__, while any infinite set that is not in bijective correspondence with $\\mathbb N$ is call **[uncountably infinite](https://arbital.com/p/2w0)**. All countably infinite sets have the same cardinality, whereas there are multiple distinct uncountably infinite cardinalities.\n\nsummary(technical): The cardinality (or size) $|X|$ of a set $X$ is the number of elements in $X.$ For example, letting $X = \\{a, b, c, d\\}, |X|=4.$\n\n\n\nThe **cardinality** of a [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz) is a formalization of the \"number of elements\" in the set.\n\nSet cardinality is an [https://arbital.com/p/-53y](https://arbital.com/p/-53y). Two sets have the same cardinality if (and only if) there exists a [bijection](https://arbital.com/p/499) between them.\n\n## Definition of equivalence classes\n\n### Finite sets\n\nA set $S$ has a cardinality of a [https://arbital.com/p/-45h](https://arbital.com/p/-45h) $n$ if there exists a bijection between $S$ and the set of natural numbers from $1$ to $n$. For example, the set $\\{9, 15, 12, 20\\}$ has a bijection with $\\{1, 2, 3, 4\\}$, which is simply mapping the $m$th element in the first set to $m$; therefore it has a cardinality of $4$.\n\nWe can see that this equivalence class is [well-defined](https://arbital.com/p/) — if there exist two sets $S$ and $T$, and there exist bijective functions $f : S \\to \\{1, 2, 3, \\ldots, n\\}$ and $g : \\{1, 2, 3, \\ldots, n\\} \\to T$, then $g \\circ f$ is a bijection between $S$ and $T$, and so the two sets also have the same cardinality as each other, which is $n$.\n\nThe cardinality of a finite set is always a natural number, never a fraction or decimal.\n\n### Infinite sets\n\nAssuming the [axiom of choice](https://arbital.com/p/69b), the cardinalities of infinite sets are represented by the [https://arbital.com/p/aleph_numbers](https://arbital.com/p/aleph_numbers). A set has a cardinality of $\\aleph_0$ if there exists a bijection between that set and the set of *all* natural numbers. This particular class of sets is also called the class of [https://arbital.com/p/-countably_infinite_sets](https://arbital.com/p/-countably_infinite_sets).\n\nLarger infinities (which are [uncountable](https://arbital.com/p/2w0)) are represented by higher Aleph numbers, which are $\\aleph_1, \\aleph_2, \\aleph_3,$ and so on through the [ordinals](https://arbital.com/p/ordinal).\n\n**In the absence of the Axiom of Choice**\n\nWithout the axiom of choice, not every set may be [well-ordered](https://arbital.com/p/55r), so not every set bijects with an [https://arbital.com/p/-ordinal](https://arbital.com/p/-ordinal), and so not every set bijects with an aleph.\nInstead, we may use the rather cunning [https://arbital.com/p/Scott_trick](https://arbital.com/p/Scott_trick).\n\n%%todo: Examples and exercises (possibly as lenses) %%\n\n%%todo: Split off a more accessible cardinality page that explains the difference between finite, countably infinite, and uncountably infinite cardinalities without mentioning alephs, ordinals, or the axiom of choice.%%", "date_published": "2016-10-05T17:51:17Z", "authors": ["Dylan Hendrickson", "Eric Rogstad", "Patrick Stevens", "Eric Bruylant", "Joe Zeng"], "summaries": ["The **cardinality** of a [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz) is a formalization of the \"number of [elements](https://arbital.com/p/5xy)\" in the set."], "tags": ["Needs splitting by mastery"], "alias": "4w5"}
{"id": "940cef4103c37e7de52bd72add973eee", "title": "Needs clickbait", "url": "https://arbital.com/p/needs_clickbait_meta_tag", "source": "arbital", "source_type": "text", "text": "A meta tag for pages which don't have [clickbait](https://arbital.com/p/597). Clickbait style guidelines:\n\nYes: \"First letter is capital.\" \nYes: \"Pique user's interest in the topic.\" \nYes: \"Most clickbaits are just one sentence telling user what they might find out if they read the page.\" \nYes: \"Can a clickbait ask the reader a leading question?\" \nYes: \"Period or other punctuation at the end.\" \nYes: Omitting clickbait if the title is descriptive enough. \nNo: \"Capitalize Every Word\" \nNo: \"**Markdown**\"", "date_published": "2016-08-19T21:01:16Z", "authors": ["Eric Bruylant", "Jaime Sevilla Molina", "Patrick Stevens"], "summaries": ["This page does not have [clickbait](https://arbital.com/p/597). Clickbait style guidelines:\n\nYes: \"First letter is capital.\" \nYes: \"Pique user's interest in the topic.\" \nYes: \"Most clickbaits are just one sentence telling user what they might find out if they read the page.\" \nYes: \"Can a clickbait ask the reader a leading question?\" \nYes: \"Period or other punctuation at the end.\" \nYes: Omitting clickbait if the title is descriptive enough. \nNo: \"Capitalize Every Word\" \nNo: \"**Markdown**\""], "tags": ["Meta tags which request an edit to the page"], "alias": "4w9"}
{"id": "49f2161a29c71520b6b4a5105be2af23", "title": "Interpretations of \"probability\"", "url": "https://arbital.com/p/probability_interpretations", "source": "arbital", "source_type": "text", "text": "What does it *mean* to say that a flipped coin has a 50% probability of landing heads?\n\nHistorically, there are two popular types of answers to this question, the \"[frequentist](https://arbital.com/p/frequentist_probability)\" and \"[subjective](https://arbital.com/p/4vr)\" (aka \"[Bayesian](https://arbital.com/p/1r8)\") answers, which give rise to [radically different approaches to experimental statistics](https://arbital.com/p/4xx). There is also a third \"[propensity](https://arbital.com/p/propensity)\" viewpoint which is largely discredited (assuming the coin is deterministic). Roughly, the three approaches answer the above question as follows:\n\n- __The propensity interpretation:__ Some probabilities are just out there in the world. It's a brute fact about coins that they come up heads half the time. When we flip a coin, it has a fundamental *propensity* of 0.5 for the coin to show heads. When we say the coin has a 50% probability of being heads, we're talking directly about this propensity.\n- __The frequentist interpretation:__ When we say the coin has a 50% probability of being heads after this flip, we mean that there's a class of events similar to this coin flip, and across that class, coins come up heads about half the time. That is, the _frequency_ of the coin coming up heads is 50% inside the event class, which might be \"all other times this particular coin has been tossed\" or \"all times that a similar coin has been tossed\", and so on.\n- __The subjective interpretation:__ Uncertainty is in the mind, not the environment. If I flip a coin and slap it against my wrist, it's already landed either heads or tails. The fact that I don't know whether it landed heads or tails is a fact about me, not a fact about the coin. The claim \"I think this coin is heads with probability 50%\" is an _expression of my own ignorance,_ and 50% probability means that I'd bet at 1 : 1 odds (or better) that the coin came up heads.\n\n\nFor a visualization of the differences between these three viewpoints, see [https://arbital.com/p/4yj](https://arbital.com/p/4yj). For examples of the difference, see [https://arbital.com/p/4yn](https://arbital.com/p/4yn). See also the [Stanford Encyclopedia of Philosophy article on interpretations of probability](https://arbital.com/p/http://plato.stanford.edu/entries/probability-interpret/).\n\nThe propensity view is perhaps the most intuitive view, as for many people, it just feels like the coin is intrinsically random. However, this view is difficult to reconcile with the idea that once we've flipped the coin, it has already landed heads or tails. If the event in question is decided [deterministically](https://arbital.com/p/determinism), the propensity view can be seen as an instance of the [https://arbital.com/p/-4yk](https://arbital.com/p/-4yk): When we mentally consider the coin flip, it feels 50% likely to be heads, so we find it very easy to imagine a *world* in which the coin is _fundamentally_ 50%-heads-ish. But that feeling is actually a fact about _us,_ not a fact about the coin; and the coin has no physical 0.5-heads-propensity hidden in there somewhere — it's just a coin.\n\nThe other two interpretations are both self-consistent, and give rise to pragmatically different statistical techniques, and there has been much debate as to which is preferable. The subjective interpretation is more generally applicable, as it allows one to assign probabilities (interpreted as betting odds) to one-off events.\n\n# Frequentism vs subjectivism\n\nAs an example of the difference between frequentism and subjectivism, consider the question: \"What is the probability that Hillary Clinton will win the 2016 US presidential election?\", as analyzed in the summer of 2016.\n\nA stereotypical (straw) frequentist would say, \"The 2016 presidential election only happens once. We can't *observe* a frequency with which Clinton wins presidential elections. So we can't do any statistics or assign any probabilities here.\"\n\nA stereotypical subjectivist would say: \"Well, prediction markets tend to be pretty [well-calibrated](https://arbital.com/p/well_calibrated) about this sort of thing, in the sense that when prediction markets assign 20% probability to an event, it happens around 1 time in 5. And the prediction markets are currently betting on Hillary at about 3 : 1 odds. Thus, I'm comfortable saying she has about a 75% chance of winning. If someone offered me 20 : 1 odds _against_ Clinton — they get \\$1 if she loses, I get \\$20 if she wins — then I'd take the bet. I suppose you could refuse to take that bet on the grounds that you Just Can't Talk About Probabilities of One-off Events, but then you'd be pointlessly passing up a really good bet.\"\n\nA stereotypical (non-straw) frequentist would reply: \"I'd take that bet too, of course. But my taking that bet *is not based on rigorous epistemology,* and we shouldn't allow that sort of thinking in experimental science and other important venues. You can do subjective reasoning about probabilities when making bets, but we should exclude subjective reasoning in our scientific journals, and that's what frequentist statistics is designed for. Your paper should not conclude \"and therefore, having observed thus-and-such data about carbon dioxide levels, I'd personally bet at 9 : 1 odds that anthropogenic global warming is real,\" because you can't build scientific consensus on opinions.\"\n\n...and then it starts getting complicated. The subjectivist responds \"First of all, I agree you shouldn't put posterior odds into papers, and second of all, it's not like your method is truly objective — the choice of \"similar events\" is arbitrary, abusable, and has given rise to [p-hacking](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging) and the [replication crisis](https://arbital.com/p/https://en.wikipedia.org/wiki/Replication_crisis).\" The frequentists say \"well your choice of prior is even more subjective, and I'd like to see you do better in an environment where peer pressure pushes people to abuse statistics and exaggerate their results,\" and then [down the rabbit hole we go](https://arbital.com/p/4xx).\n\nThe subjectivist interpretation of probability is common among artificial intelligence researchers (who often design computer systems that manipulate subjective probability distributions), wall street traders (who need to be able to make bets even in relatively unique situations), and common intuition (where people feel like they can say there's a 30% chance of rain tomorrow without worrying about the fact that tomorrow only happens once). Nevertheless, the frequentist interpretation is commonly taught in introductory statistics classes, and is the gold standard for most scientific journals.\n\nA common frequentist stance is that it is virtuous to have a large toolbox of statistical tools at your disposal. Subjectivist tools have their place in that toolbox, but they don't deserve any particular primacy (and they aren't generally accepted when it comes time to publish in a scientific journal).\n\nAn aggressive subjectivist stance is that frequentists have invented some interesting tools, and many of them are useful, but that refusing to consider subjective probabilities is toxic. Frequentist statistics were invented in a (failed) attempt to keep subjectivity out of science in a time before humanity really understood the laws of probability theory. Now we have [theorems](https://arbital.com/p/1lz) about how to manage subjective probabilities correctly, and how to factor personal beliefs out from the objective evidence provided by the data, and if you ignore these theorems you'll get in trouble. The frequentist interpretation is broken, and that's why science has p-hacking and a replication crisis even as all the wall-street traders and AI scientists use the Bayesian interpretation. This \"let's compromise and agree that everyone's viewpoint is valid\" thing is all well and good, but how much worse do things need to get before we say \"oops\" and start acknowledging the subjective probability interpretation across all fields of science?\n\nThe most common stance among scientists and researchers is much more agnostic, along the lines of \"use whatever statistical techniques work best at the time, and use frequentist techniques when publishing in journals because that's what everyone's been doing for decades upon decades upon decades, and that's what everyone's expecting.\"\n\nSee also [https://arbital.com/p/frequentist_probability](https://arbital.com/p/frequentist_probability), [https://arbital.com/p/4vr](https://arbital.com/p/4vr), and [https://arbital.com/p/4xx](https://arbital.com/p/4xx).\n\n# Which interpretation is most useful?\n\nProbably the subjective interpretation, because it subsumes the propensity and frequentist interpretations as special cases, while being more flexible than both.\n\nWhen the frequentist \"similar event\" class is clear, the subjectivist can take those frequencies (often called [base rates](https://arbital.com/p/base_rate) in this context) into account. But unlike the frequentist, she can also [combine those base rates with other evidence that she's seen](https://arbital.com/p/1lz), and assign probabilities to one-off events, and make money in prediction markets and/or stock markets (when she knows something that the market doesn't).\n\nWhen the laws of physics actually do \"contain uncertainty\", such as when they say that there are multiple different observations you might make next with differing likelihoods (as the [Schrodinger equation](https://arbital.com/p/schrodinger_equation) often will), a subjectivist can combine her propensity-style uncertainty with her personal uncertainty in order to generate her aggregate subjective probabilities. But unlike a propensity theorist, she's not forced to think that _all_ uncertainty is physical uncertainty: She can act like a propensity theorist with respect to Schrodinger-equation-induced uncertainty, while still believing that her uncertainty about a coin that has already been flipped and slapped against her wrist is in her head, rather than in the coin.\n\nThis fully general stance is consistent with the belief that frequentist tools are useful for answering frequentist questions: The fact that you can _personally_ assign probabilities to one-off events (and, e.g., evaluate how good a certain trade is on a prediction market or a stock market) does not mean that tools labeled \"Bayesian\" are always better than tools labeled \"frequentist\". Whatever interpretation of \"probability\" you use, you're encouraged to use whatever statistical tool works best for you at any given time, regardless of what \"camp\" the tool comes from. Don't let the fact that you think it's possible to assign probabilities to one-off events prevent you from using useful frequentist tools!", "date_published": "2016-07-01T05:22:14Z", "authors": ["Eric Rogstad", "Nate Soares"], "summaries": ["What does it *mean* to say that a fair coin has a 50% probability of landing heads? There are three common views:\n\n- __The propensity interpretation:__ Some probabilities are just out there in the world. It's a brute fact about coins that they come up heads half the time; we'll call this the coin's physical \"propensity towards heads.\" When we say the coin has a 50% probability of being heads, we're talking directly about this propensity.\n- __The frequentist interpretation:__ When we say the coin has a 50% probability of being heads after this flip, we mean that there's a class of events similar to this coin flip, and across that class, coins come up heads about half the time. That is, the _frequency_ of the coin coming up heads is 50% inside the event class (which might be \"all other times this particular coin has been tossed\" or \"all times that a similar coin has been tossed\" etc).\n- __The subjective interpretation:__ Uncertainty is in the mind, not the environment. If I flip a coin and slap it against my wrist, it's already landed either heads or tails. The fact that I don't know whether it landed heads or tails is a fact about me, not a fact about the coin. The claim \"I think this coin is heads with probability 50%\" is an _expression of my own ignorance,_ which means that I'd bet at 1 : 1 odds (or better) that the coin came up heads.\n\nWhen the event in question is deterministic, the propensity view is commonly considered an example of the [https://arbital.com/p/-4yk](https://arbital.com/p/-4yk). The frequentist view is self-consistent, but says that one-off events (such as the probability of it raining where you are _tomorrow in particular_) cannot be assigned \"probabilities\", because there are no objective frequencies to draw upon. The subjectivist view extends the frequentist view, and allows you to also consider probabilities (interpreted as betting odds) on one-off events. [https://arbital.com/p/1bv](https://arbital.com/p/1bv) provides tools for reasoning about subjective probabilities in a rigorous, consistent way."], "tags": [], "alias": "4y9"}
{"id": "67a6af1791df94475bf209d4ae57ac14", "title": "A-Class", "url": "https://arbital.com/p/a_class_meta_tag", "source": "arbital", "source_type": "text", "text": "A-Class pages are comprehensive, intuitive, engaging, well-organized, and, where it would be useful, include excellent supporting material (e.g. examples, images, and exercises). You would enjoy learning from it, and send it to your friends. To reach A-Class pages must receive detailed feedback from both the target audience to make sure it's a good experience to learn from and [reviewers](https://arbital.com/p/5ft) with a strong understanding of the topic.\n\nFor higher level audiences asking on [Slack](https://arbital.com/p/4ph) is a good way to get feedback, for [https://arbital.com/p/1r3](https://arbital.com/p/1r3) or [https://arbital.com/p/1r5](https://arbital.com/p/1r5) sharing with your friends (in real life of via social media) is a better bet. To get the reviewers attention ask in #reviews on Slack.\n\nA-Class pages are:\n\n1. Comprehensive and authoritative, covering all areas of the topic without notable omissions or inaccuracies.\n2. Written to an engaging, enjoyable, professional standard.\n3. Well-organized, with useful overall and local structure.\n4. Aimed at an audience, with a level of technicality, jargon, and notation appropriate for them.\n5. Intuitively explained, making use of examples, visual aids, requisites, and anything else which aids reader comprehension.\n6. Very well-[summarized](https://arbital.com/p/1kl), in ways appropriate for all likely audiences.\n\nThis tag should only be added by [reviewers](https://arbital.com/p/5ft). \n\n**[Quality scale](https://arbital.com/p/4yg)**\n\n* [https://arbital.com/p/4ym](https://arbital.com/p/4ym)\n* [https://arbital.com/p/4gs](https://arbital.com/p/4gs)\n* [https://arbital.com/p/72](https://arbital.com/p/72)\n* [https://arbital.com/p/3rk](https://arbital.com/p/3rk)\n* [https://arbital.com/p/4y7](https://arbital.com/p/4y7)\n* [https://arbital.com/p/4yd](https://arbital.com/p/4yd)\n* [https://arbital.com/p/4yf](https://arbital.com/p/4yf)\n* [https://arbital.com/p/4yl](https://arbital.com/p/4yl)", "date_published": "2016-08-19T21:47:21Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Eliezer Yudkowsky", "Alexei Andreev"], "summaries": [], "tags": [], "alias": "4yf"}
{"id": "105b7db7c64a5807d4ba91657ea41c5e", "title": "Correspondence visualizations for different interpretations of \"probability\"", "url": "https://arbital.com/p/probability_interpretations_correspondence", "source": "arbital", "source_type": "text", "text": "[Recall](https://arbital.com/p/4y9) that there are three common interpretations of what it means to say that a coin has a 50% probability of landing heads:\n\n- __The propensity interpretation:__ Some probabilities are just out there in the world. It's a brute fact about coins that they come up heads half the time; we'll call this the coin's physical \"propensity towards heads.\" When we say the coin has a 50% probability of being heads, we're talking directly about this propensity.\n- __The frequentist interpretation:__ When we say the coin has a 50% probability of being heads after this flip, we mean that there's a class of events similar to this coin flip, and across that class, coins come up heads about half the time. That is, the _frequency_ of the coin coming up heads is 50% inside the event class (which might be \"all other times this particular coin has been tossed\" or \"all times that a similar coin has been tossed\" etc).\n- __The subjective interpretation:__ Uncertainty is in the mind, not the environment. If I flip a coin and slap it against my wrist, it's already landed either heads or tails. The fact that I don't know whether it landed heads or tails is a fact about me, not a fact about the coin. The claim \"I think this coin is heads with probability 50%\" is an _expression of my own ignorance,_ which means that I'd bet at 1 : 1 odds (or better) that the coin came up heads.\n\nOne way to visualize the difference between these approaches is by visualizing what they say about when a model of the world should count as a good model. If a person's model of the world is definite, then it's easy enough to tell whether or not their model is good or bad: We just check what it says against the facts. For example, if a person's model of the world says \"the tree is 3m tall\", then this model is [correct](https://arbital.com/p/correspondence_theory_of_truth) if (and only if) the tree is 3 meters tall.\n\n\n\nDefinite claims in the model are called \"true\" when they correspond to reality, and \"false\" when they don't. If you want to navigate using a map, you had better ensure that the lines drawn on the map correspond to the territory.\n\nBut how do you draw a correspondence between a map and a territory when the map is probabilistic? If your model says that a biased coin has a 70% chance of coming up heads, what's the correspondence between your model and reality? If the coin is actually heads, was the model's claim true? 70% true? What would that mean?\n\n\n\nThe advocate of __propensity__ theory says that it's just a brute fact about the world that the world contains ontologically basic uncertainty. A model which says the coin is 70% likely to land heads is true if and only the actual physical propensity of the coin is 0.7 in favor of heads.\n\n\n\nThis interpretation is useful when the laws of physics _do_ say that there are multiple different observations you may make next (with different likelihoods), as is sometimes the case (e.g., in quantum physics). However, when the event is deterministic — e.g., when it's a coin that has been tossed and slapped down and is already either heads or tails — then this view is largely regarded as foolish, and an example of the [https://arbital.com/p/-4yk](https://arbital.com/p/-4yk): The coin is just a coin, and has no special internal structure (nor special physical status) that makes it _fundamentally_ contain a little 0.7 somewhere inside it. It's already either heads or tails, and while it may _feel_ like the coin is fundamentally uncertain, that's a feature of your brain, not a feature of the coin.\n\nHow, then, should we draw a correspondence between a probabilistic map and a deterministic territory (in which the coin is already definitely either heads or tails?)\n\nA __frequentist__ draws a correspondence between a single probability-statement in the model, and multiple events in reality. If the map says \"that coin over there is 70% likely to be heads\", and the actual territory contains 10 places where 10 maps say something similar, and in 7 of those 10 cases the coin is heads, then a frequentist says that the claim is true.\n\n\n\nThus, the frequentist preserves black-and-white correspondence: The model is either right or wrong, the 70% claim is either true or false. When the map says \"That coin is 30% likely to be tails,\" that (according to a frequentist) means \"look at all the cases similar to this case where my map says the coin is 30% likely to be tails; across all those places in the territory, 3/10ths of them have a tails-coin in them.\" That claim is definitive, given the set of \"similar cases.\"\n\nBy contrast, a __subjectivist__ generalizes the idea of \"correctness\" to allow for shades of gray. They say, \"My uncertainty about the coin is a fact about _me,_ not a fact about the coin; I don't need to point to other 'similar cases' in order to express uncertainty about _this_ case. I know that the world right in front of me is either a heads-world or a tails-world, and I have a [https://arbital.com/p/-probability_distribution](https://arbital.com/p/-probability_distribution) puts 70% probability on heads.\" They then draw a correspondence between their probability distribution and the world in front of them, and declare that the more probability their model assigns to the correct answer, the better their model is.\n\n\n\nIf the world _is_ a heads-world, and the probabilistic map assigned 70% probability to \"heads,\" then the subjectivist calls that map \"70% accurate.\" If, across all cases where their map says something has 70% probability, the territory is actually that way 7/10ths of the time, then the Bayesian calls the map \"[https://arbital.com/p/-well_calibrated](https://arbital.com/p/-well_calibrated)\". They then seek methods to make their maps more accurate, and better calibrated. They don't see a need to interpret probabilistic maps as making definitive claims; they're happy to interpret them as making estimations that can be graded on a sliding scale of accuracy.\n\n## Debate\n\nIn short, the frequentist interpretation tries to find a way to say the model is definitively \"true\" or \"false\" (by identifying a collection of similar events), whereas the subjectivist interpretation extends the notion of \"correctness\" to allow for shades of gray.\n\nFrequentists sometimes object to the subjectivist interpretation, saying that frequentist correspondence is the only type that has any hope of being truly objective. Under Bayesian correspondence, who can say whether the map should say 70% or 75%, given that the probabilistic claim is not objectively true or false either way? They claim that these subjective assessments of \"partial accuracy\" may be intuitively satisfying, but they have no place in science. Scientific reports ought to be restricted to frequentist statements, which are definitively either true or false, in order to increase the objectivity of science.\n\nSubjectivists reply that the frequentist approach is hardly objective, as it depends entirely on the choice of \"similar cases\". In practice, people can (and do!) [abuse frequentist statistics](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging) by choosing the class of similar cases that makes their result look as impressive as possible (a technique known as \"p-hacking\"). Furthermore, the manipulation of subjective probabilities is subject to the [iron laws](https://arbital.com/p/1lz) of probability theory (which are the [only way to avoid inconsistencies and pathologies](https://arbital.com/p/) when managing your uncertainty about the world), so it's not like subjective probabilities are the wild west or something. Also, science has things to say about situations even when there isn't a huge class of objective frequencies we can observe, and science should let us collect and analyze evidence even then.\n\nFor more on this debate, see [https://arbital.com/p/4xx](https://arbital.com/p/4xx).", "date_published": "2016-07-10T10:58:40Z", "authors": ["Eric Rogstad", "Nate Soares"], "summaries": ["Let's say you have a model which says a particular coin is 70% likely to be heads. How should we assess that model?\n\n- According to the [propensity](https://arbital.com/p/propensity) interpretation, the coin has a fundamental intrinsic comes-up-headsness property, and the model is correct if that property is set to 0.7. (This theory is widely considered discredited, and is an example of the [https://arbital.com/p/-4yk](https://arbital.com/p/-4yk)).\n- According to the [frequentist](https://arbital.com/p/frequentist_probability) interpretation, the model is saying that there are a whole bunch of different places where some similar model is saying the same thing as this one (i.e., \"the coin is 70% heads\"), and all models in that reference class are true if, in 70% of those different places, the coin is heads.\n- According to the [subjectivist](https://arbital.com/p/4vr) interpretation, the model is saying that the one dang coin is 70 dang percent likely to be heads, and the model is either 70% accurate (if the coin is in fact heads) or 30% accurate (if it's tails).\n\nIn other words, the propensity and frequency interpretations try to find ways to say that the model is definitively \"true\" or \"false\" (one by postulating that uncertainty is an ontologically basic part of the world, the other by identifying a collection of similar events), whereas the subjective interpretation extends the notion of \"correctness\" to allow for shades of gray."], "tags": [], "alias": "4yj"}
{"id": "73be33176baaa110babdd55f7ee8ee51", "title": "Probability interpretations: Examples", "url": "https://arbital.com/p/probability_interpretations_examples", "source": "arbital", "source_type": "text", "text": "## Betting on one-time events\n\nConsider evaluating, in June of 2016, the question: \"What is the probability of Hillary Clinton winning the 2016 US presidential election?\"\n\nOn the **propensity** view, Hillary has some fundamental chance of winning the election. To ask about the probability is to ask about this objective chance. If we see a prediction market in which prices move after each new poll — so that it says 60% one day, and 80% a week later — then clearly the prediction market isn't giving us very strong information about this objective chance, since it doesn't seem very likely that Clinton's *real* chance of winning is swinging so rapidly.\n\nOn the **frequentist** view, we cannot formally or rigorously say anything about the 2016 presidential election, because it only happens once. We can't *observe* a frequency with which Clinton wins presidential elections. A frequentist might concede that they would cheerfully buy for \\$1 a ticket that pays \\$20 if Clinton wins, considering this a favorable bet in an *informal* sense, while insisting that this sort of reasoning isn't sufficiently rigorous, and therefore isn't suitable for being included in science journals.\n\nOn the **subjective** view, saying that Hillary has an 80% chance of winning the election summarizes our *knowledge about* the election or our *state of uncertainty* given what we currently know. It makes sense for the prediction market prices to change in response to new polls, because our current state of knowledge is changing.\n\n## A coin with an unknown bias\n\nSuppose we have a coin, weighted so that it lands heads somewhere between 0% and 100% of the time, but we don't know the coin's actual bias.\n\nThe coin is then flipped three times where we can see it. It comes up heads twice, and tails once: HHT.\n\nThe coin is then flipped again, where nobody can see it yet. An honest and trustworthy experimenter lets you spin a wheel-of-gambling-odds,%note:The reason for spinning the wheel-of-gambling-odds is to reduce the worry that the experimenter might know more about the coin than you, and be offering you a deliberately rigged bet.% and the wheel lands on (2 : 1). The experimenter asks if you'd enter into a gamble where you win \\$2 if the unseen coin flip is tails, and pay \\$1 if the unseen coin flip is heads.\n\nOn a **propensity** view, the coin has some objective probability between 0 and 1 of being heads, but we just don't know what this probability is. Seeing HHT tells us that the coin isn't all-heads or all-tails, but we're still just guessing — we don't really know the answer, and can't say whether the bet is a fair bet.\n\nOn a **frequentist** view, the coin would (if flipped repeatedly) produce some long-run frequency $f$ of heads that is between 0 and 1. If we kept flipping the coin long enough, the actual proportion $p$ of observed heads is guaranteed to approach $f$ arbitrarily closely, eventually. We can't say that the *next* coin flip is guaranteed to be H or T, but we can make an objectively true statement that $p$ will approach $f$ to within epsilon if we continue to flip the coin long enough.\n\nTo decide whether or not to take the bet, a frequentist might try to apply an [unbiased estimator](https://arbital.com/p/unbiased_estimator) to the data we have so far. An \"unbiased estimator\" is a rule for taking an observation and producing an estimate $e$ of $f$, such that the [expected value](https://arbital.com/p/4b5) of $e$ is $f$. In other words, a frequentist wants a rule such that, if the hidden bias of the coin was in fact to yield 75% heads, and we repeat many times the operation of flipping the coin a few times and then asking a new frequentist to estimate the coin's bias using this rule, the *average* value of the estimated bias will be 0.75. This is a property of the _estimation rule_ which is objective. We can't hope for a rule that will always, in any particular case, yield the true $f$ from just a few coin flips; but we can have a rule which will provably have an *average* estimate of $f$, if the experiment is repeated many times.\n\nIn this case, a simple unbiased estimator is to guess that the coin's bias $f$ is equal to the observed proportion of heads, or 2/3. In other words, if we repeat this experiment many many times, and whenever we see $p$ heads in 3 tosses we guess that the coin's bias is $\\frac{p}{3}$, then this rule definitely is an unbiased estimator. This estimator says that a bet of \\$2 vs. $\\1 is fair, meaning that it doesn't yield an expected profit, so we have no reason to take the bet.\n\nOn a **subjectivist** view, we start out personally unsure of where the bias $f$ lies within the interval [1](https://arbital.com/p/0,). Unless we have any knowledge or suspicion leading us to think otherwise, the coin is just as likely to have a bias between 33% and 34%, as to have a bias between 66% and 67%; there's no reason to think it's more likely to be in one range or the other.\n\nEach coin flip we see is then [evidence](https://arbital.com/p/22x) about the value of $f,$ since a flip H happens with different probabilities depending on the different values of $f,$ and we update our beliefs about $f$ using [Bayes' rule](https://arbital.com/p/1zj). For example, H is twice as likely if $f=\\frac{2}{3}$ than if $f=\\frac{1}{3}$ so by [Bayes's Rule](https://arbital.com/p/1zm) we should now think $f$ is twice as likely to lie near $\\frac{2}{3}$ as it is to lie near $\\frac{1}{3}$.\n\nWhen we start with a uniform [prior](https://arbital.com/p/219), observe multiple flips of a coin with an unknown bias, see M heads and N tails, and then try to estimate the odds of the next flip coming up heads, the result is [Laplace's Rule of Succession](https://arbital.com/p/21c) which estimates (M + 1) : (N + 1) for a probability of $\\frac{M + 1}{M + N + 2}.$\n\nIn this case, after observing HHT, we estimate odds of 2 : 3 for tails vs. heads on the next flip. This makes a gamble that wins \\$2 on tails and loses \\$1 on heads a profitable gamble in expectation, so we take the bet.\n\nOur choice of a [uniform prior](https://arbital.com/p/219) over $f$ was a little dubious — it's the obvious way to express total ignorance about the bias of the coin, but obviousness isn't everything. (For example, maybe we actually believe that a fair coin is more likely than a coin biased 50.0000023% towards heads.) However, all the reasoning after the choice of prior was rigorous according to the laws of [probability theory](https://arbital.com/p/1bv), which is the [only method of manipulating quantified uncertainty](https://arbital.com/p/probability_coherence_theorems) that obeys obvious-seeming rules about how subjective uncertainty should behave.\n\n## Probability that the 98,765th decimal digit of $\\pi$ is $0$.\n\nWhat is the probability that the 98,765th digit in the decimal expansion of $\\pi$ is $0$?\n\nThe **propensity** and **frequentist** views regard as nonsense the notion that we could talk about the *probability* of a mathematical fact. Either the 98,765th decimal digit of $\\pi$ is $0$ or it's not. If we're running *repeated* experiments with a random number generator, and looking at different digits of $\\pi,$ then it might make sense to say that the random number generator has a 10% probability of picking numbers whose corresponding decimal digit of $\\pi$ is $0$. But if we're just picking a non-random number like 98,765, there's no sense in which we could say that the 98,765th digit of $\\pi$ has a 10% propensity to be $0$, or that this digit is $0$ with 10% frequency in the long run.\n\nThe **subjectivist** considers probabilities to just refer to their own uncertainty. So if a subjectivist has picked the number 98,765 without yet knowing the corresponding digit of $\\pi,$ and hasn't made any observation that is known to them to be entangled with the 98,765th digit of $\\pi,$ and they're pretty sure their friend hasn't yet looked up the 98,765th digit of $\\pi$ either, and their friend offers a whimsical gamble that costs \\$1 if the digit is non-zero and pays \\$20 if the digit is zero, the Bayesian takes the bet.\n\nNote that this demonstrates a difference between the subjectivist interpretation of \"probability\" and Bayesian probability theory. A perfect Bayesian reasoner that knows the rules of logic and the definition of $\\pi$ must, by the axioms of probability theory, assign probability either 0 or 1 to the claim \"the 98,765th digit of $\\pi$ is a $0$\" (depending on whether or not it is). This is one of the reasons why [perfect Bayesian reasoning is intractable](https://arbital.com/p/bayes_intractable). A subjectivist that is not a perfect Bayesian nevertheless claims that they are personally uncertain about the value of the 98,765th digit of $\\pi.$ Formalizing the rules of subjective probabilities about mathematical facts (in the way that [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv) formalized the rules for manipulating subjective probabilities about empirical facts, such as which way a coin came up) is an open problem; this in known as the problem of [https://arbital.com/p/-logical_uncertainty](https://arbital.com/p/-logical_uncertainty).", "date_published": "2016-06-30T22:13:46Z", "authors": ["Nate Soares", "Alexei Andreev"], "summaries": ["Consider evaluating, in June of 2016, the question: \"What is the probability of Hillary Clinton winning the 2016 US presidential election?\"\n\n- On the **propensity** view, Hillary has some fundamental chance of winning the election. To ask about the probability is to ask about this objective chance.\n- On the **subjective** view, saying that Hillary has an 80% chance of winning the election summarizes our *knowledge about* the election, or, equivalently, our *state of uncertainty* given what we currently know.\n- On the **frequentist** view, we cannot formally or rigorously say anything about the 2016 presidential election, because it only happens once."], "tags": [], "alias": "4yn"}
{"id": "4de5865ee6fb1c45542105fee7e63ac6", "title": "Cauchy's theorem on subgroup existence: intuitive version", "url": "https://arbital.com/p/cauchy_theorem_on_subgroup_existence_intuitive", "source": "arbital", "source_type": "text", "text": "Cauchy's Theorem states that if $G$ is a finite [https://arbital.com/p/-group](https://arbital.com/p/-group), and $p$ is a [prime](https://arbital.com/p/4mf) dividing the [order](https://arbital.com/p/3gg) of $G$, then $G$ has an element of order $p$.\nEquivalently, it has a subgroup of order $p$.\n\nThis theorem is very important as a partial converse to [Lagrange's theorem](https://arbital.com/p/4jn): while Lagrange's theorem gives us *restrictions* on what subgroups can exist, Cauchy's theorem tells us that certain subgroups *definitely do* exist.\nIn this direction, Cauchy's theorem is generalised by the [Sylow theorems](https://arbital.com/p/sylow_theorems_on_subgroup_existence).\n\n# Proof\n\nLet $G$ be a group, and write $e$ for its identity.\nWe will try and find an element of order $p$.\n\nWhat does it mean for an element $x \\not = e$ to have order $p$? Nothing more nor less than that $x^p = e$.\n(This is true because $p$ is prime; if $p$ were not prime, we would additionally have to stipulate that $x^i$ is not $e$ for any smaller $i < p$.)\n\nNow follows the magic step which casts the problem in the \"correct\" light.\n\nConsider all possible combinations of $p$ elements from the group.\nFor example, if $p=5$ and the group elements are $\\{ a, b, c, d, e\\}$ with $e$ the identity, then $(e, e, a, b, a)$ and $(e,a,b,a,e)$ are two different such combinations.\nThen $x \\not = e$ has order $p$ if and only if the combination $(x, x, \\dots, x)$ multiplies out to give $e$.\n\nThe rest of the proof will be simply following what this magic step has told us to do.\n\nSince we want to find some $x$ where $(x, x, \\dots, x)$ multiplies to give $e$, it makes sense only to consider the combinations which multiply out to give $e$ in the first place.\nSo, for instance, we will exclude the tuple $(e, e, a, b, a)$ from consideration if it is the case that $eeaba$ is the identity (equivalently, that $aba = e$).\nFor convenience, we will label the set of all the allowed combinations: let us call it $X$.\n\nOf the tuples in $X$, we are looking for one with a very specific property: every entry is equal.\nFor example, the tuple $(a,b,c,b,b)$ is no use to us, because it doesn't tell us an element $x$ such that $x^p = e$; it only tells us that $abcbb = e$.\nSo we will be done if we can find a tuple in $X$ which has every element equal (and which is not the identity).\n\nWe don't really have anything to go on here, but it will surely help to know the size of $X$: it is $|G|^{p-1}$, since every element of $X$ is determined exactly by its first $p-1$ places (after which the last is fixed).\nThere are no restrictions on the first $p-1$ places, though: we can find a $p$th element to complete any collection of $p-1$.\n\n%%hidden(Example):\nIf $p=5$, and we have the first four elements $(a, a, b, e, \\cdot)$, then we know the fifth element *must* be $b^{-1} a^{-2}$.\nIndeed, $aabe(a^{-1} a^{-2}) = e$, and [inverse elements of a group are unique](https://arbital.com/p/), so nothing else can fill the last slot.\n%%\n\nSo we have $X$, of size $|G|^{p-1}$; we have that $p$ divides $|G|$ (and so it divides $|X|$); and we also have an element $(e,e,\\dots,e)$ of $X$.\nBut notice that if $(a_1, a_2, \\dots, a_p)$ is in $X$, then so is $(a_2, a_3, \\dots, a_p, a_1)$; and so on through all the rotations of an element.\nWe can actually group up the elements of $X$ into buckets, where two tuples are in the same bucket if (and only if) one can be rotated to make the other.\n\nHow big is a bucket? If a bucket contains something of the form $(a, a, \\dots, a)$, then of course that tuple can only be rotated to give one thing, so the bucket is of size $1$.\nBut if the bucket contains any tuple $T$ which is not \"the same element repeated $p$ times\", then there must be exactly $p$ things in the bucket: namely the $p$ things we get by rotating $T$. (Exercise: verify this!)\n\n%%hidden(Show solution):\nThe bucket certainly does contain every rotation of $T$: we defined the bucket such that if a tuple is in the bucket, then so are all its rotations.\nMoreover, everything in the bucket is a rotation of $T$, because two tuples are in the bucket if and only if we can rotate one to get the other; equivalently, $A$ is in the bucket if and only if we can rotate it to get $T$.\n\nHow many such rotations are there? We claimed that there were $p$ of them.\nIndeed, the rotations are precisely $$(a_1, a_2, \\dots, a_p), (a_2, a_3, \\dots, a_p, a_1), \\dots, (a_{p-1}, a_p, a_1, \\dots, a_{p-2}), (a_p, a_1, a_2, \\dots, a_{p-1})$$\nand there are $p$ of those; are they all distinct?\nYes, they are, and this is because $p$ is prime (it fails when $p=8$, for instance, because $(1,1,2,2,1,1,2,2)$ can be rotated four places to give itself).\nIndeed, if we could rotate the tuple $T$ nontrivially (that is, strictly between $1$ and $p$ times) to get itself, then the integer $n$, the \"number of places we rotated\", must divide the integer \"length of the tuple\". \nBut that tells us that $n$ divides the prime $p$, so $n$ is $1$ or $p$ itself, and so either the tuple is actually \"the same element repeated $p$ times\" (in the case $n=1$) %%note: But we're only considering $T$ which is not of this form!%%, or the rotation was the \"stupid\" rotation by $p$ places (in the case $n=p$).\n%%\n\nOK, so our buckets are of size $p$ exactly, or size $1$ exactly.\nWe're dividing up the members of $X$ - that is, $|G|^{p-1}$ things - into buckets of these size; and we already have one bucket of size $1$, namely $(e,e,\\dots,e)$.\nBut if there were no more buckets of size $1$, then we would have $p$ dividing $|G|^{p-1}$ and also dividing the size of all but one of the buckets.\nMixing together all the other buckets to obtain an uber-bucket of size $|G|^{p-1} - 1$, the total must be divisible by $p$, since each individual bucket was %%note:Compare with the fact that the sum of even numbers is always even; that is the case that $p=2$.%%; so $|G|^{p-1} - 1$ is divisible by $p$.\nBut that is a contradiction, since $|G|^{p-1}$ is also divisible by $p$.\n\nSo there is another bucket of size $1$.\nA bucket of size $1$ contains something of the form $(a,a,\\dots,a)$: that is, it has shown us an element of order $p$.", "date_published": "2016-06-30T13:51:06Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Needs clickbait"], "alias": "4ys"}
{"id": "4063cdc24631b2a69234f2db69c87e45", "title": "Report likelihoods, not p-values", "url": "https://arbital.com/p/likelihoods_not_pvalues", "source": "arbital", "source_type": "text", "text": "This page advocates for a change in the way that statistics is done in standard scientific journals. The key idea is to report [likelihood functions](https://arbital.com/p/56s) instead of p-values, and this could have many benefits.\n\n_(Note: This page is a personal [opinion page](https://arbital.com/p/60p).)_\n\n[https://arbital.com/p/toc:](https://arbital.com/p/toc:)\n\n## What's the difference?\n\nThe status quo across scientific journals is to test data for \"[statistical significance](https://arbital.com/p/statistically_significant)\" using functions such as [p-values](https://arbital.com/p/p_value). A p-value is a number calculated from a hypothesis (called the \"null hypothesis\"), an experiment, a result, and a [https://arbital.com/p/-summary_statistic](https://arbital.com/p/-summary_statistic). For example, if the null hypothesis is \"this coin is fair,\" and the experiment is \"flip it 6 times\", and the result is HHHHHT, and the summary statistic is \"the sequence has at least five H values,\" then the p-value is 0.11, which means \"if the coin were fair, and we did this experiment a lot, then only 11% of the sequences generated would have at least five H values.\"%%note:This does _not_ mean that the coin is 89% likely to be biased! For example, if the only alternative is that the coin is biased towards tails, then HHHHHT is evidence that it's fair. This is a common source of confusion with p-values.%% If the p-value is lower than an arbitrary threshold (usually $p < 0.05$) then the result is called \"statistically significant\" and the null hypothesis is \"rejected.\"\n\nThis page advocates that scientific articles should report _likelihood functions_ instead of p-values. A likelihood function for a piece of evidence $e$ is a function $\\mathcal L$ which says, for each hypothesis $H$ in some set of hypotheses, the probability that $H$ assigned to $e$, written [$\\mathcal L_e](https://arbital.com/p/51n).%%note: Many authors write $\\mathcal L(H \\mid e)$ instead. We think this is confusing, as then $\\mathcal L(H \\mid e) = \\mathbb P(e \\mid H),$ and it's hard enough for students of statistics to keep \"probability of $H$ given $e$\" and \"probability of $e$ given $H$\" straight as it is if the notation _isn't_ swapped around every so often.%% For example, if $e$ is \"this coin, flipped 6 times, generated HHHHHT\", and the set of hypotheses are $H_{0.25} =$ \"the coin only produces heads 25% of the time\" and $H_{0.5}$ = \"the coin is fair\", then $\\mathcal L_e(H_{0.25})$ $=$ $0.25^5 \\cdot 0.75$ $\\approx 0.07\\%$ and $\\mathcal L_e(H_{0.5})$ $=$ $0.5^6$ $\\approx 1.56\\%,$ for a [likelihood ratio](https://arbital.com/p/1rq) of about $21 : 1$ in favor of the coin being fair (as opposed to biased 75% towards tails).\n\nIn fact, with a single likelihood function, we can report the amount of support $e$ gives to _every_ hypothesis $H_b$ of the form \"the coin has bias $b$ towards heads\":%note:To learn how this graph was generated, see [Bayes' rule: Functional form](https://arbital.com/p/1zj).%\n\n\n\nNote that this likelihood function is _not_ telling us the probability that the coin is actually biased, it is _only_ telling us how much the evidence supports each hypothesis. For example, this graph says that HHHHHT provides about 3.8 times as much evidence for $H_{0.75}$ over $H_{0.5}$, and about 81 times as much evidence for $H_{0.75}$ over $H_{0.25}.$\n\nNote also that the likelihood function doesn't necessarily contain the _right_ hypothesis; for example, the function above shows the support of $e$ for every possible bias on the coin, but it doesn't consider hypotheses like \"the coin alternates between H and T\". Likelihood functions, like p-values, are essentially a mere summary of the raw data — there is no substitute for the raw data when it comes to allowing people to test hypotheses that the original researchers did not consider. (In other words, even if you report likelihoods instead of p-values, it's still virtuous to share your raw data.)\n\nWhere p-values let you measure (roughly) how well the data supports a _single_ \"null hypothesis\", with an arbitrary 0.05 \"not well enough\" cutoff, the likelihood function shows the support of the evidence for lots and lots of different hypotheses at once, without any need for an arbitrary cutoff.\n\n## Why report likelihoods instead of p-values?\n\n__1. Likelihood functions are less arbitrary than p-values.__ To report a likelihood function, all you have to do is pick which hypothesis class to generate the likelihood function for. That's your only degree of freedom. This introduces one source of arbitrariness, and if someone wants to check some other hypothesis they still need access to the raw data, but it is better than the p-value case, where you only report a number for a single \"null\" hypothesis.\n\nFurthermore, in the p-value case, you have to pick not only a null hypothesis but also an experiment and a summary statistic, and these degrees of freedom can have a huge impact on the final report. These extra degrees of freedom are both unnecessary ([to carry out a probabilistic update, all you need are your own personal beliefs and a likelihood function](https://arbital.com/p/1lz)) and exploitable, and empirically, they're actively harming scientific research.\n\n__2. Reporting likelihoods would solve p-hacking.__ If you're using p-values, then you can game the statistics via your choice of experiment and summary statistics. In the example with the coin above, if you say your experiment and summary statistic are \"flip the coin 6 times and count the number of heads\" then the p-value of HHHHHT with respect to $H_{0.5}$ is 0.11, whereas if you say your experiment and summary statistic are \"flip the coin until it comes up tails and count the number of heads\" then the p-value of HHHHHT with respect to $H_{0.5}$ is 0.03, which is \"significant.\" This is called \"[p-hacking](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging)\", and it's a serious problem in modern science.\n\nIn a likelihood function, _the amount of support an evidence gives to a hypothesis does not depend on which experiment the researcher had in mind._ Likelihood functions depend only on the data you actually saw, and the hypotheses you chose to report. The only way to cheat a likelihood function is to lie about the data you collected, or refuse to report likelihoods for a particular hypothesis.\n\nIf your paper fails to report likelihoods for some obvious hypotheses, then (a) that's precisely analogous to you choosing the wrong null hypothesis to consider; (b) it's just as easily noticeable as when your paper considers the wrong null hypothesis; and (c) it can be easily rectified given access to the raw data. By contrast, p-hacking can be subtle and hard to detect after the fact.\n\n__3. Likelihood functions are very difficult to game.__ There is no analog of p-hacking for likelihood functions. This is a theorem of probability theory known as [https://arbital.com/p/-conservation_of_expected_evidence](https://arbital.com/p/-conservation_of_expected_evidence), which says that likelihood functions can't be gamed unless you're falsifying or omitting data (or screwing up the likelihood calculations).%note:Disclaimer: the theorem says likelihood functions can't be gamed, but we still shouldn't underestimate the guile of dishonest researchers struggling to make their results look important. Likelihood functions have not been put through the gauntlet of real scientific practice; p-values have. That said, when p-values were put through that gauntlet, they [failed](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging) in a [spectacular fashion](https://arbital.com/p/https://en.wikipedia.org/wiki/Replication_crisis). When rebuilding, it's probably better to start from foundations that provably cannot be gamed.%\n\n\n__4. Likelihood functions would help stop the \"vanishing effect sizes\" phenomenon.__ The [decline effect](https://arbital.com/p/https://en.wikipedia.org/wiki/Decline_effect) occurs when studies which reject a null hypothesis $H_0$ have effect sizes that get smaller and smaller and smaller over time (the more someone tries to replicate the result). This is [usually evidence](https://arbital.com/p/http://slatestarcodex.com/2014/04/28/the-control-group-is-out-of-control/) that there is no actual effect, and that the initial \"large effects\" were a result of publication bias.\n\nLikelihood functions help avoid the decline effect by _treating different effect sizes differently._ The likelihood function for coins of different biases shows that the evidence HHHHHT gives a different amount of support to $H_{0.52},$ $H_{0.61}$, and $H_{0.8}$ (which correspond to small, medium, and large effect sizes, respectively). If three different studies find low support for $H_{0.5},$ and one of them gives all of its support to the large effect, another gives all its support to the medium effect, and the third gives all of its support to the smallest effect, then likelihood functions reveal that something fishy is going on (because they're all peaked in different places).\n\nIf instead we only use p-values, and always decide whether or not to \"keep\" or \"reject\" the null hypothesis (without specifying how much support goes to different alternatives), then it's hard to notice that the studies are actually contradictory (and that something very fishy is going on). Instead, it's very tempting to exclaim \"3 out of 3 studies reject $H_{0.5}$!\" and move on.\n\n__5. Likelihood functions would help stop publication bias.__ When using p-values, if the data yields a p-value of 0.11 using a null hypothesis $H_0$, the study is considered \"insignificant,\" and many journals have a strong bias towards positive results. When reporting likelihood functions, there is no arbitrary \"significance\" threshold. A study that reports a relative likelihoods of $21 : 1$ in favor of $H_a$ vs $H_0,$ that's exactly the same _strength of evidence_ as a study that reports $21 : 1$ odds against $H_a$ vs $H_0.$ It's all just evidence, and it can all be added to the corpus, there's no arbitrary \"significance\" threshold.\n\n__6. Likelihood functions make it trivially easy to combine studies.__ When combining studies that used p-values, researchers have to perform complex meta-analyses with dozens of parameters to tune, and they often find [exactly what they were expecting to find](https://arbital.com/p/https://en.wikipedia.org/wiki/Confirmation_bias). By contrast, the way you combine multiple studies that reported likelihood functions is... (drumroll)\n...you just multiply the likelihood functions together. If study A reports that $H_{0.75}$ was favored over $H_{0.5}$ with a [https://arbital.com/p/-1rq](https://arbital.com/p/-1rq) of $3.8 : 1$, and study B reports that $H_{0.75}$ was favored over $H_{0.5}$ at $5 : 1$, then the combined likelihood functions of both studies favors $H_{0.75}$ over $H_{0.5}$ at $(3.8 \\cdot 5) : 1$ $=$ $19 : 1.$\n\nWant to combine a hundred studies on the same subject? Multiply a hundred functions together. Done. No parameter tuning, no degrees of freedom through which bias can be introduced — just multiply.\n\n__7. Likelihood functions make it obvious when something has gone wrong.__ If, when you multiply all the likelihood functions together, _all_ hypotheses have extraordinarily low likelihoods, then something has gone wrong. Either a mistake has been made somewhere, or fraud has been committed, or the true hypothesis wasn't in the hypothesis class you're considering.\n\nThe actual hypothesis that explains all the data will have decently high likelihood across all the data. If none of the hypotheses fit that description, then either you aren't considering the right hypothesis yet, or some of the studies went wrong. (Try looking for one study that has a likelihood function _very very different_ from all the other studies, and investigate that one.)\n\nLikelihood functions won't do your science for you — you still have to generate good hypotheses, and be honest in your data reporting — but they _do_ make it obvious when something went wrong. (Specifically, [each hypothesis can tell you how low its likelihood is expected to be on the data](https://arbital.com/p/227), and if _every_ hypothesis has a likelihood far lower than expected, then something's fishy.)\n\n---\n\nA scientific community using likelihood functions would produce scientific research that's easier to use. If everyone's reporting likelihood functions, then all you personally need to do in order to figure out what to believe is take your own personal (subjective) prior probabilities and multiply them by all the likelihood functions in order to get your own personal (subjective) posterior probabilities.\n\nFor example, let's say you personally think the coin is probably fair, with $10 : 1$ odds of being fair as opposed to 75% biased in favor of heads. Now let's say that study A reports a likelihood function which favors $H_{0.75}$ over $H_{0.5}$ with a likelihood ratio of $3.8 : 1.$, and study B reports a $5 : 1$ likelihood ratio in the same direction. Multiplying all these together, your personal posterior beliefs should be $19 : 10$ in favor of $H_{0.75}$ over $H_{0.5}$. This is simply [Bayes' rule](https://arbital.com/p/1lz). Reporting likelihoods instead of p-values lets science remain objective, while allowing everyone to find their own personal posterior probabilities via a simple application of Bayes' theorem.\n\n## Why should we think this would work?\n\nThis may all sound too good to be true. Can one simple change really solve that many problems in modern science?\n\nFirst of all, you can be assured that reporting likelihoods instead of p-values would not \"solve\" all the problems above, and it would surely not solve all problems with modern experimental science. Open access to raw data, preregistration of studies, a culture that rewards replication, and many other ideas are also crucial ingredients to a scientific community that zeroes in on truth.\n\nHowever, reporting likelihoods would _help_ solve lots of different problems in modern experimental science. This may come as a surprise. Aren't likelihood functions just one more statistical technique, just another tool for the toolbox? Why should we think that one single tool can solve that many problems?\n\nThe reason lies in [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv). According to the axioms of probability theory, there is only one good way to account for evidence when updating your beliefs, and that way is via likelihood functions. Any other method is subject to inconsistencies and pathologies, as per the [coherence theorems of probability theory](https://arbital.com/p/probability_coherence_theorems).\n\nIf you're manipulating equations like $2 + 2 = 4,$ and you're using methods that may or may not let you throw in an extra 3 on the right hand side (depending on the arithmetician's state of mind), then it's no surprise that you'll occasionally get yourself into trouble and deduce that $2 + 2 = 7.$ The laws of arithmetic show that there is only one correct set of tools for manipulating equations if you want to avoid inconsistency.\n\nSimilarly, the laws of probability theory show that there is only one correct set of tools for manipulating _uncertainty_ if you want to avoid inconsistency. According to [those rules](https://arbital.com/p/1lz), the right way to represent evidence is through likelihood functions.\n\nThese laws (and a solid understanding of them) are younger than the experimental science community, and the statistical tools of that community predate a modern understanding of probability theory. Thus, it makes a lot of sense that the existing literature uses different tools. However, now that humanity _does_ possess a solid understanding of probability theory, it should come as no surprise that many diverse pathologies in statistics can be cleaned up by switching to a policy of reporting likelihoods instead of p-values.\n\n## What are the drawbacks?\n\nThe main drawback is inertia. Experimental science today reports p-values almost entirely across the board. Modern statistical toolsets have built-in support for p-values (and other related statistical tools) but very little support for reporting likelihood functions. Experimental scientists are trained mainly in [https://arbital.com/p/-frequentist_statistics](https://arbital.com/p/-frequentist_statistics), and thus most are much more familiar with p-value-type tools than likelihood-function-type tools. Making the switch would be painful.\n\nBarring the switching costs, though, making the switch could well be a strict improvement over modern techniques, and would help solve some of the biggest [problems](https://arbital.com/p/http://slatestarcodex.com/2014/04/28/the-control-group-is-out-of-control/) [facing](https://arbital.com/p/http://www.stat.columbia.edu/~gelman/research/published/asa_pvalues.pdf) [science](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging) [today](https://arbital.com/p/https://en.wikipedia.org/wiki/Publication_bias).\n\nSee also the [Likelihoods not p-values FAQ](https://arbital.com/p/505) and [https://arbital.com/p/4xx](https://arbital.com/p/4xx).", "date_published": "2017-04-29T03:00:59Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Nate Soares", "austin stone"], "summaries": ["If scientists reported likelihood functions instead of p-values, this could help science avoid [p-hacking](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging), [publication bias](https://arbital.com/p/https://en.wikipedia.org/wiki/Publication_bias), [the decline effect](https://arbital.com/p/https://en.wikipedia.org/wiki/Decline_effect), and other hazards of standard statistical techniques. Furthermore, it could help make it easier to combine results from multiple studies and perfrom meta-analyses, while making statistics intuitively easier to understand. (This is a bold claim, but a claim which is largely supported by [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv).)"], "tags": ["Opinion page"], "alias": "4zd"}
{"id": "23f6e190921849c6fbe7a1507ea0f677", "title": "Cyclic Group Intro (Math 0)", "url": "https://arbital.com/p/cyclic_group_intro_math_0", "source": "arbital", "source_type": "text", "text": "##Clocks##\nSay we had a clock. We decide that we can \"add\" two numbers on the clock starting at the first number, then moving by the number of steps given by the second number. For example the sum of $5$ and $6$ is given by starting at $5$, and then moving by $6$ spaces, and ending up at $11$. But the sum of $7$ and $9$ is given by starting at $7$ and moving $9$ spaces to end up at $4$. This number can be calculated by just adding the numbers normally, then if you end up at more than $12$, subtract $12$. For example, $7+9 = 16$, and $16- 12 = 4$.\n\nNotice that if you add $12$ to anything, you just go once around the clock, and so ending up not doing anything. You can calculate it with $4$ as $4 + 12 = 16$, and $16 - 12 = 4$. Hence $12$ is the [https://arbital.com/p/-54p](https://arbital.com/p/-54p) for our clock. Hmm, usually the identity should be something like $0$. More on this in a moment.\n\nAlso, notice that no matter where you are on the clock, there's always an amount by which you can turn to end up at $12$. For example, if you are at $5$, you can add $7$ to get to $12$. This is called the [inverse](https://arbital.com/p/-inverse_mathematics). This number can easily be calculated by subtracting your number from $12$. For example, $12 - 5 = 7$ so the inverse of $5$ is $7$. Now what if we calculate the inverse of $12$? Then we get $0$. In fact, $12$ and $0$ are the same thing on our clock. Actually, we may as well scratch the $12$ off of our clock-face and just write $0$. It makes things a lot easier. Now, instead of considering $12$ the identity, we can consider $0$ the identity instead. If we are at any number, and we step $0$ times (in other words, we don't step at all), then we stay at the number we were.\n\nNotice now that instead of saying when we add $5$ and $7$ on our clock we get $12$, we actually want to say we get $0$. So now we change our rule at the top a little bit. To calculate the sum of two numbers, we add them, and if we get $12$ *or more* we subtract $12$. So now $4$ plus $2$ is still $6$, and $7$ plus $9$ is still obtained by calculating $7+9 = 16$ and $16-12 = 4$. But now, instead of calculating that $7 +5 = 12$ we now further calculate that $12 - 12 = 0$ and so say that $7$ plus $5$ is $0$ on our clock.\n\nOur clock is [closed](https://arbital.com/p/-3gy) under its addition, it has an [https://arbital.com/p/-54p](https://arbital.com/p/-54p) and [inverses](https://arbital.com/p/-inverse_mathematics). Hence, our clock is a [group](https://arbital.com/p/-3gd). From here on out we can write our group [operation](https://arbital.com/p/-operation_mathematics) as $\\bullet$. So $7 \\bullet 9 = 4$.\n\nNow, there is no reason our clock needs to have $12$ numbers. We could do exactly the same with, say, $15$ numbers. Then our operation is defined by adding the numbers, and if they add up to at least $15$, we subtract $15$. Now $5 \\bullet 7 = 12$ but $7 \\bullet 9 = 1$ since $7 + 9 = 16$ and $16 - 15 = 1$. Notice that now the inverse of $5$ is not $7$, but is instead found by calculating $15 - 5 = 10$. So the inverse of $5$ is $10$ (since $5 + 10 = 15$, so $5 \\bullet 10 = 0$). \n\nWe can write our inverses by adding a negative sign to the front of the number. So on our $15$ number clock, the inverse of $5$ is written $-5$ and so $-5 = 10$, but on our normal clock, the inverse of $5$ is $7$ so $-5 = 7$.\n\n##Cyclic Groups##\n\nThis motivates the term **cyclic**; if we go far enough, we get back to where we started. In fact, a group is called cyclic if it has a singe [generator](https://arbital.com/p/-generator_mathematics). This is an element that can be used to generate all the other elements by being added to itself enough times. On any clock, the generator is $1$. We can get any number by going $1 \\bullet 1 \\bullet 1 \\bullet \\cdots \\bullet 1$ the right number of times. \n\nTechnically, for mathematical reasons, we actually want to require that we can get every element even if we combine the generator with itself a *negative* number of times. What does this mean? Well, the generator $1$ has an inverse $-1$. In our normal clock $-1 = 11$. In our $15$ number clock $-1 = 14$. The point is that we could just as well get all the numbers by adding $-1$ to itself a bunch of times.\n\nWe can also get $0$ by \"not adding $1$ to itself at all\" or, put another way \"adding $1$ to itself $0$ times\". This may seem strange, but if you think of taking $1$, and then adding $-1$ once, it's like not taking $1$ at all and you get $0$. Don't worry too much about this. The point is that you can technically get $0$ from $1$.\n\nIn fact, the [finite groups](https://arbital.com/p/-finite_group) that you can get from adding one number to itself a bunch of times are all [isomorphic](https://arbital.com/p/-4f4) to some clock. That is to say, no matter what group you took that has this property, if you just renamed the elements, you would end up with something that looked like a clock in the way described above.\n\nTake for example a coin. It has two sides: heads and tails. Call them $h$ and $t$ for short. We can define an operation on the sides as follows, which we're going to refer to as \"adding\" sides: if you add heads to anything, you flip the coin. If you add tails to anything, you leave the coin alone. So, if you are on the heads side, and you \"add heads\", you flip the coin to tails. If you are on the heads side and you \"add tails\" you leave the coin on the heads side. If you are on the tails side and you \"add heads\", you flip the coin to heads, and if you are on the tails side and you \"add tails\" you just stay where you are. We can write the operation of 'adding\" the sides as \"$\\bullet$\" and then say $h \\bullet h = t$, and $h \\bullet t = h$, and $t \\bullet h = h$, and $t \\bullet t = t$. But hey! This is exactly the same we'd get if we called the head side $1$ and the tails side $0$, and considered a clock with just two numbers. Then $1 \\bullet 1 = 0$, and $1 \\bullet 0 = 1$, and $0 \\bullet 1 = 1$ and $0 \\bullet 0 = 0$. If you just had heads, you could get tails by just adding heads to itself, so the coin group is cyclic.\n\nFor any finite group where you can get every element starting from a single one, you can write it as a clock by relabeling the elements. Try it!\n\n##Infinite Cyclic Groups##\nHowever, things change if we allow our group to be [infinite](https://arbital.com/p/-infinity). Then we still require that every element can be obtained by adding $1$ to itself some amount of times, either positive or negative or zero. But maybe there is no way to wrap around. Consider all the [integers](https://arbital.com/p/-48l), that is to say, every number, positive, negative and zero. These actually do form a group. Then every number you can think of, you can get to by adding $1$ a bunch of times. But no matter how many $1$s you add on the positive side, you can never wrap around and get a negative number. However, the integers are still called a cyclic group, even though it's not really cyclic (I know, not a great name) because the official definition is that every element can be obtained by adding a specific something a bunch of times. \n\nAgain, the only infinite cyclic group that you get, up to [relabeling](https://arbital.com/p/-4f4), is the integers. \n\nSo the only cyclic groups are the clocks and the integers!", "date_published": "2016-08-01T08:20:02Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Mark Chimes"], "summaries": ["Every [https://arbital.com/p/-finite](https://arbital.com/p/-finite) **cyclic [group](https://arbital.com/p/-3gd)** can be thought of like a clock (except maybe with more or less numbers than twelve). Every [https://arbital.com/p/-infinite](https://arbital.com/p/-infinite) cyclic group can be thought of as the [integers](https://arbital.com/p/48l)."], "tags": ["Math 0", "B-Class"], "alias": "4zh"}
{"id": "6aa42e978213b085c6c9b4cd4995fa67", "title": "Orbit-Stabiliser theorem: External Resources", "url": "https://arbital.com/p/orbit_stabiliser_theorem_external_resources", "source": "arbital", "source_type": "text", "text": "**[https://arbital.com/p/1r5](https://arbital.com/p/1r5)**\n\n* [Group actions II: the orbit-stabilizer theorem](https://gowers.wordpress.com/2011/11/09/group-actions-ii-the-orbit-stabilizer-theorem) (Tim Gowers) - How many rotational symmetries does a cube have?", "date_published": "2016-07-02T21:58:00Z", "authors": ["Eric Bruylant", "Mark Chimes", "Patrick Stevens"], "summaries": ["[External resources](https://arbital.com/p/arbital_resources) about the [https://arbital.com/p/4l8](https://arbital.com/p/4l8)."], "tags": ["List", "External resources", "Stub"], "alias": "4zk"}
{"id": "3acda9a4c2c59a8eb59226b74cb7f337", "title": "Uncountability (Math 3)", "url": "https://arbital.com/p/uncountability_math_3", "source": "arbital", "source_type": "text", "text": "A [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz) $X$ is *uncountable* if there is no [bijection](https://arbital.com/p/499) between $X$ and [$\\mathbb{N}$](https://arbital.com/p/45h). Equivalently, there is no [injection](https://arbital.com/p/4b7) from $X$ to $\\mathbb{N}$.\n\n## Foundational Considerations ##\n\nIn set theories without the [axiom of choice](https://arbital.com/p/69b), such as [Zermelo Frankel set theory](https://arbital.com/p/ZF) without choice (ZF), it can be [consistent](https://arbital.com/p/5km) that there is a [https://arbital.com/p/-cardinal_number](https://arbital.com/p/-cardinal_number) $\\kappa$ that is incomparable to $\\aleph_0$. That is, there is no injection from $\\kappa$ to $\\aleph_0$ nor from $\\aleph_0$ to $\\kappa$. In this case, cardinality is not a [total order](https://arbital.com/p/540), so it doesn't make sense to think of uncountability as \"larger\" than $\\aleph_0$. In the presence of choice, [cardinality is a total order](https://arbital.com/p/5sh), so an uncountable set can be thought of as \"larger\" than a countable set.\n\nCountability in one [https://arbital.com/p/-model](https://arbital.com/p/-model) is not necessarily countability in another. By [Skolem's Paradox](https://arbital.com/p/skolems_paradox), there is a model of set theory $M$ where its [power set](https://arbital.com/p/6gl) of the naturals, denoted $2^\\mathbb N_M \\in M$ is countable when considered outside the model. Of course, it is a [theorem](https://arbital.com/p/6fk) that $2^\\mathbb N _M$ is uncountable, but that is within the model. That is, there is a bijection $f : \\mathbb N \\to 2^\\mathbb N_M$ that is not inside the model $M$ (when $f$ is considered as a set, its graph), and there is no such bijection inside $M$. This means that (un)countability is not [absolute](https://arbital.com/p/absoluteness).\n\n## See also\n\nIf you enjoyed this explanation, consider exploring some of [Arbital's](https://arbital.com/p/3d) other [featured content](https://arbital.com/p/6gg)!\n\nArbital is made by people like you, if you think you can explain a mathematical concept then consider [https://arbital.com/p/-4d6](https://arbital.com/p/-4d6)!", "date_published": "2016-10-26T19:09:44Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Daniel Satanove", "Patrick Stevens"], "summaries": [], "tags": ["C-Class"], "alias": "4zp"}
{"id": "76af0983a58f46d430daeeeb1c3e21b8", "title": "Rational number", "url": "https://arbital.com/p/rational_number", "source": "arbital", "source_type": "text", "text": "The rational numbers are either whole numbers or fractions of whole numbers, like $0,$ $1$, $2$, $\\frac{1}{2}$, $\\frac{97}{3}$, $-17$, $\\frac{-85}{1993},$ and so on. The [set](https://arbital.com/p/3jz) of rational numbers is written $\\mathbb Q.$\n\n[Irrational numbers](https://arbital.com/p/54z) like [$\\pi$](https://arbital.com/p/49r) and [$e$](https://arbital.com/p/e) are _not_ included in $\\mathbb Q;$ the rational numbers are only those numbers which can be written as $\\frac{a}{b}$ for integers $a$ and $b$ (where $b \\neq 0$).\n\nFormally, $\\mathbb{Q}$ is the [https://arbital.com/p/-3gz](https://arbital.com/p/-3gz) of the [https://arbital.com/p/-field_of_fractions](https://arbital.com/p/-field_of_fractions) of $\\mathbb{Z}$ (the [ring](https://arbital.com/p/3gq) of [integers](https://arbital.com/p/48l)).\nThat is, each $q \\in \\mathbb Q$ is an expression $\\frac{a}{b}$, where $b$ is a nonzero integer and $a$ is an integer, together with certain rules for addition and multiplication.\nThe rational numbers are the last intermediate stage on the way to constructing the [real numbers](https://arbital.com/p/4bc), but they are also very interesting and important in their own right.\n\nOne intuition about the rational numbers is that once we've created the [real numbers](https://arbital.com/p/4bc), then a real number $x$ is a rational number if and only if it may be written as $\\frac{a}{b}$, where $a, b$ are integers and $b$ is not $0$.\n\n\n# Examples\n\n- The integer $1$ is a rational number, because it may be written as $\\frac{1}{1}$ (or, indeed, as $\\frac{2}{2}$ or $\\frac{-1}{-1}$, or $\\frac{a}{a}$ for any nonzero integer $a$).\n- The number [$\\pi$](https://arbital.com/p/49r) is not rational ([proof](https://arbital.com/p/513)).\n- The number $\\sqrt{2}$ (being the unique positive real which, when multiplied by itself, yields $2$) is not rational ([proof](https://arbital.com/p/548)).\n\n# Properties\n\n- There are infinitely many rationals. (Indeed, every integer is rational, because the integer $n$ may be written as $\\frac{n}{1}$, and there are infinitely many integers.)\n- There are countably many rationals ([proof](https://arbital.com/p/511)). Therefore, because there are [uncountably](https://arbital.com/p/2w0) many real numbers, [https://arbital.com/p/-almost_all](https://arbital.com/p/-almost_all) real numbers are not rational.\n- The rationals are [dense](https://arbital.com/p/dense_metric_space) in the reals.\n- The rationals form a [field](https://arbital.com/p/481) ([proof](https://arbital.com/p/4zr)). Indeed, they are a subfield of the real numbers.\n\n# Construction\n\nInstead of taking the reals and selecting a certain collection which we label the \"rationals\", [it is possible to construct the rationals](https://arbital.com/p/construction_of_complexes_from_naturals) given access only to the natural numbers; and from the rationals we may construct the reals.\nIn some sense, this approach is cleaner than starting with the reals and producing the rationals, because the natural numbers are very intuitive objects but the real numbers are less so.\nWe can be closer to satisfying some deep existential unease if we can build the reals out of the much-simpler naturals.\n\nAs an analogy, being able to produce a block of wood given access to a wooden table is much less satisfying than the other way round, and we run into blocks of wood \"in the wild\" so we are pretty convinced that there actually are such things as blocks of wood.\nOn the other hand, we almost never see wooden tables in nature, so we can't be quite as sure that they're real until we've built one ourselves.\n\nSimilarly, everyone recognises broadly what a counting number is, and they're out there in the wild, but the rational numbers are somewhat less \"natural\" and their existence is less intuitive.", "date_published": "2016-07-21T15:56:29Z", "authors": ["Eric Bruylant", "Eric Rogstad", "Patrick Stevens", "Nate Soares", "Adom Hartell", "Joe Zeng"], "summaries": ["A rational number is a number which may be written as $\\frac{a}{b}$, where $a$ is an [https://arbital.com/p/-48l](https://arbital.com/p/-48l) and $b$ is a nonzero integer."], "tags": ["Start"], "alias": "4zq"}
{"id": "6b52007312faa2b6e9e1315abb2121e0", "title": "The rationals form a field", "url": "https://arbital.com/p/rationals_are_a_field", "source": "arbital", "source_type": "text", "text": "The set $\\mathbb{Q}$ of [rational numbers](https://arbital.com/p/4zq) is a [field](https://arbital.com/p/481).\n\n# Proof\n\n$\\mathbb{Q}$ is a ([commutative](https://arbital.com/p/3jb)) [ring](https://arbital.com/p/3gq) with additive identity $\\frac{0}{1}$ (which we will write as $0$ for short) and multiplicative identity $\\frac{1}{1}$ (which we will write as $1$ for short): we check the axioms individually.\n\n- $+$ is commutative: $\\frac{a}{b} + \\frac{c}{d} = \\frac{ad+bc}{bd}$, which by commutativity of addition and multiplication in $\\mathbb{Z}$ is $\\frac{cb+da}{db} = \\frac{c}{d} + \\frac{a}{b}$\n- $0$ is an identity for $+$: have $\\frac{a}{b}+0 = \\frac{a}{b} + \\frac{0}{1} = \\frac{a \\times 1 + 0 \\times b}{b \\times 1}$, which is $\\frac{a}{b}$ because $1$ is a multiplicative identity in $\\mathbb{Z}$ and $0 \\times n = 0$ for every integer $n$.\n- Every rational has an additive inverse: $\\frac{a}{b}$ has additive inverse $\\frac{-a}{b}$.\n- $+$ is [associative](https://arbital.com/p/3h4): $$\\left(\\frac{a_1}{b_1}+\\frac{a_2}{b_2}\\right)+\\frac{a_3}{b_3} = \\frac{a_1 b_2 + b_1 a_2}{b_1 b_2} + \\frac{a_3}{b_3} = \\frac{a_1 b_2 b_3 + b_1 a_2 b_3 + a_3 b_1 b_2}{b_1 b_2 b_3}$$\nwhich we can easily check is equal to $\\frac{a_1}{b_1}+\\left(\\frac{a_2}{b_2}+\\frac{a_3}{b_3}\\right)$. \n- $\\times$ is associative, trivially: $$\\left(\\frac{a_1}{b_1} \\frac{a_2}{b_2}\\right) \\frac{a_3}{b_3} = \\frac{a_1 a_2}{b_1 b_2} \\frac{a_3}{b_3} = \\frac{a_1 a_2 a_3}{b_1 b_2 b_3} = \\frac{a_1}{b_1} \\left(\\frac{a_2 a_3}{b_2 b_3}\\right) = \\frac{a_1}{b_1} \\left(\\frac{a_2}{b_2} \\frac{a_3}{b_3}\\right)$$\n- $\\times$ is commutative, again trivially: $$\\frac{a}{b} \\frac{c}{d} = \\frac{ac}{bd} = \\frac{ca}{db} = \\frac{c}{d} \\frac{a}{b}$$\n- $1$ is an identity for $\\times$: $$\\frac{a}{b} \\times 1 = \\frac{a}{b} \\times \\frac{1}{1} = \\frac{a \\times 1}{b \\times 1} = \\frac{a}{b}$$ by the fact that $1$ is an identity for $\\times$ in $\\mathbb{Z}$.\n- $+$ distributes over $\\times$: $$\\frac{a}{b} \\left(\\frac{x_1}{y_1}+\\frac{x_2}{y_2}\\right) = \\frac{a}{b} \\frac{x_1 y_2 + x_2 y_1}{y_1 y_2} = \\frac{a \\left(x_1 y_2 + x_2 y_1\\right)}{b y_1 y_2}$$\nwhile $$\\frac{a}{b} \\frac{x_1}{y_1} + \\frac{a}{b} \\frac{x_2}{y_2} = \\frac{a x_1}{b y_1} + \\frac{a x_2}{b y_2} = \\frac{a x_1 b y_2 + b y_1 a x_2}{b^2 y_1 y_2} = \\frac{a x_1 y_2 + a y_1 x_2}{b y_1 y_2}$$\nso we are done by distributivity of $+$ over $\\times$ in $\\mathbb{Z}$.\n\nSo far we have shown that $\\mathbb{Q}$ is a ring; to show that it is a field, we need all nonzero fractions to have inverses under multiplication.\nBut if $\\frac{a}{b}$ is not $0$ (equivalently, $a \\not = 0$), then $\\frac{a}{b}$ has inverse $\\frac{b}{a}$, which does indeed exist since $a \\not = 0$.\n\nThis completes the proof.", "date_published": "2016-07-06T16:28:22Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": [], "tags": ["Proof"], "alias": "4zr"}
{"id": "5700a268e4e08c04e40e5a86813b3acb", "title": "Complex number", "url": "https://arbital.com/p/complex_number", "source": "arbital", "source_type": "text", "text": "A complex number is a number of the form $z = a + b\\textrm{i}$, where $\\textrm{i}$ is the imaginary unit defined as $\\textrm{i}=\\sqrt{-1}$.\n\nDoesn't make sense? Let's backtrack a little.\n\n## Motivation - expanding the definition of numbers##\nSay we start with a child's understanding of numbers as counting numbers, or [natural numbers](https://arbital.com/p/45h). Using only natural numbers we can add and multiply whichever numbers we want, but there are some restrictions on subtraction: ask a child to subtract 5 from 3 and they'll say \"you can't do that!\".\n\nIn order to allow the computation of numbers like $5-3$, we need to define a new kind of number: negative numbers. The set of natural numbers, negative numbers and $0$ is called whole numbers, or [integers](https://arbital.com/p/48l). Unlike natural numbers, with integers we can add and subtract whatever we want. (In math terms, we say the set of integers is [closed](https://arbital.com/p/3gy) to addition.)\n\nWe still have a problem, though: what about division? Going back to our child analogy, say we're talking to a third grader. If you ask them to divide 8 by 2, no problem. But what if you as them to divide 2 by 8? \"You can't do that!\"\n\nJust like with negative numbers, we need to define a new kind of number: fractions, like $\\frac{1}{2}, \\frac{5}{3}$ or $-\\frac{6}{7}$. By adding fractions to the integers, we get a more general set of numbers - [rational numbers](https://arbital.com/p/4zq). With rational numbers we can multiply and divide however we want - the set of rational numbers is closed to multiplication. Well, except dividing by zero. No one has figured out how to do that without breaking mathematics, so that's awkward.\n\nNow say our child is a seventh grader, so they know about square roots. They can apply the square root operation to any perfect square, no problem - $\\sqrt{9}=3$. But what if you ask them to find the square root of a non-perfect square, like $\\sqrt{2}$? \"You can't do that!\"\n\nOnce again, we need to expand our set of numbers to include a new kind of number: irrational numbers. Unlike we discussed with negative numbers or fractions, irrationals can do a whole lot more that just define square roots. In fact, irrationals are so cool that according to some sources the Pythagoreans couldn't deal with their existence and ended up killing [the guy who invented them](https://arbital.com/p/https://en.wikipedia.org/wiki/Hippasus).\n\nAdding irrational numbers to our set of rational numbers, we get the [real numbers](https://arbital.com/p/4bc). Mathematically, we say that the reals are a [complete](https://arbital.com/p/real_number_completeness) [ordered](https://arbital.com/p/540) [field](https://arbital.com/p/481), which is actually one of the ways of defining them. \n\nHowever, there's still one problem: the real numbers are not an [https://arbital.com/p/-algebraically_closed_field](https://arbital.com/p/-algebraically_closed_field). In a practical sense, this means we still don't have closure under the square root operation $\\sqrt{}$, because we can't define the square roots of negative numbers.\n\nBy now you've probably noticed the pattern, though: any time someone says \"you can't do that!\", mathematicians invent a new kind of number to prove them wrong.\n\n## Introducing: imaginary numbers ##\nIn order to allow using the square root operation on negative numbers, we once again have to define a new kind of number: imaginary numbers. \n\nActually, we have to define just one number - the imaginary unit, $\\textrm{i}$. $\\textrm{i}$ is defined as a solution to the quadratic equation $x^2+1=0$. In other words, we can define $\\textrm{i}$ as equaling $\\sqrt{-1}$.\n\nAll the other imaginary numbers (square roots of negatives) follow directly from this definition of $\\textrm{i}$: for any negative number $-a$, we can define $\\sqrt{-a}=\\textrm{i}\\sqrt{a}$.\n\n##To be continued...##\nThis article is unfinished, and will later include an explanation of the complex plane as well as the algebraic properties of complex numbers.", "date_published": "2016-07-05T20:05:56Z", "authors": ["Eric Bruylant", "Yelene Seratna", "Patrick Stevens", "Eliana Ruby"], "summaries": [], "tags": ["Start", "Work in progress", "Needs clickbait"], "alias": "4zw"}
{"id": "b4e7182924523b7ab47409a4ccb3353a", "title": "Rational numbers: Intro (Math 0)", "url": "https://arbital.com/p/rational_numbers_intro_math_0", "source": "arbital", "source_type": "text", "text": "*In order to get the most out of this page, you probably want a good grasp of the [integers](https://arbital.com/p/53r) first.*\n\n\"Rational number\" is a phrase mathematicians use for the idea of a \"fraction\".\nHere, we'll go through what a fraction is and why we should care about them.\n\n# What is a fraction?\n\nSo far, we've met the [integers](https://arbital.com/p/48l): whole numbers, which can be either bigger than $0$ or less than $0$ (or the very special $0$ itself).\nThe [natural numbers](https://arbital.com/p/45h) can count the number of cows I have in my possession; the integers can also count the number of cows I have after I've given away cows from having nothing, resulting in anti-cows.\n\nIn this article, though, we'll stop talking about cows and start talking about apples instead. The reason will become clear in a moment.\n\nSuppose I have two apples. %%note:I'm terrible at drawing, so my apples look suspiciously like circles.%%\n\n\n\nWhat if I chopped one of the apples into two equally-sized pieces? (And now you know why we stopped talking about cows.)\n\nNow what I have is a whole apple, and… another apple which is in two pieces.\n\n\n\nLet's imagine now that I chop one of the pieces itself into two pieces, and for good measure I chop my remaining whole apple into three pieces.\n\n\n\nI still have the same amount of apple as I started with - I haven't eaten any of the pieces or anything - but now it's all in funny-sized chunks.\n\nNow I'll eat one of the smallest chunks. How many apples do I have now?\n\n\n\nI certainly don't just have one apple, because three of the chunks I've got in front of me will together make an apple; and I've also got some chunks left over once I've done that.\nBut I can't have two apples either, because I *started* with two and then I ate a bit.\nMathematicians like to be able to compare things, and if I forced you to make a comparison, you could say that I have \"more than one apple\" but \"fewer than two apples\".\n\nIf you're happy with that, then it's a reasonable thing to ask: \"exactly how much apple do I have?\".\nAnd the mathematician will give an answer of \"one apple and three quarters\".\n\"One and three quarters\" is an example of a **rational number** or **fraction**: it expresses a quantity that came from dividing some number of things into some number of equal parts, then possibly removing some of the parts.\n%%note:I've left out the point that just as you moved from the [counting numbers](https://arbital.com/p/45h) to the [integers](https://arbital.com/p/48l), thereby allowing you to owe someone some apples, so we can also have a negative rational number of apples. We'll get to that in time.%%\n\n# The basic building block\n\nFrom a certain point of view, the building block of the *natural* numbers is just the number $1$: all natural numbers can be made by just adding together the number $1$ some number of times. (If I have a heap of apples, I can build it up just from single apples.)\nThe building block of the integers is also the number $1$, because if you gave me some apples %%note:which perhaps I've now eaten%% so that I owe you some apples, you might as well have given them to me one by one.\n\nNow the *rationals* have building blocks too, but this time there are lots and lots of them, because if you give me any kind of \"building block\" - some quantity of apple - I can always just chop it into two pieces and make a smaller \"building block\".\n(This wasn't true when we were confined just to whole apples, as in the natural numbers! If I can't divide up an apple, then I can't make any quantity of apples smaller than one apple. %%note:Except no apples at all.%%)\n\nIt turns out that a good choice of building blocks is \"one piece, when we divide an apple into several equally-sized pieces\".\nIf we took our apple, and divided it into five equal pieces, then the corresponding building-block is \"one fifth of an apple\": five of these building blocks makes one apple.\nTo a mathematician, we have just made the rational number which is written $\\frac{1}{5}$.\n\nSimilarly, if we divided our apple instead into six equal pieces, and take just one of the pieces, then we have made the rational number which is written $\\frac{1}{6}$.\n\nThe (positive) rational numbers are just whatever we could make by taking lots of copies of building blocks.\n\n# Examples\n\n- $1$ is a rational number. It can be made with the building block that is just $1$ itself, which is what we get if we take an apple and divide it into just one piece - that is, making no cuts at all. Or, if you're a bit squeamish about not making any cuts, $1$ can be made out of two halves: two copies of the building block that results when we take an apple and cut it into two equal pieces, taking just one of the pieces. (We write $\\frac{1}{2}$ for that half-sized building block.)\n- $2$ is a rational number: it can be made out of two lots of the $1$-building-block, or indeed out of four lots of the $\\frac{1}{2}$-building-block.\n- $\\frac{1}{2}$ is a rational number: it is just the half-sized building block itself.\n- If we took the apple and instead cut it into three pieces, we obtain a building block which we write as $\\frac{1}{3}$; so $\\frac{1}{3}$ is a rational number.\n- Two copies of the $\\frac{1}{3}$-building-block makes the rational number which we write $\\frac{2}{3}$.\n- Five copies of the $\\frac{1}{3}$-building-block makes somewhat more than one apple. Indeed, three of the building blocks can be put together to make one full apple, and then we've got two building blocks left over. We write the rational number represented by five $\\frac{1}{3}$-building-blocks as $\\frac{5}{3}$.\n\n# Notation\n\nNow you've seen the notation $\\frac{\\cdot}{\\cdot}$ used a few times, where there are numbers in the places of the dots.\nYou might be able to guess how this notation works in general now: if we take the blocks resulting when we divide an apple into \"dividey-number\"-many pieces, and then take \"lots\" of those pieces, then we obtain a rational number which we write as $\\frac{\\text{lots}}{\\text{dividey-number}}$.\nMathematicians use the words \"numerator\" and \"denominator\" for what I called \"lots\" and \"dividey-number\"; so it would be $\\frac{\\text{numerator}}{\\text{denominator}}$ to a mathematician.\n\n# Exercises\n\nCan you give some examples of how we can make the number $3$ from smaller building blocks? (There are lots and lots of ways you could correctly answer this question.)\n%%hidden(Show a possible solution):\nYou already know about one way from when we talked about the natural numbers: just take three copies of the $1$-block. (That is, three apples is three single apples put together.)\n\nAnother way would be to take six half-sized blocks: $\\frac{6}{2}$ is another way to write $3$.\n\nYet another way is to take fifteen fifth-sized blocks: $\\frac{15}{5}$ is another way to write $3$.\n\nIf you want to mix things up, you could take four half-sized blocks and three third-sized blocks: $\\frac{4}{2}$ and $\\frac{3}{3}$ together make $3$.\n\n%%\n\nIf you felt deeply uneasy about the last of my possible solutions above, there is a good and perfectly valid reason why you might have done; we will get to that eventually. If that was you, just forget I mentioned that last one for now.\nIf you were comfortable with it, that's also normal.\n\nHow about making the number $\\frac{1}{2}$ from smaller blocks?\n%%hidden(Show a possible solution):\nOf course, you could start by taking just one $\\frac{1}{2}$ block.\n\nFor a more interesting answer, you could take three copies of the sixth-sized block: $\\frac{3}{6}$ is the same as $\\frac{1}{2}$. \n\n\n\nAlternatively, five copies of the tenth-sized block: $\\frac{5}{10}$ is the same as $\\frac{1}{2}$.\n\n\n%%\n\nThe way I've drawn the pictures might be suggestive: in some sense, when I've given different answers just now, they all look like \"the same answer\" but with different lines drawn on.\nThat's because the rational numbers (\"fractions\", remember) correspond to answers to the question \"how much?\".\nWhile there is always more than one way to build a given rational number out of the building blocks, the way that we build the number doesn't affect the ultimate answer to the question \"how much?\".\n$\\frac{5}{10}$ and $\\frac{1}{2}$ and $\\frac{3}{6}$ are all simply different ways of writing the same underlying quantity: the number which represents the fundamental concept of \"chop something into two equal pieces\".\nThey each express different ways of making the same amount (for instance, out of five $\\frac{1}{10}$-blocks, or one $\\frac{1}{2}$-block), but the amount itself hasn't changed.\n\n# Going more general\n\nRemember, from when we treated the integers using cows, that I can give you a cow (even if I haven't got one) by creating a cow/anti-cow pair and then giving you the cow, leaving me with an anti-cow.\nWe count the number of anti-cows that I have by giving them a *negative* number.\n\nWe can do the same here with chunks of apple.\nIf I wanted to give you half an apple, but I didn't have any apples, I could create a half-apple/half-anti-apple pair, and then give you the half-apple; this would leave me with a half-anti-apple.\n\nWe count anti-apples in the same way as we count anti-cows: they are *negative*.\n\nSee the page on [subtraction](https://arbital.com/p/56x) for a much more comprehensive explanation; this page is more of a whistle-stop tour.\n\n# Limitations\n\nWe've had the idea of building-blocks: as $\\frac{1}{n}$, where $n$ was a natural number.\nWhy should $n$ be just a natural number, though?\nWe've already seen the integers; why can't it be one of those? %%note:That is, why not let it be negative?%%\n\nAs it turns out, we *can* let $n$ be an integer, but we don't actually get anything new if we do.\nWe're going to pretend for the moment that $n$ has to be positive, because it gets a bit weird trying to divide things into three anti-chunks; this approach doesn't restrict us in any way, but if you are of a certain frame of mind, it might just look like a strange and artificial boundary to draw.\n\nHowever, you must note that $n$ cannot be $0$ (whatever your stance on dividing things into anti-chunks).\nWhile there is a way to finesse the idea of an anti-chunk %%note: And if you sit and think really hard for a long time, you might even come up with it yourself!%%, there is simply no way to make it possible to divide an apple into $0$ equal pieces.\nThat is, $\\frac{1}{0}$ is not a rational number (and you should be very wary of calling it anything that suggests it's like a number - like \"infinity\" - and under no account may you do arithmetic on it).\n\n# Summary\n\nSo far, you've met what a rational number is! We haven't gone through how to do things with them yet, but hopefully you now understand vaguely what they're there for: they express the idea of \"dividing something up into parts\", or \"sharing things out among people\" (if I have two apples to split fairly among three people, I can be fair by chopping each apple into three $\\frac{1}{3}$-sized building blocks, and then giving each person two of the blocks).\n\n[Next up](https://arbital.com/p/514), we will see how we can combine rational numbers together, eventually making a very convenient shorthand. %%note:The study of this shorthand is known as \"arithmetic\".%%", "date_published": "2016-08-01T08:04:20Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": ["The rational numbers are \"fractions\". While the [natural numbers](https://arbital.com/p/45h) measure the answer to the question \"how many apples?\" when we have some apples, the rationals measure the answer to the question \"how much apple?\" when we can cut the apples up in different ways and remove pieces before counting them."], "tags": [], "alias": "4zx"}
{"id": "ee45c28c053ee8746d1ad09517bb17a5", "title": "Report likelihoods not p-values: FAQ", "url": "https://arbital.com/p/likelihood_not_pvalue_faq", "source": "arbital", "source_type": "text", "text": "This page answers frequently asked questions about the [https://arbital.com/p/4zd](https://arbital.com/p/4zd) proposal for experimental science.\n\n_(Note: This page is a personal [opinion page](https://arbital.com/p/60p).)_\n\n[https://arbital.com/p/toc:](https://arbital.com/p/toc:)\n\n### What does this proposal entail?\n\nLet's say you have a coin and you don't know whether it's biased or not. You flip it six times and it comes up HHHHHT.\n\nTo report a [p-value](https://arbital.com/p/p_value), you have to first declare which experiment you were doing — were you flipping it six times no matter what you saw and counting the number of heads, or were you flipping it until it came up tails and seeing how long it took? Then you have to declare a \"null hypothesis,\" such as \"the coin is fair.\" Only then can you get a p-value, which in this case, is either 0.11 (if you were going to toss the coin 6 times regardless) or 0.03 (if you were going to toss until it came up heads). The p-value of 0.11 means \"if the null hypothesis _were_ true, then data that has as many H values as the observed data would only occur 11% of the time, if the declared experiment were repeated many many times.\"\n\nTo report a [likelihood](https://arbital.com/p/bayes_likelihood), you don't have to do any of that \"declare your experiment\" stuff, and you don't have to single out one special hypothesis. You just pick a whole bunch of hypotheses that seem plausible, such as the set of hypotheses $H_b$ = \"the coin has a bias of $b$ towards heads\" for $b$ between 0% and 100%. Then you look at the actual data, and report how likely that data is according to each hypothesis. In this example, that yields a graph which looks like this:\n\n\n\nThis graph says that HHHHHT is about 1.56% likely under the hypothesis $H_{0.5}$ saying that the coin is fair, and about 5.93% likely under the hypothesis $H_{0.75}$ that the coin comes up heads 75% of the time, and only 0.17% likely under the hypothesis $H_{0.3}$ that the coin only comes up tails 30% of the time.\n\nThat's all you have to do. You don't need to make any arbitrary choice about which experiment you were going to run. You don't need to ask yourself what you \"would have seen\" in other cases. You just look at the actual data, and report how likely each hypothesis in your hypothesis class said that data should be.\n\n(If you want to compare how well the evidence supports one hypothesis or another, you just use the graph to get a [likelihood ratio](https://arbital.com/p/1rq) between any two hypotheses. For example, this graph reports that the data HHHHHT supports the hypothesis $H_{0.75}$ over $H_{0.5}$ at odds of $\\frac{0.0593}{0.0156}$ = 3.8 to 1.)\n\nFor more of an explanation, see [https://arbital.com/p/4zd](https://arbital.com/p/4zd).\n\n### Why would reporting likelihoods be a good idea?\n\nExperimental scientists reporting likelihoods instead of p-values would likely help address many problems facing modern science, including [p-hacking](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging), [the vanishing effect size problem](https://arbital.com/p/http://slatestarcodex.com/2014/04/28/the-control-group-is-out-of-control/), and [publication bias](https://arbital.com/p/https://en.wikipedia.org/wiki/Publication_bias).\n\nIt would also make it easier for scientists to combine the results from multiple studies, and it would make it much much easier to conduct meta-analyses.\n\nIt would also make scientific statistics more intuitive, and easier to understand.\n\n### Likelihood functions are a Bayesian tool. Aren't Bayesian statistics subjective? Shouldn't science be objective?\n\nLikelihood functions are purely objective. In fact, there's only one degree of freedom in a likelihood function, and that's the choice of hypothesis class. This choice is no more arbitrary than the choice of a \"null hypothesis\" in standard statistics, and indeed, it's significantly less arbitrary (you can pick a large class of hypotheses, rather than just one; and none of them needs to be singled out as subjectively \"special\").\n\nThis is in stark contrast with p-values, which require that you pick an \"experimental design\" in advance, or that you talk about what data you \"could have seen\" if the experiment turned out differently. Likelihood functions only depend on the hypothesis class that you're considering, and the data that you actually saw. (This is one of the reasons why likelihood functions would solve p-hacking.)\n\nLikelihood functions are often used by Bayesian statisticians, and Bayesian statisticians do indeed use [subjective probabilities](https://arbital.com/p/4vr), which has led some people to believe that reporting likelihood functions would somehow allow hated subjectivity to seep into the hallowed halls of science.\n\nHowever, it's the [priors](https://arbital.com/p/1rm) that are subjective in Bayesian statistics, not likelihood functions. In fact, according to the [laws of probability theory](https://arbital.com/p/1lz), likelihood functions are precisely that-which-is-left-over when you factor out all subjective beliefs from an observation of evidence. In other words, probability theory tells us that likelihoods are the best summary there is for capturing the objective evidence that a piece of data provides (assuming your goal is to help make people's beliefs more accurate).\n\n### How would reporting likelihoods solve p-hacking?\n\nP-values depend on what experiment the experimenter says they had in mind. For example, if the data is HHHHHT and the experimenter says \"I was planning to flip it six times and count the number of Hs\" then the p-value (for the fair coin hypothesis) is 0.11, which is not \"significant.\" If instead the experimenter says \"I was planing to flip it until I got a T\" then the p-value is 0.03, which _is_ \"significant.\" Experimenters can ([and do!](https://arbital.com/p/http://amstat.tandfonline.com/doi/pdf/10.1080/00031305.2016.1154108)) misuse or abuse this degree of freedom to make their results appear more significant than they actually are. This is known as \"p-hacking.\"\n\nIn fact, when running complicated experiments, this can (and does!) happen to honest well-meaning researchers. Some experimenters are dishonest, and many others simply lack the time and patience to understand the subtleties of good experimental design. We don't need to put that burden on experimenters. We don't need to use statistical tools that depend on which experiment the experimenter had in mind. We can instead report the likelihood that each hypothesis assigned to the actual data.\n\nLikelihood functions don't have this \"experiment\" degree of freedom. They don't care what experiment you thought you were doing. They only care about the data you actually saw. To use likelihood functions correctly, all you have to do is look at stuff and then not lie about what you saw. Given the set of hypotheses you want to report likelihoods for, the likelihood function is completely determined by the data.\n\n### But what if the experimenter tries to game the rules by choosing how much data to collect?\n\nThat's a problem if you're reporting p-values, but it's not a problem if you're reporting likelihood functions.\n\nLet's say there's a coin that you think is fair, that I think might be biased 55% towards heads. If you're right, then every toss is going to (in [https://arbital.com/p/-4b5](https://arbital.com/p/-4b5)) provide more evidence for \"fair\" than \"biased.\" But sometimes (rarely), even if the coin is fair, you will flip it and it will generate a sequence that supports the \"bias\" hypothesis more than the \"fair\" hypothesis.\n\nHow often will this happen? It depends on how exactly you ask the question. If you can flip the coin at most 300 times, then there's about a 1.4% chance that at some point the sequence generated will support the hypothesis \"the coin is biased 55% towards heads\" 20x more than it supports the hypothesis \"the coin is fair.\" (You can verify this yourself, and tweak the parameters, using [this code](https://arbital.com/p/https://gist.github.com/Soares/941bdb13233fd0838f1882d148c9ac14).)\n\nThis is an objective fact about coin tosses. If you look at a sequence of Hs and Ts generated by a fair coin, then some tiny fraction of the time, after some number $n$ of flips, it will support the \"biased 55% towards heads\" hypothesis 20x more than it supports the \"fair\" hypothesis. This is true no matter how or why you decided to look at those $n$ coin flips. It's true if you were always planning to look at $n$ coin flips since the day you were born. It's true if each coin flip costs $1 to look at, so you decided to only look until the evidence supported one hypothesis at least 20x better than the other. It's true if you have a heavy personal desire to see the coin come up biased, and were planning to keep flipping until the evidence supports \"bias\" 20x more than it supports \"fair\". It doesn't _matter_ why you looked at the sequence of Hs and Ts. The amount by which it supports \"biased\" vs \"fair\" is objective. If the coin really is fair, then the more you flip it the more the evidence will push towards \"fair.\" It will only support \"bias\" a small unlucky fraction of the time, and that fraction is completely independent from your thoughts and intentions .\n\nLikelihoods are objective. They don't depend on your state of mind.\n\nP-values, on the other hand, run into some difficulties. A p-value is about a single hypothesis (such as \"fair\") in isolation. If the coin is fair, then [all sequences of coin tosses are equally likely](https://arbital.com/p/fair_coin_equally_likely), so you need something more than the data in order to decide whether the data is \"significant evidence\" about fairness one way or the other. Which means you have to choose a \"reference class\" of ways the coin \"could have come up.\" Which means you need to tell us which experiment you \"intended\" to run. And down the rabbit hole we go.\n\nThe p-value you report depends on how many coin tosses you say you were going to look at. If you lie about where you intended to stop, the p-value breaks. If you're out in the field collecting data, and the data just subconsciously begins to feel overwhelming, and so you stop collecting evidence (or if the data just subconsciously feels insufficient and so you collect more) then the p-value breaks. How badly to p-values break? If you can toss the coin at most 300 times, then by choosing when to stop looking, you can get a p < 0.05 significant result _21% of the time,_ and that's assuming you are required to look at at least 30 flips. If you're allowed to use small sample sizes, the number is more like 25%. You can verify this yourself, and tweak the parameters, using [this code](https://arbital.com/p/https://gist.github.com/Soares/4955bb9268129476262b28e32b8ec979).\n\nIt's no wonder that p-values are so often misused! To use p-values correctly, an experimenter has to meticulously report their intentions about the experimental design before collecting data, and then has to hold utterly unfaltering to that experiment design as the data comes in (even if it becomes clear that their experimental design was naive, and that there were crucial considerations that they failed to take into account). Using p-values correctly requires good intentions, constant vigilance, and inflexibility.\n\nContrast this with likelihood functions. Likelihood functions don't depend on your intentions. If you start collecting data until it looks overwhelming and then stop, that's great. If you start collecting data and it looks underwhelming so you keep collecting more, that's great too. Every new piece of data you do collect will support the true hypothesis more than any other hypothesis, in expectation — that's the whole point of collecting data. Likelihood functions don't depend upon your state of mind.\n\n\n\n### What if the experimenter uses some other technique to bias the result?\n\nThey can't. Or, at least, it's a theorem of [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv) that they can't. This law is known as [conservation of expected evidence](https://arbital.com/p/-conservation_expected_evidence), and it says that for any hypothesis $H$ and any piece of evidence $e$, $\\mathbb P(H) = \\mathbb P(H \\mid e) \\mathbb P(e) + \\mathbb P(H \\mid \\lnot e) \\mathbb P(\\lnot e),$ where $\\mathbb P$ stands for my personal subjective probabilities.\n\nImagine that I'm going to take your likelihood function $\\mathcal L$ and blindly combine it with my personal beliefs using [Bayes' rule](https://arbital.com/p/1lz). The question is, can you use $\\mathcal L$ to manipulate my beliefs? The answer is clearly \"yes\" if you're willing to lie about what data you saw. But what if you're honestly reporting all the data you _actually_ saw? Then can you manipulate my beliefs, perhaps by being strategic about what data you look at and how long you look at it?\n\nClearly, the answer to that question is \"sort of.\" If you have a fair coin, and you want to convince me it's biased, and you toss it 10 times, and it (by sheer luck) comes up HHHHHHHHHH, then that's a lot of evidence in favor of it being biased. But you can't use the \"hope the coin comes up heads 10 times in a row by sheer luck\" strategy to _reliably_ bias my beliefs; and if you try just flipping the coin 10 times and hoping to get lucky, then on average, you're going to produce data that convinces me that the coin is fair. The real question is, can you bias my beliefs _in expectation?_\n\nIf the answer is \"yes,\" then there will be times when I should ignore $\\mathcal L$ even if you honestly reported what you saw. If the answer is \"no,\" then there will be no such times — for every $e$ that would shift my beliefs heavily towards $H$ (such that you could say \"Aha! How naive! If you look at this data and see it is $e$, then you will believe $H$, just as I intended\"), there is an equal and opposite chance of alternative data which would push my beliefs _away_ from $H.$ So, can you set up a data collection mechanism that pushes me towards $H$ in expectation?\n\nAnd the answer to that question is _no,_ and this is a trivial theorem of probability theory. No matter what subjective belief state $\\mathbb P$ I use, if you honestly report the objective likelihood $\\mathcal L$ of the data you actually saw, and I update $\\mathbb P$ by [multiplying it by $\\mathcal L$](https://arbital.com/p/1lz), there is no way (according to $\\mathbb P$) for you to bias my probability of $H$ on average — no matter how strategically you decide which data to look at or how long to look. For more on this theorem and its implications, see [Conservation of Expected Evidence](https://arbital.com/p/conservation_expected_evidence).\n\nThere's a difference between metrics that can't be exploited in theory and metrics that can't be exploited in practice, and if a malicious experimenter really wanted to abuse likelihood functions, they could probably find some clever method. (At the least, they can always lie and make things up.) However, p-values aren't even provably inexploitable — they're so easy to exploit that sometimes well-meaning honest researchers exploit them _by accident_, and these exploits are already commonplace and harmful. When building better metrics, starting with metrics that are provably inexploitable is a good start.\n\n### What if you pick the wrong hypothesis class?\n\nIf you don't report likelihoods for the hypotheses that someone cares about, then that person won't find your likelihood function very helpful. The same problem exists when you report p-values (what if you pick the wrong null and alternative hypotheses?). Likelihood functions make the problem a little better, by making it easy to report how well the data supports a wide variety of hypotheses (instead of just ~2), but at the end of the day, there's no substitute for the raw data.\n\nLikelihoods are a summary of the data you saw. They're a useful summary, especially if you report likelihoods for a broad set of plausible hypotheses. They're a much better summary than many other alternatives, such as p-values. But they're still a summary, and there's just no substitute for the raw data.\n\n### How does reporting likelihoods help prevent publication bias?\n\nWhen you're reporting p-values, there's a stark difference between p-values that favor the null hypothesis (which are deemed \"insignificant\") and p-values that reject the null hypothesis (which are deemed \"significant\"). This \"significance\" occurs at arbitrary thresholds (e.g. p < 0.05), and significance is counted only in one direction (to be significant, you must reject the null hypothesis). Both these features contribute to publication bias: Journals only want to accept experiments that claim \"significance\" and reject the null hypothesis.\n\nWhen you're reporting [likelihood functions](https://arbital.com/p/56s), a 20 : 1 [ratio](https://arbital.com/p/1rq) is a 20 : 1 ratio is a 20 : 1 ratio. It doesn't matter if your likelihood function is peaked near \"the coin is fair\" or whether it's peaked near \"the coin is biased 82% towards heads.\" If the ratio between the likelihood of one hypothesis and the likelihood of another hypothesis is 20 : 1 then the data provides the same strength of evidence either way. Likelihood functions don't single out one \"null\" hypothesis and incentivize people to only report data that pushes away from that null hypothesis; they just talk about the relationship between the data and _all_ the interesting hypotheses.\n\nFurthermore, there's no arbitrary significance threshold for likelihood functions. If you didn't have a ton of data, your likelihood function will be pretty spread out, but it won't be useless. If you find $5 : 1$ odds in favor of $H_1$ over $H_2$, and I independently find $6 : 1$ odds in favor of $H_1$ over $H_2$, and our friend independently finds $3 : 1$ odds in favor of $H_1$ over $H_2,$ then our studies as a whole constitute evidence that favors $H_1$ over $H_2$ by a factor of $90 : 1$ — hardly insignificant! With likelihood ratios (and no arbitrary \"significance\" cutoffs) progress can be made in small steps.\n\nOf course, this wouldn't solve the problem of publication bias in full, not by a long shot. There would still be incentives to report cool and interesting results, and the scientific community might still ask for results to pass some sort of \"significance\" threshold before accepting them for publication. However, reporting likelihoods would be a good start.\n\n### How does reporting likelihoods help address vanishing effect sizes?\n\nIn a field where an effect does not actually exist, we will often observe an initial study that finds a very large effect, followed by a number of attempts at replication that find smaller and smaller and smaller effects (until someone postulates that the effect doesn't exist, and does a meta-analysis to look for p-hacking and publication bias). This is known as the [decline effect](https://arbital.com/p/https://en.wikipedia.org/wiki/Decline_effect); see also [_The control group is out of control_](https://arbital.com/p/http://slatestarcodex.com/2014/04/28/the-control-group-is-out-of-control/).\n\nThe decline effect is possible in part because p-values look only at whether the evidence says we should \"accept\" or \"reject\" a special null hypothesis, without any consideration for what that evidence says about the alternative hypotheses. Let's say we have three studies, all of which reject the null hypothesis \"the coin is fair.\" The first study rejects the null hypothesis with a 95% confidence interval of [0.9](https://arbital.com/p/0.7,) bias in favor of heads, but it was a small study and some of the experimenters were a bit sloppy. The second study is a bit bigger and a bit better organized, and rejects the null hypothesis with a 95% confidence interval of [0.62](https://arbital.com/p/0.53,). The third study is high-powered, long-running, and rejects the null hypothesis with a 95% confidence interval of [0.511](https://arbital.com/p/0.503,). It's easy to say \"look, three separate studies rejected the null hypothesis!\"\n\nBut if you look at the likelihood functions, you'll see that _something very fishy is going on_ — none of the studies actually agree with each other! The effect sizes are incompatible. Likelihood functions make this phenomenon easy to detect, because they tell you how much the data supports _all_ the relevant hypotheses (not just the null hypothesis). If you combine the three likelihood functions, you'll see that _none_ of the confidence intervals fare very well. Likelihood functions make it obvious when different studies contradict each other directly, which makes it much harder to summarize contradictory data down to \"three studies rejected the null hypothesis\".\n\n### What if I want to reject the null hypothesis without needing to have any particular alternative in mind?\n\nMaybe you don't want to report likelihoods for a large hypothesis class, because you are pretty sure you can't generate a hypothesis class that contains the correct hypothesis. \"I don't want to have to make up a bunch of alternatives,\" you protest, \"I just want to show that the null hypothesis is _wrong,_ in isolation.\"\n\nFortunately for you, that's possible using likelihood functions! The tool you're looking for is the notion of [strict confusion](https://arbital.com/p/227). A hypothesis $H$ will tell you how low its likelihood is supposed to get, and if its likelihood goes a lot lower than that value, then you can be pretty confident that you've got the wrong hypothesis.\n\nFor example, let's say that your one and only hypothesis is $H_{0.9}$ = \"the coin is biased 90% towards heads.\" Now let's say you flip the coin twenty times, and you see the sequence THTTHTTTHTTTTHTTTTTH. The [log-likelihood](https://arbital.com/p/log_likelihood) that $H_{0.9}$ _expected_ to get on a sequence of 20 coin tosses was about -9.37 [bits](https://arbital.com/p/evidence_bit),%%note: According to $H_{0.9},$ each coin toss carries $0.9 \\log_2(0.9) + 0.1 \\log_2(0.1) \\approx -0.469$ bits of evidence, so after 20 coin tosses, $H_{0.9}$ expects about $20 \\cdot 0.469 \\approx 9.37$ bits of [surprise](https://arbital.com/p/bayes_surprise). For more on why log likelihood is a convenient tool for measuring \"evidence\" and \"surprise,\" see [Bayes' rule: log odds form](https://arbital.com/p/1zh).%% for a likelihood score of about $2^{-9.37} \\approx$ $1.5 \\cdot 10^{-3},$ on average. The likelihood that $H_{0.9}$ actually gets on that sequence is -50.59 bits, for a likelihood score of about $5.9 \\cdot 10^{-16},$ which is _thirteen orders of magnitude less likely than expected._ You don't need to be clever enough to come up with an alternative hypothesis that explains the data in order to know that $H_{0.9}$ is not the right hypothesis for you.\n\nIn fact, likelihood functions make it easy to show that _lots_ of different hypotheses are strictly confused — you don't need to have a good hypothesis in your hypothesis class in order for reporting likelihood functions to be a useful service.\n\n### How does reporting likelihoods make it easier to combine multiple studies?\n\nWant to combine two studies that reported likelihood functions? Easy! Just multiply the likelihood functions together. If the first study reported 10 : 1 odds in favor of \"fair coin\" over \"biased 55% towards heads,\" and the second study reported 12 : 1 odds in favor of \"fair coin\" over \"biased 55% towards heads,\" then the combined studies support the \"fair coin\" hypothesis over the \"biased 55% towards heads\" hypothesis at a likelihood ratio of 120 : 1.\n\nIs it really that easy? Yes! That's one of the benefits of using a representation of evidence supported by a large edifice of [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv) — they're trivially easy to compose. You have to ensure that the studies are independent first, because otherwise you'll double-count the data. (If the combined likelihood ratios get really extreme, you should be suspicious about whether they were actually independent.) This isn't exactly a new problem in experimental science; we can just add it to the list of reasons why replication studies had better be independent of the original study. Also, you can only multiply the likelihood functions together on places where they're both defined: If one study doesn't report the likelihood for a hypothesis that you care about, you might need access to the raw data in order to extend their likelihood function. But if the studies are independent and both report likelihood functions for the relevant hypotheses, then all you need to do is multiply.\n\n(Don't try this with p-values. A p < 0.05 study and a p < 0.01 study don't combine into anything remotely like a p < 0.0005 study.)\n\n### How does reporting likelihoods make it easier to conduct meta-analyses?\n\nWhen studies report p-values, performing a meta-analysis is a complicated procedure that requires dozens of parameters to be finely tuned, and (lo and behold) bias somehow seeps in, and meta-analyses often find whatever the analyzer set out to find. When studies report likelihood functions, performing a meta-analysis is trivial and doesn't depend on you to tune a dozen parameters. Just multiply all the likelihood functions together.\n\nIf you want to be extra virtuous, you can check for anomalies, such as one likelihood function that's tightly peaked in a place that disagrees with all the other peaks. You can also check for [strict confusion](https://arbital.com/p/227), to get a sense for how likely it is that the correct hypothesis is contained within the hypothesis class that you considered. But mostly, all you've got to do is multiply the likelihood functions together.\n\n### How does reporting likelihood functions make it easier to detect fishy studies?\n\nWith likelihood functions, it's much easier to find the studies that don't match up with each other — look for the likelihood function that has its peak in a different place than all the other peaks. That study deserves scrutiny: either those experimenters had something special going on in the background of their experiment, or something strange happened in their data collection and reporting process.\n\nFurthermore, likelihoods combined with the notion of [strict confusion](https://arbital.com/p/227) make it easy to notice when something has gone seriously wrong. As per the above answers, you can combine multiple studies by multiplying their likelihood functions together. What happens if the likelihood function is super small everywhere? That means that either (a) some of the data is fishy, or (b) you haven't considered the right hypothesis yet.\n\nWhen you _have_ considered the right hypothesis, it will have decently high likelihood under _all_ the data. There's only one real world underlying all our data, after all — it's not like different experimenters are measuring different underlying universes. If you multiply all the likelihood functions together and _all_ the hypotheses turn out looking wildly unlikely, then you've got some work to do — you haven't yet considered the right hypothesis.\n\nWhen reporting p-values, contradictory studies feel like the norm. Nobody even _tries_ to make all the studies fit together, as if they were all measuring the same world. With likelihood functions, we could actually aspire towards a world where scientific studies on the same topic are _all_ combined. A world where people try to find hypotheses that fit _all_ the data at once, and where a single study's data being out of place (and making all the hypotheses currently under consideration become [https://arbital.com/p/-227](https://arbital.com/p/-227)) is a big glaring \"look over here!\" signal. A world where it feels like studies are _supposed_ to fit together, where if scientists haven't been able to find a hypothesis that explains all the raw data, then they know they have their work cut out for them.\n\nWhatever the right hypothesis is, it will almost surely not be strictly confused under the actual data. Of course, when you come up with a completely new hypothesis (such as \"the coin most of us have been using is fair but study #317 accidentally used a different coin\") you're going to need access to the raw data of some of the previous studies in order to extend their likelihood functions and see how well they do on this new hypothesis. As always, there's just no substitute for raw data.\n\n### Why would this make statistics easier to do and understand?\n\np < 0.05 does not mean \"the null hypothesis is less than 5% likely\" (though that's what young students of statistics often _want_ it to mean). What the null hypothesis means is \"given a particular experimental design (e.g., toss the coin 100 times and count the heads) and the data (e.g., the sequence of 100 coin flips), if the null hypothesis _were_ true, then data that matches my chosen statistic (e.g., the number of heads) would only occur 5% of the time, if we repeated this experiment over and over and over.\"\n\nWhy the complexity? Statistics is designed to keep subjective beliefs out of the hallowed halls of science. Your science paper shouldn't be able to conclude \"and, therefore, I personally believe that the coin is very likely to be biased, and I'd bet on that at 20 : 1 odds.\" Still, much of this complexity is unnecessary. Likelihood functions achieve the same goal of objectivity, but without all the complexity.\n\n[$\\mathcal L_e](https://arbital.com/p/51n) $< 0.05$ _also_ doesn't mean \"$H$ is less than 5% likely\", it means \"H assigned less than 0.05 probability to $e$ happening.\" The student still needs to learn to keep \"probability of $e$ given $H$\" and \"probability of $H$ given $e$\" distinctly separate in their heads. However, likelihood functions do have a _simpler_ interpretation: $\\mathcal L_e(H)$ is the probability of the actual data $e$ occurring if $H$ were in fact true. No need to talk about experimental design, no need to choose a summary statistic, no need to talk about what \"would have happened.\" Just look at how much probability each hypothesis assigned to the actual data; that's your likelihood function.\n\nIf you're going to report p-values, you need to be meticulous in considering the complexities and subtleties of experiment design, on pain of creating p-values that are broken in non-obvious ways (thereby contributing to the [replication crisis](https://arbital.com/p/https://en.wikipedia.org/wiki/Replication_crisis)). When reading results, you need to take the experimenter's intentions into account. None of this is necessary with likelihoods.\n\nTo understand $\\mathcal L_e(H),$ all you need to know is how likely $e$ was according to $H.$ Done.\n\n### Isn't this just one additional possible tool in the toolbox? Why switch entirely away from p-values?\n\nThis may all sound too good to be true. Can one simple change really solve that many problems in modern science?\n\nFirst of all, you can be assured that reporting likelihoods instead of p-values would not \"solve\" all the problems above, and it would surely not solve all problems with modern experimental science. Open access to raw data, preregistration of studies, a culture that rewards replication, and many other ideas are also crucial ingredients to a scientific community that zeroes in on truth.\n\nHowever, reporting likelihoods would help solve lots of different problems in modern experimental science. This may come as a surprise. Aren't likelihood functions just one more statistical technique, just another tool for the toolbox? Why should we think that one single tool can solve that many problems?\n\nThe reason lies in [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv). According to the axioms of probability theory, there is only one good way to account for evidence when updating your beliefs, and that way is via likelihood functions. Any other method is subject to inconsistencies and pathologies, as per the [coherence theorems of probability theory](https://arbital.com/p/probability_coherence_theorems).\n\nIf you're manipulating equations like $2 + 2 = 4,$ and you're using methods that may or may not let you throw in an extra 3 on the right hand side (depending on the arithmetician's state of mind), then it's no surprise that you'll occasionally get yourself into trouble and deduce that $2 + 2 = 7.$ The laws of arithmetic show that there is only one correct set of tools for manipulating equations if you want to avoid inconsistency.\n\nSimilarly, the laws of probability theory show that there is only one correct set of tools for manipulating _uncertainty_ if you want to avoid inconsistency. According to [those rules](https://arbital.com/p/1lz), the right way to represent evidence is through likelihood functions.\n\nThese laws (and a solid understanding of them) are younger than the experimental science community, and the statistical tools of that community predate a modern understanding of probability theory. Thus, it makes a lot of sense that the existing literature uses different tools. However, now that humanity _does_ possess a solid understanding of probability theory, it should come as no surprise that many diverse pathologies in statistics can be cleaned up by switching to a policy of reporting likelihoods instead of p-values.\n\n\n### If it's so great why aren't we doing it already?\n\n[Probability theory](https://arbital.com/p/1bv) (and a solid understanding of all that it implies) is younger than the experimental science community, and the statistical tools of that community predate a modern understanding of probability theory. In particular, modern statistical tools were designed in an attempt to keep subjective reasoning out of the hallowed halls of science. You shouldn't be able to publish a scientific paper which concludes \"and therefore, I personally believe that this coin is biased towards heads, and would bet on that at 20 : 1 odds.\" Those aren't the foundations upon which science can be built.\n\nLikelihood functions are strongly associated with Bayesian statistics, and Bayesian statistical tools tend to manipulate subjective probabilities. Thus, it wasn't entirely clear how to use tools such as likelihood functions without letting subjectivity bleed into science.\n\nNowadays, we have a better understanding of how to separate out subjective probabilities from objective claims, and it's known that likelihood functions don't carry any subjective baggage with them. In fact, they carry _less_ subjective baggage than p-values do: A likelihood function depends only on the data that you _actually saw,_ whereas p-values depend on your experimental design and your intentions.\n\nThere are good historical reasons why the existing scientific community is using p-values, but now that humanity _does_ possess a solid theoretical understanding of probability theory (and how to factor subjective probabilities out from objective claims), it's no surprise that a wide array of diverse problems in modern statistics can be cleaned up by reporting likelihoods instead of p-values.\n\n### Has this ever been tried?\n\nNo. Not yet. To our knowledge, most scientists haven't even considered this proposal — and for good reason! There are a lot of big fish to fry when it comes to addressing the [replication crisis](https://arbital.com/p/https://en.wikipedia.org/wiki/Replication_crisis), [p-hacking](https://arbital.com/p/https://en.wikipedia.org/wiki/Data_dredging), [the problem of vanishing effect sizes](https://arbital.com/p/http://slatestarcodex.com/2014/04/28/the-control-group-is-out-of-control/), [publication bias](https://arbital.com/p/https://en.wikipedia.org/wiki/Publication_bias), and other problems facing science today. The scientific community at large is huge, decentralized, and has a lot of inertia. Most activists who are trying to shift it already have their hands full advocating for very important policies such as open access journals and pre-registration of trials. So it makes sense that nobody's advocating hard for reporting likelihoods instead of p-values — yet.\n\nNevertheless, there are good reasons to believe that reporting likelihoods instead of p-values would help solve many of the issues in modern experimental science.", "date_published": "2017-09-09T14:57:58Z", "authors": ["[ ]", "Nate Soares", "Diane Ritter", "Eric Bruylant", "Travis Rivera"], "summaries": [], "tags": ["Opinion page"], "alias": "505"}
{"id": "0055dc109f9532014f2fb4a9457d3310", "title": "Real number (as Cauchy sequence)", "url": "https://arbital.com/p/real_number_as_cauchy_sequence", "source": "arbital", "source_type": "text", "text": "summary(Technical): The real numbers can be constructed as a field consisting of all Cauchy sequences of rationals, quotiented by the equivalence relation given by \"two sequences are equivalent if and only if they eventually get arbitrarily close to each other\".\n\nConsider the set of all [Cauchy sequences](https://arbital.com/p/53b) of [rational numbers](https://arbital.com/p/4zq): concretely, the set $$X = \\{ (a_n)_{n=1}^{\\infty} : a_n \\in \\mathbb{Q}, (\\forall \\epsilon \\in \\mathbb{Q}^{>0}) (\\exists N \\in \\mathbb{N})(\\forall n, m \\in \\mathbb{N}^{>N})(|a_n - a_m| < \\epsilon) \\}$$\n\nDefine an [https://arbital.com/p/-53y](https://arbital.com/p/-53y) on this set, by $(a_n) \\sim (b_n)$ if and only if, for every rational $\\epsilon > 0$, there is a [https://arbital.com/p/-45h](https://arbital.com/p/-45h) $N$ such that for all $n \\in \\mathbb{N}$ bigger than $N$, we have $|a_n - b_n| < \\epsilon$.\nThis is an equivalence relation (exercise).\n%%hidden(Show solution):\n- It is symmetric, because $|a_n - b_n| = |b_n - a_n|$.\n- It is reflexive, because $|a_n - a_n| = 0$ for every $n$, and this is $< \\epsilon$.\n- It is transitive, because if $|a_n - b_n| < \\frac{\\epsilon}{2}$ for sufficiently large $n$, and $|b_n - c_n| < \\frac{\\epsilon}{2}$ for sufficiently large $n$, then $|a_n - b_n| + |b_n - c_n| < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon$ for sufficiently large $n$; so by the [https://arbital.com/p/-triangle_inequality](https://arbital.com/p/-triangle_inequality), $|a_n - c_n| < \\epsilon$ for sufficiently large $n$.\n%%\n\nWrite $[https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ for the equivalence class of $(a_n)_{n=1}^{\\infty}$. (This is a slight abuse of notation, omitting the brackets that indicate that $a_n$ is actually a sequence rather than a rational number.)\n\nThe set of *real numbers* is the set of equivalence classes of $X$ under this equivalence relation, endowed with the following [totally ordered](https://arbital.com/p/540) [field](https://arbital.com/p/481) structure:\n\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/b_n](https://arbital.com/p/b_n) := [+ b_n](https://arbital.com/p/a_n)$\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n) := [\\times b_n](https://arbital.com/p/a_n)$\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ if and only if $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ or there is some $N$ such that for all $n > N$, $a_n \\leq b_n$.\n\nThis field structure is well-defined ([proof](https://arbital.com/p/51h)).\n\n# Examples\n\n- Any rational number $r$ may be viewed as a real number, being the class $[https://arbital.com/p/r](https://arbital.com/p/r)$ (formally, the equivalence class of the sequence $(r, r, \\dots)$).\n- The real number $\\pi$ is indeed a real number under this definition; it is represented by, for instance, $(3, 3.1, 3.14, 3.141, \\dots)$. It is also represented as $(100, 3, 3.1, 3.14, \\dots)$, along with many other possibilities.", "date_published": "2016-07-05T20:06:38Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": ["If we want to construct the real numbers in terms of simpler objects (such as the rationals), one way to do it is to take our putative real number and consider sequences of rational numbers which in some sense \"get closer and closer\" to the real number."], "tags": [], "alias": "50d"}
{"id": "6baddf9f3aa6597d9668ba3c7cd09a61", "title": "Real number (as Dedekind cut)", "url": "https://arbital.com/p/real_number_as_dedekind_cut", "source": "arbital", "source_type": "text", "text": "%%comment: Mnemonics for defined macros: \\Ql = Q left, \\Qr = Q right, \\Qls = Q left strict, \\Qrs = Q right strict.%%\n\nThe rational numbers have a problem that makes them unsuitable for use in calculus — they have \"gaps\" in them. This may not be obvious or even make sense at first, because between any two rational numbers you can always find infinitely many other rational numbers. How could there be *gaps* in a set like that? $\\newcommand{\\rats}{\\mathbb{Q}} \\newcommand{\\Ql}{\\rats^\\le} \\newcommand{\\Qr}{\\rats^\\ge} \\newcommand{\\Qls}{\\rats^<} \\newcommand{\\Qrs}{\\rats^>}$\n$\\newcommand{\\set}[https://arbital.com/p/1](https://arbital.com/p/1){\\left\\{#1\\right\\}} \\newcommand{\\sothat}{\\ |\\ }$ \n\nBut using the construction of [Dedekind cuts](https://arbital.com/p/dedekind_cut), we can suss out these gaps into plain view. A Dedekind cut of a [https://arbital.com/p/-540](https://arbital.com/p/-540) $S$ is a pair of sets $(A, B)$ such that:\n\n1. Every element of $S$ is in exactly one of $A$ or $B$. (That is, $(A, B)$ is a [partition](https://arbital.com/p/set_partition) of $S$.)\n2. Every element of $A$ is less than every element of $B$.\n3. Neither $A$ nor $B$ is [empty](https://arbital.com/p/empty_set). (We'll see why this restriction matters in a moment.)\n\nOne example of such a cut might be the set where $A$ is the negative rational numbers and $B$ is the nonnegative rational numbers (positive or zero). We see that it satisfies the three properties of a Dedekind cut:\n\n1. Every rational number is either negative or nonnegative, but not both.\n2. Every rational number which is negative is less than a rational number that is nonnegative.\n3. There exists at least one negative rational number (e.g. $-1$) and one nonnegative rational number (e.g. $1$).\n\nIn fact, Dedekind cuts are intended to represent sets of rational numbers that are less than or greater than a specific real number (once we've defined them). To represent this, let's call them $\\Ql$ and $\\Qr$.\n\nKnowing this, why does it matter that neither set in a Dedekind cut is empty?\n\n%%hidden(Show solution): \nIf $\\Ql$ were empty, then we'd have a real number less than _all_ the rational numbers, which is $-\\infty$, which we don't want to define as a real number. Similarly, if $\\Qr$ were empty, then we'd get $+\\infty$.\n%%\n\n## Completion of a space\n\nIf a space is [complete](https://arbital.com/p/) (doesn't have any gaps in it), then in any Dedekind cut $(\\Ql, \\Qr)$, either $\\Ql$ will have a greatest element or $\\Qr$ will have a least element. (We can't have both at the same time — why?)\n\n%%hidden(Show solution):\nSuppose $\\Ql$ had a greatest element $q_u$ and $\\Qr$ had a least element $q_v$. We can't have $q_u = q_v$, because the same number would be in both sets. So then because the rational numbers are a dense space, there must exist a rational number $r$ so that $q_u < r < q_v$. Then $r$ is not in either $\\Ql$ or $\\Qr$, contradicting property 1 of a Dedekind cut.\n%%\n\nBut in the rational numbers, we can find a Dedekind cut where neither $\\Ql$ nor $\\Qr$ have a greatest or least element respectively.\n\nConsider the pair of sets $(\\Ql, \\Qr)$ where $\\Ql = \\set{x \\in \\rats \\mid x^3 \\le 2}$ and $\\Qr = \\set{x \\in \\rats \\mid x^3 \\ge 2}$.\n\n1. Every rational number has a cube either greater than 2 or less than 2, \n2. Because $f(x) = x^3$ is a [monotonically increasing](https://arbital.com/p/) function, we have that $p < q \\iff p^3 < q^3$, which means that every element in $\\Ql$ is less than every element in $\\Qr$.\n\nSo $(\\Ql, \\Qr)$ is a Dedekind cut. However, there is no rational number whose cube is *equal to* $2$, so $\\Ql$ has no greatest element and $\\Qr$ has no least element.\n\nThis represents a gap in the numbers, because we can invent a new number to place in that gap (in this case $\\sqrt[https://arbital.com/p/3](https://arbital.com/p/3){2}$), which is \"between\" any two numbers in $\\Ql$ and $\\Qr$.\n\n## Definition of real numbers\n\nBefore we move on, we will define one more structure that makes the construction more elegant. Define a *one-sided Dedekind cut* as any Dedekind cut $(\\Ql, \\Qr)$ with the additional property that the set $\\Ql$ has no greatest element (in which case we now call it $\\Qls$). The case where $\\Ql$ has a greatest element $q_g$ can be trivially transformed into the equivalent case on the other side by moving $q_g$ to $\\Qr$ where it is automatically the least element due to being less than any other element in $\\Qr$.\n\nThen we define the real numbers as the set of one-sided Dedekind cuts of the rational numbers.\n\n* A rational number $r$ is mapped to itself by the Dedekind cut where $r$ itself is the least element of $\\Qr$. (If the cuts weren't one-sided, $r$ would also be mapped to the set where $r$ was the greatest element of $\\Ql$, which would make the mapping non-unique.)\n* An irrational number $q$ is newly defined by the Dedekind cut where all the elements of $\\Qls$ are less than $q$ and all the elements of $\\Qr$ are (strictly) greater than $q$.\n\nNow we can define the [total order](https://arbital.com/p/549) $\\le$ for two real numbers $a = (\\Qls_a, \\Qr_a)$ and $b = (\\Qls_b, \\Qr_b)$ as follows: $a \\le b$ when $\\Qls_a \\subseteq \\Qls_b$. \n\nUsing this, we can show that unlike in the [Cauchy sequence definition](https://arbital.com/p/50d), we don't need to define any equivalence classes — every real number is uniquely defined by a one-sided Dedekind cut.\n\n%%hidden(Proof):\nIf $a = b$, then $a \\le b$ and $b \\le a$. By the definition of the order, we have that $\\Qls_a \\subseteq \\Qls_b$ and $\\Qls_b \\subseteq \\Qls_a$, which means that $\\Qls_a = \\Qls_b$, which means that the Dedekind cuts corresponding to $a$ and $b$ are also equal.\n%%", "date_published": "2016-07-23T17:41:20Z", "authors": ["Kevin Clancy", "Dylan Hendrickson", "Patrick Stevens", "Eric Bruylant", "Joe Zeng"], "summaries": ["The real numbers can be defined using [Dedekind cuts](https://arbital.com/p/dedekind_cut), which are [partitions](https://arbital.com/p/set_partition) of the rational numbers into two sets where every element in one set is less than every element in the other set. Each real number is uniquely defined by a one-sided Dedekind cut."], "tags": ["Stub"], "alias": "50g"}
{"id": "1014914303b01d5efab54863af6b6c78", "title": "Currying", "url": "https://arbital.com/p/currying", "source": "arbital", "source_type": "text", "text": "Currying converts a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) of N [inputs](https://arbital.com/p/input) into a function of a single input, which returns a function that takes a second input, which returns a function that takes a third input, and so on, until N functions have been evaluated and N inputs have been taken, at which point the original function is evaluated on the given inputs and the result is returned.\n\nFor example:\n\nConsider a function of four parameters, $F:(X,Y,Z,N)→R$.\n\nThe curried version of the function, $curry(F)$ would have the type signature $X→(Y→(Z→(N→R)))$, and $curry(F)(4)(3)(2)(1)$ would equal $F(4,3,2,1)$.\n\nCurrying is named after the logician Haskell Curry. If you can't remember this derivation and need a mnemonic, you might like to imagine a function being cooked in a curry so thoroughly that it breaks up into small, bite-sized pieces, just as functional currying breaks a function up into smaller functions. (It might also help to note that there is a programming language also named after Haskell Curry (Haskell), that features currying prominently. All functions in Haskell are pre-curried by convention. This often makes [partially applying](https://wiki.haskell.org/Partial_application) functions effortless.)", "date_published": "2016-07-17T22:22:13Z", "authors": ["Eric Bruylant", "M Yass"], "summaries": [], "tags": [], "alias": "50p"}
{"id": "74a291797985c4b5d7910419fc0790a2", "title": "Proof", "url": "https://arbital.com/p/proof_meta_tag", "source": "arbital", "source_type": "text", "text": "Proof pages provide [formal proofs](https://arbital.com/p/proof) about a statement, but don't motivate or explain them.", "date_published": "2016-07-02T21:33:01Z", "authors": ["Eric Bruylant"], "summaries": [], "tags": [], "alias": "50q"}
{"id": "1c576c77c375e28e01bc8985dedd2c5c", "title": "External resources", "url": "https://arbital.com/p/external_resources_meta_tag", "source": "arbital", "source_type": "text", "text": "This lens links out to other great resources across the web. External resources lenses should never be the main lens, Arbital should always have at least a [https://arbital.com/p/-72](https://arbital.com/p/-72) with a popover [summary](https://arbital.com/p/1kl).", "date_published": "2016-07-02T21:53:46Z", "authors": ["Eric Bruylant"], "summaries": [], "tags": ["Stub"], "alias": "50r"}
{"id": "89ca5bd32183ebdd4da99ad1b3d7c5a3", "title": "The set of rational numbers is countable", "url": "https://arbital.com/p/rationals_are_countable", "source": "arbital", "source_type": "text", "text": "The set $\\mathbb{Q}$ of [rational numbers](https://arbital.com/p/4zq) is countable: that is, there is a [bijection](https://arbital.com/p/499) between $\\mathbb{Q}$ and the set $\\mathbb{N}$ of [natural numbers](https://arbital.com/p/45h).\n\n# Proof\n\nBy the [Schröder-Bernstein theorem](https://arbital.com/p/cantor_schroeder_bernstein_theorem), it is enough to find an [injection](https://arbital.com/p/4b7) $\\mathbb{N} \\to \\mathbb{Q}$ and an injection $\\mathbb{Q} \\to \\mathbb{N}$.\n\nThe former is easy, because $\\mathbb{N}$ is a subset of $\\mathbb{Q}$ so the identity injection $n \\mapsto \\frac{n}{1}$ works.\n\nFor the latter, we may define a function $\\mathbb{Q} \\to \\mathbb{N}$ as follows.\nTake any rational in its lowest terms, as $\\frac{p}{q}$, say. %%note:That is, the [GCD](https://arbital.com/p/greatest_common_divisor) of the numerator $p$ and denominator $q$ is $1$.%%\nAt most one of $p$ and $q$ is negative (if both are negative, we may just cancel $-1$ from the top and bottom of the fraction); by multiplying by $\\frac{-1}{-1}$ if necessary, assume without loss of generality that $q$ is positive.\nIf $p = 0$ then take $q = 1$.\n\nDefine $s$ to be $1$ if $p$ is positive, and $2$ if $p$ is negative.\n\nThen produce the natural number $2^p 3^q 5^s$.\n\nThe function $f: \\frac{p}{q} \\mapsto 2^p 3^q 5^s$ is injective, because [prime factorisations are unique](https://arbital.com/p/fundamental_theorem_of_arithmetic) so if $f\\left(\\frac{p}{q}\\right) = f \\left(\\frac{a}{b} \\right)$ (with both fractions in their lowest terms, and $q$ positive) then $|p| = |a|, q=b$ and the sign of $p$ is equal to the sign of $a$.\nHence the two fractions were the same after all.", "date_published": "2016-07-03T07:47:06Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Proof"], "alias": "511"}
{"id": "43fa9dc6204769091cd1bf8b453b88e2", "title": "Pi is irrational", "url": "https://arbital.com/p/pi_is_irrational", "source": "arbital", "source_type": "text", "text": "The number [$\\pi$](https://arbital.com/p/49r) is not [rational](https://arbital.com/p/4zq).\n\n# Proof\n\nFor any fixed real number $q$, and any [https://arbital.com/p/-45h](https://arbital.com/p/-45h) $n$, let $$A_n = \\frac{q^n}{n!} \\int_0^{\\pi} [](https://arbital.com/p/x)^n \\sin(x) dx$$\nwhere $n!$ is the [https://arbital.com/p/-5bv](https://arbital.com/p/-5bv) of $n$, $\\int$ is the [https://arbital.com/p/-definite_integral](https://arbital.com/p/-definite_integral), and $\\sin$ is the [https://arbital.com/p/-sin_function](https://arbital.com/p/-sin_function).\n\n## Preparatory work\n\nExercise: $A_n = (4n-2) q A_{n-1} - (q \\pi)^2 A_{n-2}$.\n%%hidden(Show solution):\nWe use [https://arbital.com/p/-integration_by_parts](https://arbital.com/p/-integration_by_parts).\n\n\n%%\n\nNow, $$A_0 = \\int_0^{\\pi} \\sin(x) dx = 2$$\nso $A_0$ is an integer.\n\nAlso $$A_1 = q \\int_0^{\\pi} x (\\pi-x) \\sin(x) dx$$ which by a simple calculation is $4q$.\n%%hidden(Show calculation):\nExpand the integrand and then integrate by parts repeatedly:\n$$\\frac{A_1}{q} = \\int_0^{\\pi} x (\\pi-x) \\sin(x) dx = \\pi \\int_0^{\\pi} x \\sin(x) dx - \\int_0^{\\pi} x^2 \\sin(x) dx$$\n\nThe first integral term is $$[\\cos](https://arbital.com/p/-x)_0^{\\pi} + \\int_0^{\\pi} \\cos(x) dx = \\pi$$\n\nThe second integral term is $$[\\cos](https://arbital.com/p/-x^2)_{0}^{\\pi} + \\int_0^{\\pi} 2x \\cos(x) dx$$\nwhich is $$\\pi^2 + 2 \\left( [\\sin](https://arbital.com/p/x)_0^{\\pi} - \\int_0^{\\pi} \\sin(x) dx \\right)$$\nwhich is $$\\pi^2 -4$$\n\nTherefore $$\\frac{A_1}{q} = \\pi^2 - (\\pi^2 - 4) = 4$$\n%%\n\nTherefore, if $q$ and $q \\pi$ are integers, then so is $A_n$ [inductively](https://arbital.com/p/5fz), because $(4n-2) q A_{n-1}$ is an integer and $(q \\pi)^2 A_{n-2}$ is an integer.\n\nBut also $A_n \\to 0$ as $n \\to \\infty$, because $\\int_0^{\\pi} [](https://arbital.com/p/x)^n \\sin(x) dx$ is in modulus at most $$\\pi \\times \\max_{0 \\leq x \\leq \\pi} [](https://arbital.com/p/x)^n \\sin(x) \\leq \\pi \\times \\max_{0 \\leq x \\leq \\pi} [](https://arbital.com/p/x)^n = \\pi \\times \\left[https://arbital.com/p/\\frac{\\pi^2}{4}\\right](https://arbital.com/p/\\frac{\\pi^2}{4}\\right)^n$$\nand hence $$|A_n| \\leq \\frac{1}{n!} \\left[q}{4}\\right](https://arbital.com/p/\\frac{\\pi^2)^n$$\n\nFor $n$ larger than $\\frac{\\pi^2 q}{4}$, this expression is getting smaller with $n$, and moreover it gets smaller faster and faster as $n$ increases; so its limit is $0$.\n%%hidden(Formal treatment):\nWe claim that $\\frac{r^n}{n!} \\to 0$ as $n \\to \\infty$, for any $r > 0$.\n\nIndeed, we have $$\\frac{r^{n+1}/(n+1)!}{r^n/n!} = \\frac{r}{n+1}$$\nwhich, for $n > 2r-1$, is less than $\\frac{1}{2}$.\nTherefore the ratio between successive terms is less than $\\frac{1}{2}$ for sufficiently large $n$, and so the sequence must shrink at least geometrically to $0$.\n%%\n\n## Conclusion\n\nSuppose (for [contradiction](https://arbital.com/p/46z)) that $\\pi$ is rational; then it is $\\frac{p}{q}$ for some integers $p, q$.\n\nNow $q \\pi$ is an integer (indeed, it is $p$), and $q$ is certainly an integer, so by what we showed above, $A_n$ is an integer for all $n$.\n\nBut $A_n \\to 0$ as $n \\to \\infty$, so there is some $N$ for which $|A_n| < \\frac{1}{2}$ for all $n > N$; hence for all sufficiently large $n$, $A_n$ is $0$.\nWe already know that $A_0 = 2$ and $A_1 = 4q$, neither of which is $0$; so let $N$ be the first integer such that $A_n = 0$ for all $n \\geq N$, and we can already note that $N > 1$.\n\nThen $$0 = A_{N+1} = (4N-2) q A_N - (q \\pi)^2 A_{N-1} = - (q \\pi)^2 A_{N-1}$$\nwhence $q=0$ or $\\pi = 0$ or $A_{N-1} = 0$.\n\nCertainly $q \\not = 0$ because $q$ is the denominator of a fraction; and $\\pi \\not = 0$ by whatever definition of $\\pi$ we care to use.\nBut also $A_{N-1}$ is not $0$ because then $N-1$ would be an integer $m$ such that $A_n = 0$ for all $n \\geq m$, and that contradicts the definition of $N$ as the *least* such integer.\n\nWe have obtained the required contradiction; so it must be the case that $\\pi$ is irrational.", "date_published": "2016-07-21T17:36:18Z", "authors": ["Eric Rogstad", "Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Proof"], "alias": "513"}
{"id": "34470c5622365830ce3ecabe10cb84a7", "title": "Arithmetic of rational numbers (Math 0)", "url": "https://arbital.com/p/arithmetic_of_rational_numbers_math_0", "source": "arbital", "source_type": "text", "text": "Now that we've seen [what the rational numbers are](https://arbital.com/p/4zx), we'll move on to working out how to combine them.\n\nThere are four main ways to combine rational numbers, and collectively they are called \"arithmetic\":\n\n- [Addition](https://arbital.com/p/55m)\n- [Subtraction](https://arbital.com/p/56x)\n- [Multiplication](https://arbital.com/p/59s)\n- [Division](https://arbital.com/p/5jd)\n\nAdditionally, we can *[compare](https://arbital.com/p/5pk)* rational numbers with each other; and we can show that all of these operations [play nicely together](https://arbital.com/p/5pm) in a certain sense.\n\nThis page is the start of an arc which will go through all of these ideas.\nIn the process, we'll get you familiar with the notation that we use for rational numbers; it's a very convenient shorthand once you've learnt [the conventions](https://arbital.com/p/5q5).", "date_published": "2016-08-01T06:15:58Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": ["We can take two [rational numbers](https://arbital.com/p/4zq) and [add](https://arbital.com/p/addition) them together, just as we can take two [*natural* numbers](https://arbital.com/p/45h) and add them together; and all the other operations of the integers can be viewed in the rationals too."], "tags": ["Math 0"], "alias": "514"}
{"id": "0cb1e8ec1c0fd14e91722a8ed5855c31", "title": "Field structure of rational numbers", "url": "https://arbital.com/p/field_structure_of_rational_numbers", "source": "arbital", "source_type": "text", "text": "The [rational numbers](https://arbital.com/p/4zq), being the [https://arbital.com/p/-field_of_fractions](https://arbital.com/p/-field_of_fractions) of the [integers](https://arbital.com/p/48l), have the following [field](https://arbital.com/p/481) structure:\n\n- Addition is given by $\\frac{a}{b} + \\frac{p}{q} = \\frac{aq+bp}{bq}$\n- Multiplication is given by $\\frac{a}{b} \\frac{c}{d} = \\frac{ac}{bd}$\n- The identity under addition is $\\frac{0}{1}$\n- The identity under multiplication is $\\frac{1}{1}$\n- The additive inverse of $\\frac{a}{b}$ is $\\frac{-a}{b}$\n- The multiplicative inverse of $\\frac{a}{b}$ (where $a \\not = 0$) is $\\frac{b}{a}$.\n\nIt additionally inherits a [total ordering](https://arbital.com/p/540) which respects the field structure: $0 < \\frac{c}{d}$ if and only if $c$ and $d$ are both positive or $c$ and $d$ are both negative.\nAll other information about the ordering can be derived from this fact: $\\frac{a}{b} < \\frac{c}{d}$ if and only if $0 < \\frac{c}{d} - \\frac{a}{b}$.", "date_published": "2016-07-29T18:39:26Z", "authors": ["Eric Bruylant", "Patrick Stevens"], "summaries": [], "tags": ["Formal definition"], "alias": "516"}
{"id": "74944b5836cb84c030ef46a0c1611635", "title": "Elementary Algebra", "url": "https://arbital.com/p/elementary_algebra", "source": "arbital", "source_type": "text", "text": "How do we describe relations between different things? How can we figure out new true things from true things we already know?\nHow can we find and think about patterns we notice with numbers? Algebra is a unified framework for thinking about these questions, and gives us lots of tools to help us answer them, which work in an extremely wide variety of situations. \n\n#Equations\nAlgebra is based on the arithmetic of numbers, and the relations between them. \nLet's look at a basic equation: $$2 + 2 = 4$$\nThe 'equals' sign (=) tells us that both sides of the equation are actually the same. If we have two, and add two more to it, then we'll have four. \n\nSome other relations are 'less than' (<), for example $2 < 4$. or 'greater than' (>), for example $5 > 1$. \n\n##The right order\nIn equations, parenthesis tell us the right order to do things - things inside of parenthesis have to be done before the rest. This is important because doing things in different orders can give us different answers!\n$$2 + (3 \\times 4) = 14$$\n$$(2 + 3) \\times 4 = 20$$\n\nIt's annoying to have to use parentheses all the time (though it might be helpful if you find yourself getting confused about something). It would be nice if we could just write $2 + 3 \\times 4$ and have everyone know that we meant $2 + (3 \\times 4)$ . There's a standard [order of operations](https://arbital.com/p/54s) that everyone uses so that we don't have to use too many parentheses. \n\n\n\n#Balancing the truth\nIf we know one equation, what are some ways we could get _new_ equations from it, that will still be true? \nWe could make a change to one side, but then the equation would stop being true... unless we did the same change to the other side also! For example, we know $2+2=4$. What if we add three to both sides? If we check \n$(2 + 2) + 3 = 4 + 3$, we can see that it's still true! \n\n##Substitution \nOne way of getting new true things is really important. If we know two things are the same, we can always substitute one of them for the other, and this automatically will give us an equality relation between the two things! \n\nThis is really helpful for breaking down calculations into manageable pieces. For example, if we want to calculate $2^3 \\times 2^4$, we can first expand $2^3 = 2 \\times 2 \\times 2$, and then calculate $2 \\times 2 \\times 2 = 8$. Now, we combine these last two equations to see that $2^3 = 8$. Often, people will do both of these in one step, so if you ever are having a hard time figuring out how someone got an equation, you can try breaking it down like this to see if that helps. Next, we can do the same thing to see that $2^4 = 2 \\times 2 \\times 2 \\times 2 = 16$. We can then substitute both of these back into the original expression, to get $2^3 \\times 2^4 = 8 \\times 16$. One final calculation lets us see that $2^3\\times 2^4 = 128$. \n\n#Naming numbers\nWhat are all those letters doing in math, anyway?\n\nWhen you first learn someone's name, do you know everything there is to know about them? Sometimes, we know that there's _some_ number that fits in an equation, but we don't know what particular number it is yet. It's really helpful to be able to talk about the number anyway, in fact, giving it a name is often the first step to figuring out what it really is. \n\nThis kind of name is called a **variable**.\n\n## Doing lots of things at once\nAnother way names are useful is if we want to say something about lots of numbers at once!\nFor example, you might notice that $0 \\times 3 = 0$, $0 \\times -4 = 0$, $0 \\times 1224 = 0$ and so on. In fact, it's true that zero times _any_ number is zero. We could write this as $0 \\times \\text{any number} = 0$. Or if we need to keep referring to that number, we could give it a shorter name, while mentioning it could be any number: $0 \\times x = 0$, where $x$ can be any number. This is really useful because it allows us to express patterns much more easily!\n\n#Important patterns\nThese patterns hold for all [natural numbers](https://arbital.com/p/45h), [integers](https://arbital.com/p/48l), and [rational numbers](https://arbital.com/p/4zq), including for variables that are known to be one of these types of numbers, and much more! For an in-depth exploration of these patterns and their consequences, see the page on [rings](https://arbital.com/p/3gq). \n##Commutativity\n$$ a + b = b + a$$\n$$ a \\times b = b\\times a$$\n##Identity\n$$ 0 + a = a$$\n$$ 1 \\times a = a$$\n##Associativity \n$$ (a + b) + c = a + (b + c)$$\n$$ (a \\times b ) \\times c = a \\times (b\\times c)$$\n##Distributivity\n$$ a \\times (b + c) = a\\times b + a\\times c$$\n##Additive inverse\n$$ a + (-a) = a - a = 0 $$\n\n\n#Next steps\n* [https://arbital.com/p/Solving_equations](https://arbital.com/p/Solving_equations)\n* [Functions](https://arbital.com/p/3jy)", "date_published": "2016-08-06T13:10:55Z", "authors": ["Kevin Clancy", "Alexei Andreev", "Gabriel Sjöstrand", "Adele Lopez", "Eric Bruylant"], "summaries": [], "tags": ["C-Class"], "alias": "517"}
{"id": "bac452730468dbff7b9a92750a24e345", "title": "The reals (constructed as classes of Cauchy sequences of rationals) form a field", "url": "https://arbital.com/p/reals_as_classes_of_cauchy_sequences_form_a_field", "source": "arbital", "source_type": "text", "text": "The real numbers, when [constructed as equivalence classes of Cauchy sequences](https://arbital.com/p/50d) of [rationals](https://arbital.com/p/4zq), form a [totally ordered](https://arbital.com/p/540) [field](https://arbital.com/p/481), with the inherited field structure given by \n\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/a_n+b_n](https://arbital.com/p/a_n+b_n)$\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [\\times b_n](https://arbital.com/p/a_n)$\n- $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ if and only if either $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ or for sufficiently large $n$, $a_n \\leq b_n$.\n\n# Proof\n\nFirstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives $(x_n)_{n=1}^{\\infty}$ and $(y_n)_{n=1}^{\\infty}$ of the same equivalence class $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n)$, we don't somehow get different answers.\n\n## Well-definedness of $+$\n\nWe wish to show that $[https://arbital.com/p/x_n](https://arbital.com/p/x_n)+[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n) + [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ whenever $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n)$ and $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$; this is an exercise.\n%%hidden(Show solution):\nSince $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n)$, it must be the case that both $(x_n)$ and $(y_n)$ are Cauchy sequences such that $x_n - y_n \\to 0$ as $n \\to \\infty$.\nSimilarly, $a_n - b_n \\to 0$ as $n \\to \\infty$.\n\nWe require $[https://arbital.com/p/x_n+a_n](https://arbital.com/p/x_n+a_n) = [https://arbital.com/p/y_n+b_n](https://arbital.com/p/y_n+b_n)$; that is, we require $x_n+a_n - y_n-b_n \\to 0$ as $n \\to \\infty$.\n\nBut this is true: if we fix rational $\\epsilon > 0$, we can find $N_1$ such that for all $n > N_1$, we have $|x_n - y_n| < \\frac{\\epsilon}{2}$; and we can find $N_2$ such that for all $n > N_2$, we have $|a_n - b_n| < \\frac{\\epsilon}{2}$.\nLetting $N$ be the maximum of the two $N_1, N_2$, we have that for all $n > N$, $|x_n + a_n - y_n - b_n| \\leq |x_n - y_n| + |a_n - b_n|$ by the [https://arbital.com/p/-triangle_inequality](https://arbital.com/p/-triangle_inequality), and hence $\\leq \\epsilon$.\n%% \n\n## Well-definedness of $\\times$\n\nWe wish to show that $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ whenever $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) = [https://arbital.com/p/y_n](https://arbital.com/p/y_n)$ and $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$; this is also an exercise.\n%%hidden(Show solution):\nWe require $[a_n](https://arbital.com/p/x_n) = [b_n](https://arbital.com/p/y_n)$; that is, $x_n a_n - y_n b_n \\to 0$ as $n \\to \\infty$.\n\nLet $\\epsilon > 0$ be rational.\nThen $$|x_n a_n - y_n b_n| = |x_n (a_n - b_n) + b_n (x_n - y_n)|$$ using the very handy trick of adding the expression $x_n b_n - x_n b_n$.\n\nBy the triangle inequality, this is $\\leq |x_n| |a_n - b_n| + |b_n| |x_n - y_n|$.\n\nWe now use the fact that [https://arbital.com/p/cauchy_sequences_are_bounded](https://arbital.com/p/cauchy_sequences_are_bounded), to extract some $B$ such that $|x_n| < B$ and $|b_n| < B$ for all $n$;\nthen our expression is less than $B (|a_n - b_n| + |x_n - y_n|)$.\n\nFinally, for $n$ sufficiently large we have $|a_n - b_n| < \\frac{\\epsilon}{2 B}$, and similarly for $x_n$ and $y_n$, so the result follows that $|x_n a_n - y_n b_n| < \\epsilon$.\n%%\n\n## Well-definedness of $\\leq$\n\nWe wish to show that if $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ and $[https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/d_n](https://arbital.com/p/d_n)$, then $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ implies $[https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\leq [https://arbital.com/p/d_n](https://arbital.com/p/d_n)$.\n\nSuppose $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, but suppose for [contradiction](https://arbital.com/p/46z) that $[https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ is not $\\leq [https://arbital.com/p/d_n](https://arbital.com/p/d_n)$: that is, $[https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\not = [https://arbital.com/p/d_n](https://arbital.com/p/d_n)$ and there are arbitrarily large $n$ such that $c_n > d_n$.\nThen there are two cases.\n\n- If $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ then $[https://arbital.com/p/d_n](https://arbital.com/p/d_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$, so the result follows immediately. %%note:We didn't need the extra assumption that $[https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\not \\leq [https://arbital.com/p/d_n](https://arbital.com/p/d_n)$ here.%%\n- If for all sufficiently large $n$ we have $a_n \\leq b_n$, then \n\n## Additive [commutative](https://arbital.com/p/3jb) [group structure](https://arbital.com/p/3gd) on $\\mathbb{R}$\n\nThe additive identity is $[https://arbital.com/p/0](https://arbital.com/p/0)$ (formally, the equivalence class of the sequence $(0, 0, \\dots)$).\nIndeed, $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/0](https://arbital.com/p/0) = [https://arbital.com/p/a_n+0](https://arbital.com/p/a_n+0) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$.\n\nThe additive inverse of the element $[https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ is $[https://arbital.com/p/-a_n](https://arbital.com/p/-a_n)$, because $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/-a_n](https://arbital.com/p/-a_n) = [https://arbital.com/p/a_n-a_n](https://arbital.com/p/a_n-a_n) = [https://arbital.com/p/0](https://arbital.com/p/0)$.\n\nThe operation is commutative: $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/a_n+b_n](https://arbital.com/p/a_n+b_n) = [https://arbital.com/p/b_n+a_n](https://arbital.com/p/b_n+a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$.\n\nThe operation is closed, because the sum of two Cauchy sequences is a Cauchy sequence (exercise).\n%%hidden(Show solution):\nIf $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\\epsilon > 0$.\nWe wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n+b_n - a_m - b_m| < \\epsilon$.\n\nBut $|a_n+b_n - a_m - b_m| \\leq |a_n - a_m| + |b_n - b_m|$ by the triangle inequality; so picking $N$ so that $|a_n - a_m| < \\frac{\\epsilon}{2}$ and $|b_n - b_m| < \\frac{\\epsilon}{2}$ for all $n, m > N$, the result follows.\n%%\n\nThe operation is associative: $$[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + ([https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n)) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/b_n+c_n](https://arbital.com/p/b_n+c_n) = [https://arbital.com/p/a_n+b_n+c_n](https://arbital.com/p/a_n+b_n+c_n) = [https://arbital.com/p/a_n+b_n](https://arbital.com/p/a_n+b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) = ([https://arbital.com/p/a_n](https://arbital.com/p/a_n)+[https://arbital.com/p/b_n](https://arbital.com/p/b_n))+[https://arbital.com/p/c_n](https://arbital.com/p/c_n)$$\n\n## [Ring structure](https://arbital.com/p/3gq)\n\nThe multiplicative identity is $[https://arbital.com/p/1](https://arbital.com/p/1)$ (formally, the equivalence class of the sequence $(1,1, \\dots)$).\nIndeed, $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/1](https://arbital.com/p/1) = [\\times 1](https://arbital.com/p/a_n) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$.\n\n$\\times$ is closed, because the product of two Cauchy sequences is a Cauchy sequence (exercise).\n%%hidden(Show solution):\nIf $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\\epsilon > 0$.\nWe wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n b_n - a_m b_m| < \\epsilon$.\n\nBut $$|a_n b_n - a_m b_m| = |a_n (b_n - b_m) + b_m (a_n - a_m)| \\leq |b_m| |a_n - a_m| + |a_n| |b_n - b_m|$$ by the triangle inequality.\n\nCauchy sequences are bounded, so there is $B$ such that $|a_n|$ and $|b_m|$ are both less than $B$ for all $n$ and $m$.\n\nSo picking $N$ so that $|a_n - a_m| < \\frac{\\epsilon}{2B}$ and $|b_n - b_m| < \\frac{\\epsilon}{2B}$ for all $n, m > N$, the result follows.\n%%\n\n$\\times$ is clearly commutative: $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [\\times b_n](https://arbital.com/p/a_n) = [\\times a_n](https://arbital.com/p/b_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n) \\times [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$.\n\n$\\times$ is associative: $$[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times ([https://arbital.com/p/b_n](https://arbital.com/p/b_n) \\times [https://arbital.com/p/c_n](https://arbital.com/p/c_n)) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [\\times c_n](https://arbital.com/p/b_n) = [\\times b_n \\times c_n](https://arbital.com/p/a_n) = [\\times b_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/c_n](https://arbital.com/p/c_n) = ([https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)) \\times [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$$\n\n$\\times$ distributes over $+$: we need to show that $[https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times ([https://arbital.com/p/a_n](https://arbital.com/p/a_n)+[https://arbital.com/p/b_n](https://arbital.com/p/b_n)) = [https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$.\nBut this is true: $$[https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times ([https://arbital.com/p/a_n](https://arbital.com/p/a_n)+[https://arbital.com/p/b_n](https://arbital.com/p/b_n)) = [https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/a_n+b_n](https://arbital.com/p/a_n+b_n) = [\\times ](https://arbital.com/p/x_n) = [\\times a_n + x_n \\times b_n](https://arbital.com/p/x_n) = [\\times a_n](https://arbital.com/p/x_n) + [\\times b_n](https://arbital.com/p/x_n) = [https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/x_n](https://arbital.com/p/x_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$$\n\n## Field structure\n\nTo get from a ring to a field, it is necessary and sufficient to find a multiplicative inverse for any $[https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ not equal to $[https://arbital.com/p/0](https://arbital.com/p/0)$.\n\nSince $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\not = 0$, there is some $N$ such that for all $n > N$, $a_n \\not = 0$.\nThen defining the sequence $b_i = 1$ for $i \\leq N$, and $b_i = \\frac{1}{a_i}$ for $i > N$, we obtain a sequence which induces an element $[https://arbital.com/p/b_n](https://arbital.com/p/b_n)$ of $\\mathbb{R}$; and it is easy to check that $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/1](https://arbital.com/p/1)$.\n%%hidden(Show solution):\n$[https://arbital.com/p/a_n](https://arbital.com/p/a_n) [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [b_n](https://arbital.com/p/a_n)$; but the sequence $(a_n b_n)$ is $1$ for all $n > N$, and so it lies in the same equivalence class as the sequence $(1, 1, \\dots)$.\n%%\n\n## Ordering on the field\n\nWe need to show that:\n\n- if $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, then for every $[https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ we have $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$;\n- if $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ and $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, then $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times[https://arbital.com/p/b_n](https://arbital.com/p/b_n)$.\n\nWe may assume that the inequalities are strict, because if equality holds in the assumption then everything is obvious.\n%%hidden(Show obvious bits):\nIf $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, then for every $[https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ we have $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) = [https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ by well-definedness of addition.\nTherefore $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$.\n\nIf $[https://arbital.com/p/0](https://arbital.com/p/0) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ and $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, then $[https://arbital.com/p/0](https://arbital.com/p/0) = [https://arbital.com/p/0](https://arbital.com/p/0) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n) = [https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, so it is certainly true that $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$.\n%%\n\nFor the former: suppose $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) < [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, and let $[https://arbital.com/p/c_n](https://arbital.com/p/c_n)$ be an arbitrary equivalence class.\nThen $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) = [https://arbital.com/p/a_n+c_n](https://arbital.com/p/a_n+c_n)$; $[https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) = [https://arbital.com/p/b_n+c_n](https://arbital.com/p/b_n+c_n)$; but we have $a_n + c_n \\leq b_n + c_n$ for all sufficiently large $n$, because $a_n \\leq b_n$ for sufficiently large $n$.\nTherefore $[https://arbital.com/p/a_n](https://arbital.com/p/a_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n) \\leq [https://arbital.com/p/b_n](https://arbital.com/p/b_n) + [https://arbital.com/p/c_n](https://arbital.com/p/c_n)$, as required.\n\nFor the latter: suppose $[https://arbital.com/p/0](https://arbital.com/p/0) < [https://arbital.com/p/a_n](https://arbital.com/p/a_n)$ and $[https://arbital.com/p/0](https://arbital.com/p/0) < [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$.\nThen for sufficiently large $n$, we have both $a_n$ and $b_n$ are positive; so for sufficiently large $n$, we have $a_n b_n \\geq 0$.\nBut that is just saying that $[https://arbital.com/p/0](https://arbital.com/p/0) \\leq [https://arbital.com/p/a_n](https://arbital.com/p/a_n) \\times [https://arbital.com/p/b_n](https://arbital.com/p/b_n)$, as required.", "date_published": "2016-07-05T20:06:56Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": [], "tags": ["Proof"], "alias": "51h"}
{"id": "798a8b9c3590d2d52670db453e194679", "title": "Likelihood notation", "url": "https://arbital.com/p/likelihood_notation", "source": "arbital", "source_type": "text", "text": "The likelihood of a piece of evidence $e$ according to a hypothesis $H,$ known as \"the likelihood of $e$ given $H$\", is often written either $\\mathcal L_e(H)$ or $\\mathcal L(H \\mid e).$ The latter notation is confusing, because then $\\mathcal L(H \\mid e) = \\mathbb P(e \\mid H).$ Many students of statistics find it hard enough to keep the difference between $\\mathbb P(H \\mid e)$ and $\\mathbb P(e \\mid H)$ straight in their heads if we _don't_ occasionally swap the order of the arguments when talking about similar functions, so on Arbital, we much prefer the notation $\\mathcal L_e(H) = \\mathbb P(e \\mid H).$\n\n[Make this a child of 'likelihood' when 'likelihood' exists.](https://arbital.com/p/fixme:)", "date_published": "2016-07-07T03:16:06Z", "authors": ["Nate Soares", "Alexei Andreev"], "summaries": [], "tags": ["Start", "Non-standard terminology", "Definition"], "alias": "51n"}
{"id": "38c55fd6224ae5dbb9029cd067bde90a", "title": "Modal logic", "url": "https://arbital.com/p/modal_logic", "source": "arbital", "source_type": "text", "text": "Modal logic is a [https://arbital.com/p/formal_system](https://arbital.com/p/formal_system) in which we expand [https://arbital.com/p/-propositional_calculus](https://arbital.com/p/-propositional_calculus) with two new operators meant to represent necessity ($\\square$) and possibility ($\\diamond$).%%note:There are more [general modal logics](https://arbital.com/p/) with different modal predicates out there.%%\n\nThose two operators can be defined in terms of each other. For example, we have that something is possible iff it is not necessary that its opposite is true; ie, $\\diamond A \\iff \\neg \\square \\neg A$.\n\nThus in modal logic, you can express sentences as $\\square A$ to express that it is necessary that $A$ is true. You can also go more abstract and say $\\neg\\square \\bot$ to express that it is not necessary that false is true. If we read the box operator as \"there is no mathematical proof of\" then the previous sentence becomes \"there is no mathematical proof of falsehood\", which encodes the [consistency of arithmetic](https://arbital.com/p/).\n\nThere are many systems of modal logic %note: See for example [T](https://arbital.com/p/t_modal_logic),[B](https://arbital.com/p/b_modal_logic),[S5](https://arbital.com/p/s5_modal_logic),[S4](https://arbital.com/p/s4_modal_logic),[K](https://arbital.com/p/k_modal_logic), [GLS](https://arbital.com/p/gls_modal_logic)%, which differ in the sentences they take as [https://arbital.com/p/-axioms](https://arbital.com/p/-axioms) and which [rules of inference](https://arbital.com/p/) they allow.\n\nOf particular interest are the systems called [normal systems of modal logic](https://arbital.com/p/5j8), and especially the [$Gödel-Löb$](https://arbital.com/p/5l3) system (called $GL$ for short, also known as the logic of provability). The main interest of $GL$ comes from being a [decidable](https://arbital.com/p/) logic which allows us to reason about sentences of [Peano arithmetic](https://arbital.com/p/3ft) talking about [provability](https://arbital.com/p/5gt) via [translations](https://arbital.com/p/), as proved by the [arithmetical adequacy theorems of Solomonov](https://arbital.com/p/).\n\nAnother widely studied system of modal logic is $GLS$, for which Solomonov proved [adequacy for truth](https://arbital.com/p/) in the [https://arbital.com/p/-standard_model_of_arithmetic](https://arbital.com/p/-standard_model_of_arithmetic).\n\nThe [semantics](https://arbital.com/p/semantics_mathematical) of the normal systems of modal logic come in the form of [https://arbital.com/p/Kripke_models](https://arbital.com/p/Kripke_models): digraph structures composed of worlds over which a visibility relation is defined.\n\n##Sentences\nWell formed sentences of modal logic are defined as follows:\n\n* $\\bot$ is a well-formed sentence of modal logic.\n* The sentence letters $p,q,...$ are well-formed sentences of modal logic.\n* If $A$ is a well-formed sentence, then $\\square A$ is a well-formed sentence.\n* If $A$ and $B$ are well formed sentences, then $A\\to B$ is a well-formed sentence.\n\nThe rest of [logical connectors](https://arbital.com/p/) can then be defined as abbreviations for structures made from implication and $\\bot$. For example , $\\neg A = A\\to \\bot$. The possibility operator $\\diamond$ is defined as $\\neg\\square\\neg$ as per the previous discussion.", "date_published": "2016-07-27T19:13:12Z", "authors": ["Eric Bruylant", "Jaime Sevilla Molina", "Patrick LaVictoire"], "summaries": [], "tags": ["Start"], "alias": "534"}
{"id": "6b1f182b0095ca076e7b66a3b5452e36", "title": "Integers: Intro (Math 0)", "url": "https://arbital.com/p/integers_intro_math0", "source": "arbital", "source_type": "text", "text": "The integers are an extension of the [natural numbers](https://arbital.com/p/45h) into the *negatives*.\n\n## Negative numbers\n\nNegative numbers are numbers less than zero. They are notated as numbers with a minus sign in front of them, like $-2$ or $-15$ or $-6387$.\n\nWhen many people are first introduced to the concept of negative numbers, one question they have is, \"how do negative numbers exist? How can we have less than nothing of something?\"\n\nUsually this question is answered with something about travelling along the number line and then going in the opposite direction past zero. But we can also answer this question from the traditional perspective of quantities, using _antimatter_.\n\n## Arithmetic with anti-cows\n\nMatter and antimatter are substances that annihilate each other — they cancel each other out. In the same way, positive and negative quantities cancel each other out to some extent when added together.\n\n\n\nLet's create a physical example of this antimatter phenomenon, and consider a collection of cows. [diagram of cows](https://arbital.com/p/comment:) If a natural number is how many cows we have, then negative numbers are the presence of *anti-cows*. [For the sake of easy reading, let's colour a cow red, and an anti-cow blue: diagram of red cow = cow, blue cow = anticow](https://arbital.com/p/comment:) For example, if you have $3$ anti-cows, you have $-3$ cows. If you have $128$ anti-cows, you have $-128$ cows, and so on.\n\nWhen a cow and an anti-cow come into contact, they annihilate each other and nothing is left (except technically an enormous burst of energy that could cause a magnitude 9 earthquake, but we won't get into the nitty-gritty physics of antimatter annihilation here). [diagram of cow/anti-cow annihilation](https://arbital.com/p/comment:) Moreover, each cow will annihilate exactly one anti-cow, and vice versa. \n\nThis lets us perform subtraction by adding anti-cows instead of taking away cows. If you want to perform $6 - 4$, you imagine you have a collection of six cows, and you add four anti-cows to the mix, which annihilate four of the cows leaving you with two. This is exactly the same as if you'd taken away four cows (except that you've now released enough energy to power the entire world's energy demands for seven months). [diagram of 6 - 4](https://arbital.com/p/comment:)\n\nThis form of adding negatives is slightly more flexible than regular subtraction. In regular subtraction, you couldn't take away more cows than there already were. But when you add negatives, you can keep adding anti-cows after there are no cows left, at which point you just have more and more anti-cows, giving you a more and more negative number. So if you wanted to subtract $4 - 6$, you couldn't take away six cows from four, but you could add six anti-cows, and four of them would annihilate the four cows, leaving you with two anti-cows or a final answer of $-2$. [diagram of 4 - 6](https://arbital.com/p/comment:)\n\n## Useful properties of negative numbers\n\nExtending the natural numbers into negatives gives us lots of useful properties when doing arithmetic.\n\n### Addition of negative numbers\n\nNegative numbers add up just like positive ones. If you have $-3$ and $-7$, they add up to $-10$, just like three anti-cows and seven anti-cows would come together to make ten anti-cows. [diagram of ](https://arbital.com/p/comment:)\n\nKnowing this, how would you calculate $(-6) + (-8) + (-12) + (-20)$?\n\n%%hidden(Show solution):\nSimply add up all the negative quantities together. $$6 + 8 + 12 + 20 = 46 \\to (-6) + (-8) + (-12) + (-20) = -46$$\n%%\n\n### Subtraction as additive inversion\n\nIf we represent a natural number as a number of cows, then the same number of anti-cows is called the *additive inverse* of that number, because a number and its additive inverse add up to zero (equal numbers of cows and anti-cows will all annihilate each other to give nothing), which is the \"additive [identity](https://arbital.com/p/54p)\" (or \"the number that changes nothing when you add it to something\").\n\nUsing additive inverses allows us to express subtraction in the form of addition — for example, $5 - 2$ can be rewritten as $5 + (-2)$ (which just means that instead of taking away two cows, you're adding two anti-cows). When we do this, suddenly we can rearrange subtraction operations along with addition operations even though subtraction isn't [commutative](https://arbital.com/p/3jb).\n\nThen, if you have a giant addition and subtraction problem you want to solve, you can add up all the numbers you want to add (into a big collection of cows), then add up all the numbers you want to subtract (into a big collection of anti-cows), and just do one subtraction (annihilating cow/anti-cow pairs) to get your answer. For example, $6 - 2 + 7 - 5$ is really just $6 + (-2) + 7 + (-5)$ which is equal to $6 + 7 + (-5) + (-2)$. Then, you add up the number of cows (13) and anti-cows (7), and annihilate them to get the final answer of $6$.\n\nHere's another exercise for you: How would you calculate $13 + 8 - 5 + 6 + 4 - 12 - 9 + 1$?\n\n%%hidden(Show solution):\nFirst, rewrite each subtraction operation as addition of a negative number: $$13 + 8 + (-5) + 6 + 4 + (-12) + (-9) + 1$$\n\nThen rearrange the equation to group all the positive and negative numbers (cows and anti-cows) together: $$13 + 8 + 6 + 4 + 1 + (-5) + (-12) + (-9)$$\n\nThen add up all the positive and negative numbers separately: $$(13 + 8 + 6 + 4 + 1) + ((-5) + (-12) + (-9)) = 32 + (-26)$$\n\nAnd finally, subtract it all at once: $$32 + (-26) = 32 - 26 = 6$$\n\nAnd voilà, you have six cows.\n%%\n\nWhat about $8 - 6 + 4 - 13 + 7 - 5 - 9 + 12$?\n\n%%hidden(Show solution):\nSame as before, we rewrite, regroup, add up, and subtract:\n\n$$8 + (-6) + 4 + (- 13) + 7 + (- 5) + (- 9) + 12 \\\\ \\Downarrow$$\n\n$$8 + 4 + 7 + 12 + (-6) + (-13) + (-5) + (-9) \\\\ \\Downarrow$$ \n\n$$(8 + 4 + 7 + 12) + ((-6) + (-13) + (-5) + (-9)) = 31 + (-33)$$\n\n$$31 + (-33) = 31 - 33$$\n\nBut hold on — now you have more anti-cows than cows, so you can't subtract directly! Not to worry — simply flip the subtraction problem around, and subtract the cows from the anti-cows, knowing that the final result will be negative instead.\n\n$$31 - 33 = -(33 - 31) = -2$$\n%%\n\n## The set of integers\n\nBack to defining the integers. The set of integers is simply the set of natural numbers and their additive inverses — i.e. the set of numbers of cows you can have if you can have anti-cows as well. This set branches out to infinity on both sides, and we usually write it as $\\{ \\ldots, -2, -1, 0, 1, 2, \\ldots \\}$.", "date_published": "2016-07-07T15:30:38Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": ["The integers are an extension of the [natural numbers](https://arbital.com/p/45h) into the *negatives*.\n\nIn this article we'll be exploring a novel way to think of negative numbers, using collections of cows and anti-cows. *Mooooo!*"], "tags": ["Math 0", "C-Class"], "alias": "53r"}
{"id": "093e46b309e9643f101e6f85e5bf4d7f", "title": "The reals (constructed as Dedekind cuts) form a field", "url": "https://arbital.com/p/reals_as_dedekind_cuts_form_a_field", "source": "arbital", "source_type": "text", "text": "The real numbers, when [constructed as Dedekind cuts](https://arbital.com/p/50g) over the [rationals](https://arbital.com/p/4zq), form a [field](https://arbital.com/p/481).\n\nWe shall often write the one-sided [Dedekind cut](https://arbital.com/p/dedekind_cut) $(A, B)$ %%note:Recall: \"one-sided\" means that $A$ has no greatest element.%% as simply $\\mathbf{A}$ (using bold face for Dedekind cuts); we can do this because if we already know $A$ then $B$ is completely determined.\nThis will make our notation less messy.\n\nThe field structure, together with the [total ordering](https://arbital.com/p/total_order) on it, is as follows (where we write $\\mathbf{0}$ for the Dedekind cut $(\\{ r \\in \\mathbb{Q} \\mid r < 0\\}, \\{ r \\in \\mathbb{Q} \\mid r \\geq 0 \\})$): \n\n- $(A, B) + (C, D) = (A+C, B+D)$\n- $\\mathbf{A} \\leq \\mathbf{C}$ if and only if everything in $A$ lies in $C$.\n- Multiplication is somewhat complicated.\n - If $\\mathbf{0} \\leq \\mathbf{A}$, then $\\mathbf{A} \\times \\mathbf{C} = \\{ a c \\mid a \\in A, a > 0, c \\in C \\}$. \n - If $\\mathbf{A} < \\mathbf{0}$ and $\\mathbf{0} \\leq \\mathbf{C}$, then $\\mathbf{A} \\times \\mathbf{C} = \\{ a c \\mid a \\in A, c \\in C, c > 0 \\}$.\n - If $\\mathbf{A} < \\mathbf{0}$ and $\\mathbf{C} < \\mathbf{0}$, then $\\mathbf{A} \\times \\mathbf{C} = \\{\\} $ \n\nwhere $(A, B)$ is a one-sided [Dedekind cut](https://arbital.com/p/dedekind_cut) (so that $A$ has no greatest element).\n\n(Here, the \"set sum\" $A+C$ is defined as \"everything that can be made by adding one thing from $A$ to one thing from $C$\": namely, $\\{ a+c \\mid a \\in A, c \\in C \\}$ in [https://arbital.com/p/-3lj](https://arbital.com/p/-3lj); and $A \\times C$ is similarly $\\{ a \\times c \\mid a \\in A, c \\in C \\}$.)\n\n# Proof\n\n## Well-definedness\n\nWe need to show firstly that these operations do in fact produce [Dedekind cuts](https://arbital.com/p/dedekind_cut).\n\n### Addition\nFirstly, we need everything in $A+C$ to be less than everything in $B+D$.\nThis is true: if $a+c \\in A+C$, and $b+d \\in B+D$, then since $a < b$ and $c < d$, we have $a+c < b+d$.\n\nNext, we need $A+C$ and $B+D$ together to contain all the rationals.\nThis is true: \n\nFinally, we need $(A+C, B+D)$ to be one-sided: that is, $A+C$ needs to have no top element, or equivalently, if $a+c \\in A+C$ then we can find a bigger $a' + c'$ in $A+C$.\nThis is also true: if $a+c$ is an element of $A+C$, then we can find an element $a'$ of $A$ which is bigger than $a$, and an element $c'$ of $C$ which is bigger than $C$ (since both $A$ and $C$ have no top elements, because the respective Dedekind cuts are one-sided); then $a' + c'$ is in $A+C$ and is bigger than $a+c$.\n\n### Multiplication\n\n\n### Ordering\n\n\n## Additive [commutative](https://arbital.com/p/3jb) [group structure](https://arbital.com/p/3gd)\n\n\n\n## [Ring structure](https://arbital.com/p/3gq)\n\n\n\n## [Field structure](https://arbital.com/p/481)\n\n\n\n## Ordering on the field", "date_published": "2016-07-08T15:20:19Z", "authors": ["Dylan Hendrickson", "Patrick Stevens"], "summaries": [], "tags": ["Work in progress", "Proof"], "alias": "53v"}
{"id": "28037ad90e5c24acd26c3053b3d8ea5f", "title": "Equivalence relation", "url": "https://arbital.com/p/equivalence_relation", "source": "arbital", "source_type": "text", "text": "An **equivalence relation** is a binary [https://arbital.com/p/-3nt](https://arbital.com/p/-3nt) $\\sim$ on a [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz) $S$ that can be used to say whether [elements](https://arbital.com/p/element_of_a_set) of $S$ are equivalent.\n\nAn equivalence relation satisfies the following properties:\n\n1. For any $x \\in S$, $x \\sim x$ (the [reflexive](https://arbital.com/p/reflexive_relation) property).\n2. For any $x,y \\in S$, if $x \\sim y$ then $y \\sim x$ (the [symmetric](https://arbital.com/p/symmetric_relation) property).\n3. For any $x,y,z \\in S$, if $x \\sim y$ and $y \\sim z$ then $x \\sim z$ (the [transitive](https://arbital.com/p/573) property).\n\nIntuitively, any element is equivalent to itself, equivalence isn't directional, and if two elements are equivalent to the same third element then they're equivalent.\n\n## Equivalence classes\n\nWhenever we have a set $S$ with an equivalence relation $\\sim$, we can divide $S$ into *equivalence classes*, i.e. sets of mutually equivalent elements. The equivalence class of some $x \\in S$ is the set of elements of $S$ equivalent to $x$, written $[https://arbital.com/p/x](https://arbital.com/p/x)=\\{y \\in S \\mid x \\sim y\\}$, and we say that $x$ is a \"representative\" of $[https://arbital.com/p/x](https://arbital.com/p/x)$. We call the set of equivalence classes $S/\\sim = \\{[https://arbital.com/p/x](https://arbital.com/p/x) \\mid x \\in S\\}$.\n\nFrom the definition of an equivalence relation, it's easy to show that $x \\in [https://arbital.com/p/x](https://arbital.com/p/x)$ and $[https://arbital.com/p/x](https://arbital.com/p/x)=[https://arbital.com/p/y](https://arbital.com/p/y)$ if and only if $x \\sim y$.\n\nIf you have a set already [partitioned](https://arbital.com/p/set_partition) into subsets ($S$ is the [https://arbital.com/p/-disjoint_union](https://arbital.com/p/-disjoint_union) of the elements of a collection $A$), you can go the other way and define a relation with two elements related whenever they're in the same subset ($x \\sim y$ when there is some $U \\in A$ with $x,y \\in U$). Then this is an equivalence relation, and the partition of the set is the set of equivalence classes under this relation (that is, $[https://arbital.com/p/x](https://arbital.com/p/x) \\in A$ and $A=S/\\sim$).\n\n##Defining functions on equivalence classes\n\nSuppose you have a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) $f: S \\to U$ and want to define a corresponding function $f^*: S/\\sim \\to U$, where $U$ can be any set. You define $f^*([https://arbital.com/p/x](https://arbital.com/p/x))$ to be $f(x)$. This could be a problem; what if $x \\sim y$ but $f(x) \\neq f(y)$? Then $f^*([https://arbital.com/p/x](https://arbital.com/p/x))=f^*([https://arbital.com/p/y](https://arbital.com/p/y))$ wouldn't be [well-defined](https://arbital.com/p/-well_defined). Whenever you define a function on equivalence classes in terms of representatives, you have to make sure the definition doesn't depend on which representative you happen to pick. In fact, one way you might arrive at an equivalence relation is to say that $x \\sim y$ whenever $f(x)=f(y)$.\n\nIf you have a function $f: S \\to S$ and want to define $f^*: S/\\sim \\to S/\\sim$ by $f^*([https://arbital.com/p/x](https://arbital.com/p/x))=[https://arbital.com/p/f](https://arbital.com/p/f)$, you have to verify that whenever $x \\sim y$, $[https://arbital.com/p/f](https://arbital.com/p/f)=[https://arbital.com/p/f](https://arbital.com/p/f)$, equivalently $f(x) \\sim f(y)$. \n\n#Examples\n\nConsider the [integers](https://arbital.com/p/48l) with the relation $x \\sim y$ if $n|x-y$, for some $n \\in \\mathbb N$. This is the integers [mod](https://arbital.com/p/modular_arithmetic) $n$. The [https://arbital.com/p/-addition](https://arbital.com/p/-addition) and [https://arbital.com/p/-multiplication](https://arbital.com/p/-multiplication) operations can be inherited from the integers, so it makes sense to talk about addition and multiplication mod $n$. \n\nThe [real numbers](https://arbital.com/p/4bc) can be defined as equivalence classes of [Cauchy sequences](https://arbital.com/p/50d).\n\n[https://arbital.com/p/4f4](https://arbital.com/p/4f4) is an equivalence relation (unless the objects form a [https://arbital.com/p/-proper_class](https://arbital.com/p/-proper_class)).", "date_published": "2016-07-07T16:52:44Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Joe Zeng"], "summaries": ["An **equivalence relation** is a binary [https://arbital.com/p/-3nt](https://arbital.com/p/-3nt) that is [reflexive](https://arbital.com/p/reflexive_relation), [symmetric](https://arbital.com/p/symmetric_relation), and [transitive](https://arbital.com/p/573). You can think of it as a way to say when two elements of a [https://arbital.com/p/-3jz](https://arbital.com/p/-3jz) are \"the same\" or \"equivalent,\" despite not being literally the same element. It can be used to [partition](https://arbital.com/p/set_partition) a set into equivalence classes."], "tags": ["Start"], "alias": "53y"}
{"id": "5e65505218386278b829414c51d961ec", "title": "Totally ordered set", "url": "https://arbital.com/p/totally_ordered_set", "source": "arbital", "source_type": "text", "text": "A **totally ordered set** is a pair $(S, \\le)$ of a set $S$ and a *total order* $\\le$ on $S$, which is a [https://arbital.com/p/-binary_relation](https://arbital.com/p/-binary_relation) that satisfies the following properties:\n\n1. For all $a, b \\in S$, if $a \\le b$ and $b \\le a$, then $a = b$. (the [antisymmetric](https://arbital.com/p/antisymmetric_relation) property)\n2. For all $a, b, c \\in S$, if $a \\le b$ and $b \\le c$, then $a \\le c$. (the [transitive](https://arbital.com/p/573) property)\n3. For all $a, b \\in S$, either $a \\le b$ or $b \\le a$, or both. (the [totality](https://arbital.com/p/total_relation) property)\n\nA totally ordered set is a special type of [partially ordered set](https://arbital.com/p/3rb) that satisfies the total property — in general, posets only satisfy the [reflexive](https://arbital.com/p/5dy) property, which is that $a \\le a$ for all $a \\in S$.\n\n## Examples of totally ordered sets\n\nThe [real numbers](https://arbital.com/p/4bc) are a totally ordered set. So are any of the subsets of the real numbers, such as the [rational numbers](https://arbital.com/p/4zq) or the [integers](https://arbital.com/p/48l).\n\n## Examples of not totally ordered sets\n\nThe [complex numbers](https://arbital.com/p/4zw) do not have a canonical total ordering, and especially not a total ordering that preserves all the properties of the ordering of the real numbers, although one can define a total ordering on them quite easily.", "date_published": "2016-07-21T16:07:16Z", "authors": ["Kevin Clancy", "Eric Rogstad", "Dylan Hendrickson", "Eric Bruylant", "Joe Zeng"], "summaries": ["A totally ordered set is a pair $(S, \\le)$ of a set $S$ and a *total order* $\\le$ on $S$, which is a [binary relation](https://arbital.com/p/3nt) that satisfies the following properties:\n\n1. For all $a, b \\in S$, if $a \\le b$ and $b \\le a$, then $a = b$. (the [antisymmetric](https://arbital.com/p/antisymmetric_relation) property)\n2. For all $a, b, c \\in S$, if $a \\le b$ and $b \\le c$, then $a \\le c$. (the [transitive](https://arbital.com/p/573) property)\n3. For all $a, b \\in S$, either $a \\le b$ or $b \\le a$, or both. (the [totality](https://arbital.com/p/total_relation) property)"], "tags": ["Start", "Needs summary"], "alias": "540"}
{"id": "715c30cac984cd8a3c8b9189d2a1a9b8", "title": "Intro to Number Sets", "url": "https://arbital.com/p/number_sets_intro", "source": "arbital", "source_type": "text", "text": "There are several common [sets](https://arbital.com/p/3jz) of [numbers](https://arbital.com/p/54y) that mathematicians use in their studies. In order from simple to complex, they are:\n\n1. The [natural numbers](https://arbital.com/p/506) $\\mathbb{N}$\n\n2. The [integers](https://arbital.com/p/53r) $\\mathbb{Z}$\n\n3. The [rational numbers](https://arbital.com/p/rational_number_math0) $\\mathbb{Q}$\n\n4. The [real numbers](https://arbital.com/p/real_number_math0) $\\mathbb{R}$\n\n5. The [complex numbers](https://arbital.com/p/complex_number_math0) $\\mathbb{C}$\n\nEach set is constructed in some way from the previous one, and this path will show you how they are constructed from the most basic numbers. You may have come across these terms in a math class that you attended, and may have had other definitions given to you. In this path, you will obtain a firm, complete understanding of these sets, how they are constructed, and what they mean in mathematics.\n\n\n## Why are number sets important?\n\nBefore we go any further though, it would be nice to know the motivation behind defining the number sets first.\n\nA [set](https://arbital.com/p/3jz) is a fancy name for a collection of objects. Some collections of objects have special properties — such as the set of all blue things, which are special in that they're all blue. In math, if a set of objects all have a certain property, we can make **inferences** about them — that is, there are certain things we can say about them that you can deduce logically. For example:\n\n> In a [field](https://arbital.com/p/481), every nonzero number has a multiplicative inverse.\n\nYou don't need to know what a field is yet (it's a special type of set), but now you can make inferences about them without restricting yourself to a specific example when talking about them. For example, you know that if a set _is_ a field, then every number in that set that isn't zero can divide into another number in that set (by multiplying by its \"multiplicative inverse\") and produce yet another number in that set.\n\nConversely, you can also tell when a set is or isn't a field based on whether it satisfies the properties a field has. For example, since you can't divide $3$ by $2$ (because the result is $1.5$ which is not a natural number), you now know that the natural numbers are not a field.\n\n\nNow let's turn to our first set: the natural numbers.", "date_published": "2016-08-20T14:27:36Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": [], "tags": ["Stub"], "alias": "544"}
{"id": "17ce2f3fd8dedfb2e0a362656c0b7cba", "title": "The square root of 2 is irrational", "url": "https://arbital.com/p/sqrt_2_is_irrational", "source": "arbital", "source_type": "text", "text": "$\\sqrt 2$, the unique [https://arbital.com/p/-positive](https://arbital.com/p/-positive) [https://arbital.com/p/-4bc](https://arbital.com/p/-4bc) whose square is 2, is not a [https://arbital.com/p/-4zq](https://arbital.com/p/-4zq).\n\n#Proof\n\nSuppose $\\sqrt 2$ is rational. Then $\\sqrt 2=\\frac{a}{b}$ for some integers $a$ and $b$; [https://arbital.com/p/-without_loss_of_generality](https://arbital.com/p/-without_loss_of_generality) let $\\frac{a}{b}$ be in [https://arbital.com/p/-lowest_terms](https://arbital.com/p/-lowest_terms), i.e. $\\gcd(a,b)=1$. We have\n\n$$\\sqrt 2=\\frac{a}{b}$$\n\nFrom the definition of $\\sqrt 2$,\n\n$$2=\\frac{a^2}{b^2}$$\n$$2b^2=a^2$$\n\nSo $a^2$ is a multiple of $2$. Since $2$ is [prime](https://arbital.com/p/4mf), $a$ must be a multiple of 2; let $a=2k$. Then\n\n$$2b^2=(2k)^2=4k^2$$\n$$b^2=2k^2$$\n\nSo $b^2$ is a multiple of $2$, and so is $b$. But then $2|\\gcd(a,b)$, which contradicts the assumption that $\\frac{a}{b}$ is in lowest terms! So there isn't any way to express $\\sqrt 2$ as a fraction in lowest terms, and thus there isn't a way to express $\\sqrt 2$ as a ratio of integers at all. That is, $\\sqrt 2$ is irrational.", "date_published": "2016-07-06T18:40:58Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Nate Soares"], "summaries": [], "tags": ["Proof"], "alias": "548"}
{"id": "4827a071f948b306aee595117652ff27", "title": "Order relation", "url": "https://arbital.com/p/order_relation", "source": "arbital", "source_type": "text", "text": "An **order relation** (also called an **order** or **ordering**) is a [binary relation](https://arbital.com/p/3nt) $\\le$ on a [set](https://arbital.com/p/3jz) $S$ that can be used to order the elements in that set.\n\nAn order relation satisfies the following properties:\n\n1. For all $a \\in S$, $a \\le a$. (the [reflexive](https://arbital.com/p/reflexive_relation) property)\n2. For all $a, b \\in S$, if $a \\le b$ and $b \\le a$, then $a = b$. (the [antisymmetric](https://arbital.com/p/antisymmetric_relation) property)\n3. For all $a, b, c \\in S$, if $a \\le b$ and $b \\le c$, then $a \\le c$. (the [transitive](https://arbital.com/p/transitive_relation) property)\n\nA set that has an order relation is called a [partially ordered set](https://arbital.com/p/3rb) (or \"poset\"), and $\\le$ is its *partial order*.\n\n## Totality of an order\n\nThere is also a fourth property that distinguishes between two different types of orders:\n\n4. For all $a, b \\in S$, either $a \\le b$ or $b \\le a$ or both. (the [total](https://arbital.com/p/total_relation) property)\n\nThe total property implies the reflexive property, by setting $a = b$.\n\nIf the order relation satisfies the total property, then $S$ is called a [https://arbital.com/p/-540](https://arbital.com/p/-540), and $\\le$ is its *total order*.\n\n## Well-ordering\n\nA fifth property that extends the idea of a \"total order\" is that of the [well-ordering](https://arbital.com/p/55r):\n\n5. For every subset $X$ of $S$, $X$ has a least element: an element $x$ such that for all $y \\in X$, we have $x \\leq y$.\n\nWell-orderings are very useful: they are the orderings we can perform [induction](https://arbital.com/p/mathematical_induction) over. (For more on this viewpoint, see the page on [https://arbital.com/p/structural_induction](https://arbital.com/p/structural_induction).)\n\n# Derived relations\n\nThe order relation immediately affords several other relations.\n\n## Reverse order\n\nWe can define a *reverse order* $\\ge$ as follows: $a \\ge b$ when $b \\le a$. \n\n## Strict order \n\nFrom any poset $(S, \\le)$, we can derive a *strict order* $<$, which disallows equality. For $a, b \\in S$, $a < b$ when $a \\le b$ and $a \\neq b$. This strict order is still antisymmetric and transitive, but it is no longer reflexive.\n\nWe can then also define a reverse strict order $>$ as follows: $a > b$ when $b \\le a$ and $a \\neq b$.\n\n## Incomparability\n\nIn a poset that is not totally ordered, there exist elements $a$ and $b$ where the order relation is undefined. If neither $a \\leq b$ nor $b \\leq a$ then we say that $a$ and $b$ are *incomparable*, and write $a \\parallel b$. \n\n## Cover relation\n\nFrom any poset $(S, \\leq)$, we can derive an underlying *cover relation* $\\prec$, defined such that for $a, b \\in S$, $a \\prec b$ whenever the following two conditions are satisfied:\n\n1. $a < b$.\n2. For all $s \\in S$, $a \\leq s < b$ implies that $a = s$.\n\nSimply put, $a \\prec b$ means that $b$ is the smallest element of $S$ which is strictly greater than $a$.\n$a \\prec b$ is pronounced \"$a$ is covered by $b$\", or \"$b$ covers $a$\", and $b$ is said to be a *cover* of $a$.", "date_published": "2016-07-07T14:32:44Z", "authors": ["Patrick Stevens", "Joe Zeng"], "summaries": [], "tags": ["Formal definition"], "alias": "549"}
{"id": "6c2fcfeb283d36030e42d827a914b170", "title": "Identity element", "url": "https://arbital.com/p/identity_element", "source": "arbital", "source_type": "text", "text": "An identity [element](https://arbital.com/p/5xy) in a set $S$ with a [binary operation](https://arbital.com/p/-3kb) $*$ is an element $i$ that leaves any element $a \\in S$ unchanged when combined with it in that operation.\n\nFormally, we can define an element $i$ to be an identity element if the following two statements are true:\n\n1. For all $a \\in S$, $i * a = a$. If only this statement is true then $i$ is said to be a *left identity*.\n2. For all $a \\in S$, $a * i = a$. If only this statement is true then $i$ is said to be a *right identity*.\n\nThe existence of an identity element is a property of many algebraic structures, such as [groups](https://arbital.com/p/3gd), [rings](https://arbital.com/p/3gq), and [fields](https://arbital.com/p/481).", "date_published": "2016-08-20T11:19:57Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": ["An identity [element](https://arbital.com/p/5xy) in a set $S$ with a [binary operation](https://arbital.com/p/-3kb) $*$ is an element $i$ that leaves any element $a \\in S$ unchanged when combined with it in that operation."], "tags": ["Math 2", "Formal definition"], "alias": "54p"}
{"id": "44742fd2a0effcba795df70508671669", "title": "Proof that there are infinitely many primes", "url": "https://arbital.com/p/infinitely_many_primes", "source": "arbital", "source_type": "text", "text": "The proof that there are infinitely many [prime numbers](https://arbital.com/p/4mf) is a [https://arbital.com/p/-46z](https://arbital.com/p/-46z).\n\nFirst, we note that there is indeed a prime: namely $2$.\nWe will also state a lemma: that every number greater than $1$ has a prime which divides it.\n(This is the easier half of the [https://arbital.com/p/5rh](https://arbital.com/p/5rh), and the slightly stronger statement that \"every number may be written as a product of primes\" is proved there.)\n\nNow we can proceed to the meat of the proof.\nSuppose that there were finitely many prime numbers $p_1, p_2, \\ldots, p_n$.\nSince we know $2$ is prime, we know $2$ appears in that list.\n\nThen consider the number $P = p_1p_2\\ldots p_n + 1$ — that is, the product of all primes plus 1.\nSince $2$ appeared in the list, we know $P \\geq 2+1 = 3$, and in particular $P > 1$.\n\nThe number $P$ can't be divided by any of the primes in our list, because it's 1 more than a multiple of them.\nBut there is a prime which divides $P$, because $P>1$; we stated this as a lemma.\nThis is immediately a contradiction: $P$ cannot be divided by any prime, even though all integers greater than $1$ can be divided by a prime.\n\n# Example\n\nThere is a common misconception that $p_1 p_2 \\dots p_n+1$ must be prime if $p_1, \\dots, p_n$ are all primes.\nThis isn't actually the case: if we let $p_1, \\dots, p_6$ be the first six primes $2,3,5,7,11,13$, then you can check by hand that $p_1 \\dots p_6 + 1 = 30031$; but $30031 = 59 \\times 509$.\n(These are all somewhat ad-hoc; there is no particular reason I knew that taking the first six primes would give me a composite number at the end.)\nHowever, we *have* discovered a new prime this way (in fact, two new primes!): namely $59$ and $509$.\n\nIn general, this is a terrible way to discover new primes.\nThe proof tells us that there must be some new prime dividing $30031$, without telling us anything about what those primes are, or even if there is more than one of them (that is, it doesn't tell us whether $30031$ is prime or composite).\nWithout knowing in advance that $30031$ is equal to $59 \\times 509$, it is in general very difficult to *discover* those two prime factors.\nIn fact, it's an open problem whether or not prime factorisation is \"easy\" in the specific technical sense of there being a polynomial-time algorithm to do it, though most people believe that prime factorisation is \"hard\" in this sense.", "date_published": "2016-08-15T11:09:43Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": [], "tags": ["Proof"], "alias": "54r"}
{"id": "6817747af45130faaa62d1ae0082da51", "title": "Order of operations", "url": "https://arbital.com/p/order_of_operations", "source": "arbital", "source_type": "text", "text": "The **order of operations** is a notational convention employed in arithmetical expressions to disambiguate the order in which operations should be performed.\n\nIt is also referred to PEMDAS, BEDMAS, BODMAS, or other such acronyms depending on where you live. The acronym is always 6 letters, and refers to the following things in order:\n\n1. [https://arbital.com/p/parentheses](https://arbital.com/p/parentheses) (or Brackets), in order of inside-most parentheses first\n2. [Exponents](https://arbital.com/p/exponent) (or Orders or Indices), in order of inside-most (or rightmost) exponents first\n3. [https://arbital.com/p/multiplication](https://arbital.com/p/multiplication) and [https://arbital.com/p/division](https://arbital.com/p/division), in order from left to right\n4. [https://arbital.com/p/addition](https://arbital.com/p/addition) and [https://arbital.com/p/subtraction](https://arbital.com/p/subtraction), in order from left to right\n\n\n## Motivation\n\nThe infix notation we use to write expressions of mathematical operators is inherently ambiguous. Only in some very special examples where the expression consists of only a single [associative](https://arbital.com/p/3h4) operation do all possible orders of evaluation evaluate to the same thing. When considering the expression $2 - 4 + 3$, do we do the $2 - 4$ first or the $4 + 3$ first? The result is either $1$ or $-5$ depending on which you choose. When considering the expression $7 + 8 \\times 9 - 6$, which operation do we do first? The result could be anywhere from $31$ to $129$ depending on the order you do them in.\n\nTherefore, to relieve mathematicians of placing brackets around every expression, there are some standard, intuitive conventions put in place.\n\n\n## Example expressions\n\nConsider the expression $3 + 7 \\times 2^{(6 + 8)}$. We do parentheses first: $$3 + 7 \\times 2^{14}$$ Then exponents: $$3 + 7 \\times 16384$$ Then multiplication and division: $$3 + 114688$$ And finally addition and subtraction: $$114691$$\n\n\n## Notable ambiguous cases\n\nThe expression $48 \\div 2 (9 + 3)$ is a controversial ambiguity in the order of operations that results from an ambiguity in multiplication by juxtaposition — which is, the convention that numbers next to each other in brackets should be multiplied together. For example, $2(3+5)$ is equal to $2 \\times (3 + 5)$ or $16$.\n\nIf multiplication by juxtaposition is taken as a _parenthetical_ operation due to the parentheses in the operands, then the result is 2:\n\n$$\\begin{align*}\n48 \\div 2 (9 + 3) &= 48 \\div 2 (12) \\\\\n&= 48 \\div 24 \\\\\n&= 2\n\\end{align*}$$\n\nHowever, if multiplication by juxtaposition is taken as simply a regular _multiplication_ operation, the result is 288: \n\n$$\\begin{align*}\n48 \\div 2 (9 + 3) &= 48 \\div 2 \\times 12 \\\\\n&= 24 \\times 12 \\\\\n&= 288\n\\end{align*}$$\n\nThe second step is different because the 48 and 2 are operated on first, as the left most division or multiplication operation in the expression.", "date_published": "2016-07-06T21:01:11Z", "authors": ["Eric Bruylant", "Patrick Stevens", "Joe Zeng"], "summaries": [], "tags": ["Start"], "alias": "54s"}
{"id": "a8f103c953c6b1e1a3e7208c0b87fbbb", "title": "Whole number", "url": "https://arbital.com/p/whole_number", "source": "arbital", "source_type": "text", "text": "A **whole number** is an intuitive concept of a number that is not [fractional](https://arbital.com/p/). There are three different definitions of the whole numbers, whose usage depends heavily on the author:\n\n1. The [natural numbers](https://arbital.com/p/45h).\n\n2. The natural numbers and [https://arbital.com/p/-zero](https://arbital.com/p/-zero) (in definitions of the [https://arbital.com/p/-number_sets](https://arbital.com/p/-number_sets) where the natural numbers do not contain zero).\n\n3. The [integers](https://arbital.com/p/48l).\n\nThe set of whole numbers, when it is separate from the natural numbers and the integers, is usually notated as either $\\mathbb{N}^0$ or $\\mathbb{W}$.\n\nThis page is a disambiguation page for three different definitions.", "date_published": "2016-07-06T13:00:19Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": ["An intuitive concept of a number that is not fractional. Depending on the author, the set of whole numbers is defined as either the [natural numbers](https://arbital.com/p/45h), the natural numbers and [https://arbital.com/p/-zero](https://arbital.com/p/-zero), or the [integers](https://arbital.com/p/48l)."], "tags": ["C-Class", "Disambiguation"], "alias": "54t"}
{"id": "ce188387c48e400e354b183bf510958f", "title": "Blue oysters", "url": "https://arbital.com/p/blue_oysters", "source": "arbital", "source_type": "text", "text": "You're collecting exotic oysters in Nantucket, and there are two different bays you could harvest oysters from. In both bays, 11% of the oysters contain valuable pearls and 89% are empty. In the first bay, 4% of the pearl-containing oysters are blue, and 8% of the non-pearl-containing oysters are blue. In the second bay, 13% of the pearl-containing oysters are blue, and 26% of the non-pearl-containing oysters are blue. You created a special device that helps you find blue oysters. Would you rather harvest blue oysters from the first bay or the second bay?\n\nYou're encouraged to try to solve this problem yourself, and to refrain from looking at the answer. The answer can be found in [https://arbital.com/p/1zh](https://arbital.com/p/1zh).", "date_published": "2016-07-15T21:32:59Z", "authors": ["Nate Soares", "Alexei Andreev"], "summaries": [], "tags": ["Just a requisite", "Example problem"], "alias": "54v"}
{"id": "240848a9be081e455737ed97fc0f9348", "title": "Example problem", "url": "https://arbital.com/p/example_problem", "source": "arbital", "source_type": "text", "text": "Tag for pages that provide an example problem referenced by a number of other pages.\n\nThe summary of the page should contain the entire text of the problem (so that the problem can be referenced by greenlink). The body of the page should either contain the answer (e.g., hidden behind an 'Answer' button), or should contain a link to the page that contains the answer. The body of the page should not explain the solution to the problem, but may link to the page that does.", "date_published": "2016-07-05T23:18:28Z", "authors": ["Kevin Clancy", "Nate Soares"], "summaries": [], "tags": [], "alias": "54w"}
{"id": "dfc782bbd0ea2b685c150d24973988a2", "title": "Number", "url": "https://arbital.com/p/number", "source": "arbital", "source_type": "text", "text": "A **number** is an object used to represent quantity or value.\n\nExamples of numbers include the [natural numbers](https://arbital.com/p/45h), the [integers](https://arbital.com/p/48l), the [rational numbers](https://arbital.com/p/4zq), and the [real numbers](https://arbital.com/p/4bc).", "date_published": "2016-07-07T17:05:10Z", "authors": ["Dylan Hendrickson", "Eric Bruylant", "Joe Zeng"], "summaries": [], "tags": ["Stub"], "alias": "54y"}
{"id": "c9a4a585998a1403afbd3979ccc7059d", "title": "Irrational number", "url": "https://arbital.com/p/irrational_number", "source": "arbital", "source_type": "text", "text": "An irrational number is a [https://arbital.com/p/-4bc](https://arbital.com/p/-4bc) that is not a [https://arbital.com/p/-4zq](https://arbital.com/p/-4zq). This set is generally denoted using either $\\mathbb{I}$ or $\\overline{\\mathbb{Q}}$, the latter of which represents it as the [complement](https://arbital.com/p/set_complement) of the rationals within the reals.\n\nIn the [Cauchy sequence definition](https://arbital.com/p/50d) of real numbers, the irrational numbers are the equivalence classes of Cauchy sequences of rational numbers that do not converge in the rationals. In the [Dedekind cut definition](https://arbital.com/p/50g), the irrational numbers are the one-sided Dedekind cuts where the set $\\mathbb{Q}^\\ge$ does not have a least element.\n\n## Properties of irrational numbers\n\nIrrational numbers have decimal expansions (and indeed, representations in any base $b$) that do not repeat or terminate.\n\nThe set of irrational numbers is [uncountable](https://arbital.com/p/2w0).", "date_published": "2016-07-06T09:44:11Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": [], "tags": ["Stub"], "alias": "54z"}
{"id": "5a7e576be92c126782dcde536acbdfa8", "title": "Shift towards the hypothesis of least surprise", "url": "https://arbital.com/p/flee_from_surprise", "source": "arbital", "source_type": "text", "text": "The [log-odds form of Bayes' rule](https://arbital.com/p/1zh) says that strength of belief and strength of evidence can both be measured in [bits](https://arbital.com/p/3y2). These evidence-bits can also be used to measure a quantity called \"Bayesian surprise\", which yields [One final, if this is the last thing in the path](https://arbital.com/p/fixme:)yet another intuition for understanding Bayes' rule.\n\nRoughly speaking, we can measure how surprised a hypothesis $H_i$ was by the evidence $e$ by measuring how much probability it put on $e.$ If $H_i$ put 100% of its probability mass on $e$, then $e$ is completely unsurprising (to $H_i$). If $H_i$ put 0% of its probability mass on $e$, then $e$ is _as surprising as possible._ Any measure of $\\mathbb P(e \\mid H_i),$ the probability $H_i$ assigned to $e$, that obeys this property, is worthy of the label \"surprise.\" Bayesian surprise is $-\\!\\log(\\mathbb P(e \\mid H_i)),$ which is a quantity that obeys these intuitive constraints and has some other interesting features.\n\nConsider again the [https://arbital.com/p/-54v](https://arbital.com/p/-54v) problem. Consider the hypotheses $H$ and $\\lnot H$, which say \"the oyster will contain a pearl\" and \"no it won't\", respectively. To keep the numbers easy, let's say we draw an oyster from a third bay, where $\\frac{1}{8}$ of pearl-carrying oysters are blue and $\\frac{1}{4}$ of empty oysters are blue.\n\nImagine what happens when the oyster is blue. $H$ predicted blueness with $\\frac{1}{8}$ of its probability mass, while $\\lnot H$ predicted blueness with $\\frac{1}{4}$ of its probability mass. Thus, $\\lnot H$ did better than $H,$ and goes up in probability. Previously, we've been combining both $\\mathbb P(e \\mid H)$ and $\\mathbb P(e \\mid \\lnot H)$ into unified likelihood ratios, like $\\left(\\frac{1}{8} : \\frac{1}{4}\\right)$ $=$ $(1 : 2),$ which says that the 'blue' observation carries 1 bit of evidence $H.$ However, we can also take the logs first, and combine second.\n\nBecause $H$ assigned only an eighth of its probability mass to the 'blue' observation, and because [Bayesian update works by eliminating incorrect probability mass](https://arbital.com/p/1y6), we have to adjust our belief in $H$ by $\\log_2\\left(\\frac{1}{8}\\right) = -3$ bits away from $H.$ (Each negative bit means \"throw away half of $H$'s probability mass,\" and we have to do that 3 times in order to remove the probability that $H$ failed to assign to $e$.)\n\nSimilarly, because $\\lnot H$ assigned only a quarter of its probability mass to the 'blue' observation, we have to adjust our belief in $H$ by $\\log_2\\left(\\frac{1}{4}\\right) = -2$ bits away from $\\lnot H.$\n\nThus, when the 'blue' observation comes in, we move our belief (measured in bits) 3 notches away from $H$ and then two notches back towards $H.$ On net, our belief shifts 1 notch away from $H$.\n\n\n\n_$H$ assigned 1/8th of its probability mass to blueness, so it emits $-\\!\\log_2\\left(\\frac{1}{8}\\right)=3$ bits of surprise pushing away from $H$. $\\lnot H$ assigned 1/4th of its probability mass to blueness, so it emits $-\\!\\log_2\\left(\\frac{1}{4}\\right)=2$ bits of surprise pushing away from $\\lnot H$ (and towards $H$). Thus, belief in $H$ moves 1 bit towards $\\lnot H$, on net._\n\nIf instead $H$ predicted blue with probability 4% (penalty $\\log_2(0.04) \\approx -4.64$) and $\\lnot H$ predicted blue with probability 8% (penalty $\\log_2(0.08) \\approx -3.64$), then we would have shifted a bit over 4.6 notches towards $\\lnot H$ and a bit over 3.6 notches back towards $H,$ but we would have shifted the same number of notches _on net._ This is why it's only the _relative_ difference between the number of bits docked from $H$ and the number of bits docked from $\\lnot H$ that matters.\n\nIn general, given an observation $e$ and a hypothesis $H,$ the number of bits we need to dock from our belief in $H$ is $\\log_2(\\mathbb P(e \\mid H)),$ that is, the log of the probability that $H$ assigned to $e.$ This quantity is never positive, because the logarithm of $x$ for $0 \\le x \\le 1$ is in the range $[0](https://arbital.com/p/-\\infty,)$. If we negate it, we get a non-negative quantity that relates $H$ to $e$, which is 0 when $H$ was certain that $e$ was going to happen, and which is infinite when $H$ was certain that $e$ wasn't going to happen, and which is measured in the same units as evidence and belief. Thus, this quantity is often called \"surprise,\" and intuitively, it measures how surprised the hypothesis $H$ was by $e$ (in bits).\n\nThere is some correlation between Bayesian surprise and the times when a human would feel surprised (at seeing something that they thought was unlikely), but, of course, the human emotion is quite different. (A human can feel surprised for other reasons than \"my hypotheses failed to predict the data,\" and humans are also great at ignoring evidence instead of feeling surprised.)\n\nGiven this definition of Bayesian surprise, we can view Bayes' rule as saying that _surprise repels belief._ When you make an observation $e,$ each hypothesis emits repulsive \"surprise\" signals, which shift your hypothesis. Referring again to the image above, when $H$ predicts the observation you made with $\\frac{1}{8}$ of its probability mass, and $\\lnot H$ predicts it with $\\frac{1}{4}$ of its probability mass, we can imagine $H$ emitting a surprise signal with a strength of 3 bits away from $H$ and $\\lnot H$ emitting a surprise signal with a strength of 2 bits away from $\\lnot H$. Both those signals push the belief in $H$ in different directions, and it ends up 1 bit closer to $\\lnot H$ (which emitted the weaker surprise signal).\n\nIn other words, whenever you find yourself feeling surprised by something you saw, think of the _least surprising explanation_ for that evidence — and then award that hypothesis a few bits of belief.", "date_published": "2016-11-03T01:10:16Z", "authors": ["Eric Bruylant", "Nate Soares", "Dony Christie", "Alexei Andreev"], "summaries": [], "tags": ["B-Class"], "alias": "552"}
{"id": "fbdc22bf186860f90197623b3c775443", "title": "Bayes' rule: Definition", "url": "https://arbital.com/p/bayes_rule_definition", "source": "arbital", "source_type": "text", "text": "Bayes' rule is the mathematics of [probability theory](https://arbital.com/p/1rf) governing how to update your beliefs in the light of new evidence.\n\n[https://arbital.com/p/toc:](https://arbital.com/p/toc:)\n\n## [Notation](https://arbital.com/p/1y9)\n\nIn much of what follows, we'll use the following [notation](https://arbital.com/p/1y9):\n\n- Let the hypotheses being considered be $H_1$ and $H_2$.\n- Let the evidence observed be $e_0.$\n- Let $\\mathbb P(H_i)$ denote the [prior probability](https://arbital.com/p/1rm) of $H_i$ before observing the evidence.\n- Let the [conditional probability](https://arbital.com/p/1rj) $\\mathbb P(e_0\\mid H_i)$ denote the [likelihood](https://arbital.com/p/1rq) of observing evidence $e_0$ assuming $H_i$ to be true.\n- Let the [conditional probability](https://arbital.com/p/1rj) $\\mathbb P(H_i\\mid e_0)$ denote the [posterior probability](https://arbital.com/p/1rp) of $H_i$ after observing $e_0.$\n\n## [Odds](https://arbital.com/p/1x5)/[proportional](https://arbital.com/p/1zm) form\n\nBayes' rule in the [odds form](https://arbital.com/p/1x5) or [proportional form](https://arbital.com/p/1zm) states:\n\n$$\\dfrac{\\mathbb P(H_1)}{\\mathbb P(H_2)} \\times \\dfrac{\\mathbb P(e_0\\mid H_1)}{\\mathbb P(e_0\\mid H_2)} = \\dfrac{\\mathbb P(H_1\\mid e_0)}{\\mathbb P(H_2\\mid e_0)}$$\n\nIn other words, the [prior](https://arbital.com/p/1rm) [odds](https://arbital.com/p/1rb) times the [likelihood ratio](https://arbital.com/p/1rq) yield the [posterior](https://arbital.com/p/1rp) odds. [Normalizing](https://arbital.com/p/1rk) these odds will then yield the posterior probabilities.\n\nIn [other other words](https://arbital.com/p/1zm): If you initially think $h_i$ is $\\alpha$ times as probable as $h_k$, and then see evidence that you're $\\beta$ times as likely to see if $h_i$ is true as if $h_k$ is true, you should update to thinking that $h_i$ is $\\alpha \\cdot \\beta$ times as probable as $h_k.$\n\nSuppose that Professor Plum and Miss Scarlet are two suspects in a murder, and that we start out thinking that Professor Plum is twice as likely to have committed the murder as Miss Scarlet ([prior](https://arbital.com/p/1rm) [odds](https://arbital.com/p/1rb) of 2 : 1). We then discover that the victim was poisoned. We think that Professor Plum is around one-fourth as likely to use poison as Miss Scarlet ([likelihood ratio](https://arbital.com/p/1rq) of 1 : 4). Then after observing the victim was poisoned, we should think Plum is around half as likely to have committed the murder as Scarlet: $2 \\times \\dfrac{1}{4} = \\dfrac{1}{2}.$ This reflects [posterior](https://arbital.com/p/1rp) odds of 1 : 2, or a posterior probability of 1/3, that Professor Plum did the deed.\n\n## [Proof](https://arbital.com/p/1xr)\n\nThe [proof of Bayes' rule](https://arbital.com/p/1xr) is by the definition of [conditional probability](https://arbital.com/p/1rj) $\\mathbb P(X\\wedge Y) = \\mathbb P(X\\mid Y) \\cdot \\mathbb P(Y):$\n\n$$\n\\dfrac{\\mathbb P(H_i)}{\\mathbb P(H_j)} \\times \\dfrac{\\mathbb P(e\\mid H_i)}{\\mathbb P(e\\mid H_j)}\n= \\dfrac{\\mathbb P(e \\wedge H_i)}{\\mathbb P(e \\wedge H_j)}\n= \\dfrac{\\mathbb P(e \\wedge H_i) / \\mathbb P(e)}{\\mathbb P(e \\wedge H_j) / \\mathbb P(e)}\n= \\dfrac{\\mathbb P(H_i\\mid e)}{\\mathbb P(H_j\\mid e)}\n$$\n\n## [Log odds form](https://arbital.com/p/1zh)\n\nThe [log odds form of Bayes' rule](https://arbital.com/p/1zh) states:\n\n$$\\log \\left ( \\dfrac\n {\\mathbb P(H_i)}\n {\\mathbb P(H_j)}\n\\right )\n+\n\\log \\left ( \\dfrac\n {\\mathbb P(e\\mid H_i)}\n {\\mathbb P(e\\mid H_j)}\n\\right ) \n =\n\\log \\left ( \\dfrac\n {\\mathbb P(H_i\\mid e)}\n {\\mathbb P(H_j\\mid e)}\n\\right )\n$$\n\nE.g.: \"A study of Chinese blood donors found that roughly 1 in 100,000 of them had HIV (as determined by a very reliable gold-standard test). The non-gold-standard test used for initial screening had a sensitivity of 99.7% and a specificity of 99.8%, meaning that it was 500 times as likely to return positive for infected as non-infected patients.\" Then our prior belief is -5 orders of magnitude against HIV, and if we then observe a positive test result, this is evidence of strength +2.7 orders of magnitude for HIV. Our posterior belief is -2.3 orders of magnitude, or odds of less than 1 to a 100, against HIV.\n\nIn log odds form, the same [strength of evidence](https://arbital.com/p/22x) (log [likelihood ratio](https://arbital.com/p/1rq)) always [moves us the same additive distance](https://arbital.com/p/1zh) along a line representing strength of belief (also in log odds). If we measured distance in probabilities, then the same 2 : 1 likelihood ratio might move us a different distance along the probability line depending on whether we started with prior 10% probability or 50% probability.\n\n## Visualizations\n\nGraphical of visualizing Bayes' rule include [frequency diagrams, the waterfall visualization](https://arbital.com/p/1wy), the [spotlight visualization](https://arbital.com/p/1zm), the [magnet visualization](https://arbital.com/p/1zh), and the [Venn diagram for the proof](https://arbital.com/p/1xr).\n\n## Examples\n\nExamples of Bayes' rule may be found [here](https://arbital.com/p/1wt).\n\n## [Multiple hypotheses and updates](https://arbital.com/p/1zg)\n\nThe [odds form of Bayes' rule](https://arbital.com/p/1x5) works for odds ratios between more than two hypotheses, and applying multiple pieces of evidence. Suppose there's a bathtub full of coins. 1/2 of the coins are \"fair\" and have a 50% probability of producing heads on each coinflip; 1/3 of the coins produce 25% heads; and 1/6 produce 75% heads. You pull out a coin at random, flip it 3 times, and get the result HTH. You may legitimately calculate:\n\n$$\\begin{array}{rll}\n(1/2 : 1/3 : 1/6) \\cong & (3 : 2 : 1) & \\\\\n\\times & (2 : 1 : 3) & \\\\\n\\times & (2 : 3 : 1) & \\\\\n\\times & (2 : 1 : 3) & \\\\\n= & (24 : 6 : 9) & \\cong (8 : 2 : 3)\n\\end{array}$$\n\nSince multiple pieces of evidence may not be [conditionally independent](https://arbital.com/p/conditional_independence) from one another, it is important to be aware of the [Naive Bayes assumption](https://arbital.com/p/naive_bayes_assumption) and whether you are making it.\n\n## [Probability form](https://arbital.com/p/554)\n\nAs a formula for a single probability $\\mathbb P(H_i\\mid e),$ Bayes' rule states:\n\n$$\\mathbb P(H_i\\mid e) = \\dfrac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)}$$\n\n## [Functional form](https://arbital.com/p/1zj)\n\nIn [functional form](https://arbital.com/p/1zj), Bayes' rule states:\n\n$$\\mathbb P(\\mathbf{H}\\mid e) \\propto \\mathbb P(e\\mid \\mathbf{H}) \\cdot \\mathbb P(\\mathbf{H}).$$\n\nThe posterior probability function over hypotheses given the evidence, is *proportional* to the likelihood function from the evidence to those hypotheses, times the prior probability function over those hypotheses.\n\nSince posterior probabilities over [mutually exclusive and exhaustive](https://arbital.com/p/1rd) possibilities must sum to $1,$ [normalizing](https://arbital.com/p/1rk) the product of the likelihood function and prior probability function will yield the exact posterior probability function.", "date_published": "2016-10-04T04:14:07Z", "authors": ["Eric Bruylant", "Nate Soares", "Eliezer Yudkowsky", "Alexei Andreev"], "summaries": [], "tags": ["C-Class"], "alias": "553"}
{"id": "5010e403929e8370530c1ae78384747d", "title": "Bayes' rule: Probability form", "url": "https://arbital.com/p/bayes_rule_probability", "source": "arbital", "source_type": "text", "text": "The formulation of [https://arbital.com/p/1lz](https://arbital.com/p/1lz) you are most likely to see in textbooks runs as follows:\n\n$$\\mathbb P(H_i\\mid e) = \\dfrac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)}$$\n\nWhere:\n\n- $H_i$ is the hypothesis we're interested in.\n- $e$ is the piece of evidence we observed.\n- $\\sum_k (\\text {expression containing } k)$ [means](https://arbital.com/p/summation_notation) \"Add up, for every $k$, the sum of all the (expressions containing $k$).\"\n- $\\mathbf H$ is a set of [mutually exclusive and exhaustive](https://arbital.com/p/1rd) hypotheses that include $H_i$ as one of the possibilities, and the expression $H_k$ inside the sum ranges over all the possible hypotheses in $\\mathbf H$.\n\nAs a quick example, let's say there's a bathtub full of potentially biased coins.\n\n- Coin type 1 is fair, 50% heads / 50% tails. 40% of the coins in the bathtub are type 1.\n- Coin type 2 produces 70% heads. 35% of the coins are type 2.\n- Coin type 3 produces 20% heads. 25% of the coins are type 3.\n\nWe want to know the [posterior](https://arbital.com/p/1rp) probability that a randomly drawn coin is of type 2, after flipping the coin once and seeing it produce heads once.\n\nLet $H_1, H_2, H_3$ stand for the hypotheses that the coin is of types 1, 2, and 3 respectively. Then using [conditional probability notation](https://arbital.com/p/1rj), we want to know the probability $\\mathbb P(H_2 \\mid heads).$\n\nThe probability form of Bayes' theorem says:\n\n$$\\mathbb P(H_2 \\mid heads) = \\frac{\\mathbb P(heads \\mid H_2) \\cdot \\mathbb P(H_2)}{\\sum_k \\mathbb P(heads \\mid H_k) \\cdot \\mathbb P(H_k)}$$\n\nExpanding the sum:\n\n$$\\mathbb P(H_2 \\mid heads) = \\frac{\\mathbb P(heads \\mid H_2) \\cdot \\mathbb P(H_2)}{[P](https://arbital.com/p/\\mathbb) + [P](https://arbital.com/p/\\mathbb) + [P](https://arbital.com/p/\\mathbb)}$$\n\nComputing the actual quantities:\n\n$$\\mathbb P(H_2 \\mid heads) = \\frac{0.70 \\cdot 0.35 }{[\\cdot 0.40](https://arbital.com/p/0.50) + [\\cdot 0.35](https://arbital.com/p/0.70) + [\\cdot 0.25](https://arbital.com/p/0.20)} = \\frac{0.245}{0.20 + 0.245 + 0.05} = 0.\\overline{49}$$\n\nThis calculation was big and messy. Which is fine, because the probability form of Bayes' theorem is okay for directly grinding through the numbers, but not so good for doing things in your head.\n\n# Meaning\n\nWe can think of the advice of Bayes' theorem as saying:\n\n\"Think of how much each hypothesis in $H$ contributed to our expectation of seeing the evidence $e$, including both the [likelihood](https://arbital.com/p/56v) of seeing $e$ if $H_k$ is true, and the [prior](https://arbital.com/p/1rm) probability of $H_k$. The [posterior](https://arbital.com/p/1rp) of $H_i$ after seeing $e,$ is the amount $H_i$ contributed to our expectation of seeing $e,$ within the total expectation of seeing $e$ contributed by every hypothesis in $H.$\"\n\nOr to say it at somewhat greater length:\n\nImagine each hypothesis $H_1,H_2,H_3\\ldots$ as an expert who has to distribute the probability of their predictions among all possible pieces of evidence. We can imagine this more concretely by visualizing \"probability\" as a lump of clay.\n\nThe total amount of clay is one kilogram (probability $1$). Each expert $H_k$ has been allocated a fraction $\\mathbb P(H_k)$ of that kilogram. For example, if $\\mathbb P(H_4)=\\frac{1}{5}$ then expert 4 has been allocated 200 grams of clay.\n\nWe're playing a game with the experts to determine which one is the best predictor.\n\nEach time we're about to make an observation $E,$ each expert has to divide up all their clay among the possible outcomes $e_1, e_2, \\ldots.$\n\nAfter we observe that $E = e_j,$ we take away all the clay that wasn't put onto $e_j.$ And then our new belief in all the experts is the relative amount of clay that each expert has left.\n\nSo to know how much we now believe in expert $H_4$ after observing $e_3,$ say, we need to know two things: First, the amount of clay that $H_4$ put onto $e_3,$ and second, the total amount of clay that all experts (including $H_4$) put onto $e_3.$\n\nIn turn, to know *that,* we need to know how much clay $H_4$ started with, and what fraction of its clay $H_4$ put onto $e_3.$ And similarly, to compute the total clay on $e_3,$ we need to know how much clay each expert $H_k$ started with, and what fraction of their clay $H_k$ put onto $e_3.$\n\nSo Bayes' theorem here would say:\n\n$$\\mathbb P(H_4 \\mid e_3) = \\frac{\\mathbb P(e_3 \\mid H_4) \\cdot \\mathbb P(H_4)}{\\sum_k \\mathbb P(e_3 \\mid H_k) \\cdot \\mathbb P(H_k)}$$\n\nWhat are the incentives of this game of clay?\n\nOn each round, the experts who gain the most are the experts who put the most clay on the observed $e_j,$ so if you know for certain that $e_3$ is about to be observed, your incentive is to put all your clay on $e_3.$\n\nBut putting *literally all* your clay on $e_3$ is risky; if $e_5$ is observed instead, you lose all your clay and are out of the game. Once an expert's amount of clay goes all the way to zero, there's no way for them to recover over any number of future rounds. That hypothesis is done, dead, and removed from the game. (\"Falsification,\" some people call that.) If you're not certain that $e_5$ is literally impossible, you'd be wiser to put at least a *little* clay on $e_5$ instead. That is to say: if your mind puts some probability on $e_5,$ you'd better put some clay there too!\n\n([As it happens](https://arbital.com/p/bayes_score), if at the end of the game we score each expert by the logarithm of the amount of clay they have left, then each expert is incentivized to place clay exactly proportionally to their honest probability on each successive round.)\n\nIt's an important part of the game that we make the experts put down their clay in advance. If we let the experts put down their clay afterwards, they might be tempted to cheat by putting down all their clay on whichever $e_j$ had actually been observed. But since we make the experts put down their clay in advance, they have to *divide up* their clay among the possible outcomes: to give more clay to $e_3,$ that clay has to be taken away from some other outcome, like $e_5.$ To put a very high probability on $e_3$ and gain a lot of relative credibility if $e_3$ is observed, an expert has to stick their neck out and risk losing a lot of credibility if some other outcome like $e_5$ happens instead. *If* we force the experts to make advance predictions, that is!\n\nWe can also derive from this game that the question \"does evidence $e_3$ support hypothesis $H_4$?\" depends on how well $H_4$ predicted $e_3$ _compared to the competition._ It's not enough for $H_4$ to predict $e_3$ well if every other hypothesis also predicted $e_3$ well--your amazing new theory of physics gets no points for predicting that the sky is blue. $H_k$ only goes up in probability when it predicts $e_j$ better than the alternatives. And that means we have to ask what the alternative hypotheses predicted, even if we think those hypotheses are false.\n\nIf you get in a car accident, and don't want to relinquish the hypothesis that you're a great driver, then you can find all sorts of reasons (\"the road was slippery! my car freaked out!\") why $\\mathbb P(e \\mid GoodDriver)$ is not too low. But $\\mathbb P(e \\mid BadDriver)$ is also part of the update equation, and the \"bad driver\" hypothesis *better* predicts the evidence. Thus, your first impulse, when deciding how to update your beliefs in the face of a car accident, should not be \"But my preferred hypothesis allows for this evidence!\" It should instead be \"Points to the 'bad driver' hypothesis for predicting this evidence better than the alternatives!\" (And remember, you're allowed to [increase $\\mathbb P](https://arbital.com/p/update_beliefs_incrementally), while still thinking that it's less than 50% probable.)\n\n# Proof\n\nThe proof of Bayes' theorem follows from the definition of [https://arbital.com/p/-1rj](https://arbital.com/p/-1rj):\n\n$$\\mathbb P(X \\mid Y) = \\frac{\\mathbb P(X \\wedge Y)}{\\mathbb P (Y)}$$\n\nAnd from the [law of marginal probability](https://arbital.com/p/law_marginal_probability):\n\n$$\\mathbb P(Y) = \\sum_k \\mathbb P(Y \\wedge X_k)$$\n\nTherefore:\n\n$$\n\\mathbb P(H_i \\mid e) = \\frac{\\mathbb P(H_i \\wedge e)}{\\mathbb P (e)} \\tag{defn. conditional prob.}\n$$\n\n$$\n\\mathbb P(H_i \\mid e) = \\frac{\\mathbb P(e \\wedge H_i)}{\\sum_k \\mathbb P (e \\wedge H_k)} \\tag {law of marginal prob.}\n$$\n\n$$\n\\mathbb P(H_i \\mid e) = \\frac{\\mathbb P(e \\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P (e \\mid H_k) \\cdot \\mathbb P(H_k)} \\tag {defn. conditional prob.}\n$$\n\nQED.", "date_published": "2017-08-13T04:21:13Z", "authors": ["Eric Bruylant", "Alexei Andreev", "Nadeem Mohsin", "Eric Rogstad", "Nate Soares", "Adom Hartell", "Eliezer Yudkowsky"], "summaries": ["The formulation of [https://arbital.com/p/1lz](https://arbital.com/p/1lz) you are most likely to see in textbooks says:\n\n$$\\mathbb P(H_i\\mid e) = \\dfrac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)}$$\n\nThis follows from the definition of [https://arbital.com/p/-1rj](https://arbital.com/p/-1rj) which states that $\\mathbb P(X \\mid Y) = \\frac{\\mathbb P(X \\wedge Y)}{\\mathbb P (Y)}$, and the [law of marginal probability](https://arbital.com/p/law_marginal_probability) which says that $\\mathbb P(Y) = \\sum_k \\mathbb P(Y \\wedge X_k)$.\n\nWe can think of the corresponding advice as saying, \"Think of how much each hypothesis in $H$ contributed to our expectation of seeing the evidence $e$, including both the [likelihood](https://arbital.com/p/56v) of seeing $e$ if $H_k$ is true, and the [prior](https://arbital.com/p/1rm) probability of $H_k$. The [posterior](https://arbital.com/p/1rp) of $H_i$ after seeing $e,$ is the amount $H_i$ contributed to our expectation of seeing $e,$ within the total expectation of seeing $e$ contributed by every hypothesis in $H.$"], "tags": ["B-Class"], "alias": "554"}
{"id": "f1e15f0d4cab1433e2b06b1cb4e5fc6b", "title": "Odds form to probability form", "url": "https://arbital.com/p/bayes_odds_to_probability", "source": "arbital", "source_type": "text", "text": "The odds form of Bayes' rule works for any two hypotheses $H_i$ and $H_j,$ and looks like this:\n\n$$\\frac{\\mathbb P(H_i \\mid e)}{\\mathbb P(H_j \\mid e)} = \\frac{\\mathbb P(H_i)}{\\mathbb P(H_j)} \\times \\frac{\\mathbb P(e \\mid H_i)}{\\mathbb P(e \\mid H_j)} \\tag{1}.$$\n\nThe probabilistic form of Bayes' rule requires a hypothesis set $H_1,H_2,H_3,\\ldots$ that is [https://arbital.com/p/-1rd](https://arbital.com/p/-1rd), and looks like this:\n\n$$\\mathbb P(H_i\\mid e) = \\frac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)} \\tag{2}.$$\n\nWe will now show that equation (2) follows from equation (1). Given a collection $H_1,H_2,H_3,\\ldots$ of mutually exclusive and exhaustive hypotheses and a hypothesis $H_i$ from that collection, we can form another hypothesis $\\lnot H_i$ consisting of all the hypotheses $H_1,H_2,H_3,\\ldots$ _except_ $H_i.$ Then, using $\\lnot H_i$ as $H_j$ and multiplying the fractions on the right-hand side of equation (1), we see that\n\n$$\\frac{\\mathbb P(H_i \\mid e)}{\\mathbb P(\\lnot H_i \\mid e)} = \\frac{\\mathbb P(H_i) \\cdot \\mathbb P(e \\mid H_i)}{\\mathbb P(\\lnot H_i)\\cdot \\mathbb P(e \\mid \\lnot H_i)}.$$\n\n$\\mathbb P(\\lnot H_i)\\cdot \\mathbb P(e \\mid \\lnot H_i)$ is the prior probability of $\\lnot H_i$ times the degree to which $\\lnot H_i$ predicted $e.$ Because $\\lnot H_i$ is made of a bunch of mutually exclusive hypotheses, this term can be calculated by summing $\\mathbb P(H_k) \\cdot \\mathbb P(e \\mid H_k)$ for every $H_k$ in the collection except $H_i.$ Performing that replacement, and swapping the order of multiplication, we get:\n\n$$\\frac{\\mathbb P(H_i \\mid e)}{\\mathbb P(\\lnot H_i \\mid e)} = \\frac{\\mathbb P(e \\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_{k \\neq i} \\mathbb P(e \\mid H_k) \\cdot \\mathbb P(H_k)}.$$\n\nThese are the posterior odds for $H_i$ versus $\\lnot H_i.$ Because $H_i$ and $\\lnot H_i$ are mutually exclusive and exhaustive, we can convert these odds into a probability for $H_i,$ by calculating numerator / (numerator + denominator), in the same way that $3 : 4$ odds become a 3 / (3 + 4) probability. When we do so, equation (2) drops out:\n\n$$\\mathbb P(H_i\\mid e) = \\frac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)}.$$\n\nThus, we see that the probabilistic formulation of Bayes' rule follows from the odds form, but is less general, in that it only works when the set of hypotheses being considered are mutually exclusive and exhaustive.\n\nWe also see that the probabilistic formulation converts the posterior odds into a posterior probability. When computing [multiple updates in a row](https://arbital.com/p/bayes_rule_vector), you actually only need to perform this \"normalization\" step once at the very end of your calculations — which means that the odds form of Bayes' rule is also more efficient in practice.", "date_published": "2016-07-06T04:34:25Z", "authors": ["Eric Bruylant", "Nate Soares"], "summaries": [], "tags": ["Needs clickbait", "B-Class"], "alias": "555"}
{"id": "c6a7acda9075ab7b8e4080e349b9752d", "title": "Logistic function", "url": "https://arbital.com/p/logistic_function", "source": "arbital", "source_type": "text", "text": "The logistic function is a [https://arbital.com/p/-sigmoid](https://arbital.com/p/-sigmoid) function that maps the [real numbers](https://arbital.com/p/4bc) to the unit interval $(0, 1)$ using the formula $\\displaystyle f(x) = \\frac{1}{1 + e^{-x}}$.\n\nMore generally, there exists a [family](https://arbital.com/p/family_of_functions) of logistic functions that can be written as $\\displaystyle f(x) = \\frac{M}{1 + \\alpha^{c(x_0 - x)}}$, where:\n\n* $M$ is the upper bound of the function (in which case the function maps to the interval $(0, M)$ instead). When $M = 1$, the logistic function is usually being used to calculate some [https://arbital.com/p/-1rf](https://arbital.com/p/-1rf) or [https://arbital.com/p/-4w3](https://arbital.com/p/-4w3) of a total.\n\n* $x_0$ is the inflection point of the curve, or the value of $x$ where the function's growth stops speeding up and starts slowing down.\n\n* $\\alpha$ is a variable controlling the steepness of the curve.\n\n* $c$ is a scaling factor for the distance.\n\n## Applications\n\n* The logistic function is used to model growth rates of populations in an ecosystem with a limited carrying capacity.\n\n* The inverse logistic function (with $\\alpha = 2$) is used to convert a probability to log-odds form for use in [https://arbital.com/p/1lz](https://arbital.com/p/1lz).\n\n* The logistic function (with $\\alpha = 10$ and $c = 1/400$) is used to calculate the expected probability of a player winning given a specific difference in rating in the Elo rating system.", "date_published": "2016-07-06T23:33:37Z", "authors": ["Eric Bruylant", "Joe Zeng"], "summaries": ["The logistic function is a [https://arbital.com/p/-sigmoid](https://arbital.com/p/-sigmoid) function that maps the [real numbers](https://arbital.com/p/4bc) to the unit interval $(0, 1)$ using the formula $\\displaystyle f(x) = \\frac{1}{1 + e^{-x}}$ or variants of this formula."], "tags": ["Start", "Needs parent"], "alias": "558"}
{"id": "ef3ad0bf3d8a7b0ca5dac3394d94f721", "title": "Sparking widgets", "url": "https://arbital.com/p/sparking_widgets", "source": "arbital", "source_type": "text", "text": "10% of widgets are bad and 90% are good. 4% of good widgets emit sparks, and 12% of bad widgets emit sparks. If a widget is sparking, find an easy way to calculate the chance that it's a bad widget, keeping the operations as simple as possible so that it's easy to do the calculation entirely in your head.\n\nYou're encouraged to solve the problem yourself without looking at the answer. an answer can be found in [https://arbital.com/p/bayes_extraordinary_claims](https://arbital.com/p/bayes_extraordinary_claims).", "date_published": "2016-07-06T12:58:00Z", "authors": ["Nate Soares"], "summaries": [], "tags": ["Just a requisite", "Example problem"], "alias": "559"}
{"id": "0214905a38271739bc39bac06c141973", "title": "Sock-dresser search", "url": "https://arbital.com/p/sockdresser_search", "source": "arbital", "source_type": "text", "text": "You left your socks somewhere in your room. You think there's a 4/5 chance that they're in your dresser, so you start looking through your dresser's 8 drawers. After checking 6 drawers at random, you haven't found your socks yet. What is the probability you will find your socks in the next drawer you check?\n\nYou're encouraged to solve the problem yourself before looking up the answer. The answer can be found in [https://arbital.com/p/1y6](https://arbital.com/p/1y6).", "date_published": "2016-07-06T12:11:50Z", "authors": ["Nate Soares"], "summaries": [], "tags": ["Just a requisite", "Example problem"], "alias": "55b"}
{"id": "0cd530209c6ba8a5e57b21db2b51dc42", "title": "Ordinary claims require ordinary evidence", "url": "https://arbital.com/p/bayes_ordinary_claims", "source": "arbital", "source_type": "text", "text": "A corollary to [extraordinary claims require extraordinary evidence](https://arbital.com/p/21v) is \"ordinary claims require only ordinary evidence\"%%note: Attributed to [Gwern Branwen](https://www.gwern.net/), who hasn't popularized this claim nearly as much as Carl Sagan popularized the extraordinary version.%%\n\nThis corollary is probably more important in practice, as, in our day-to-day lives, we encounter more ordinary claims than extraordinary claims, but may be more tempted to reject many of those ordinary claims by demanding extraordinary evidence (a bias known as \"motivated skepticism\").\n\n# Example: A failing new employee\n\nOne contributed example: \n\n> A few years back, a senior person at my workplace told me that a new employee wasn't getting his work done on time, and that she'd had to micromanage him to get any work out of him at all. This was an unpleasant fact for a number of reasons; I'd liked the guy, and I'd advocated for hiring him to our Board of Directors just a few weeks earlier (which is why the senior manager was talking to me). I could have demanded more evidence, I could have demanded that we give him more time to work out, I could have demanded a videotape and signed affidavits… but a new employee not working out, just *isn't that improbable*. Could I have named the exact prior odds of an employee not working out, could I have said [how much more likely](https://arbital.com/p/1zm) I was to hear that exact report of a long-term-bad employee than a long-term-good employee? No, but 'somebody hires the wrong person' happens all the time, and I'd seen it happen before. It wasn't an extraordinary claim, and I wasn't licensed to ask for extraordinary evidence. To put numbers on it, I thought the proportion of bad to good employees was on the order of 1 : 4 at the time, and the likelihood ratio for the manager's report seemed more like 10 : 1.\n\nOr to put it another way: The rule is 'extraordinary claims require extraordinary evidence', not 'inconvenient but ordinary claims require extraordinary evidence'.\n\nIn everyday life, we consider many more ordinary claims than extraordinary claims — including many ordinary claims whose inconvenience might tempt us to dismiss their adequate but ordinary evidence. In those cases, it's important to remember that ordinary claims only require ordinary evidence.", "date_published": "2016-07-06T13:33:00Z", "authors": ["Eric Bruylant", "Nate Soares", "Alexei Andreev"], "summaries": [], "tags": ["C-Class"], "alias": "55c"}
{"id": "eee9ac83608eed1e1ed3f6ec0e96ffaf", "title": "Ordered ring", "url": "https://arbital.com/p/ordered_ring", "source": "arbital", "source_type": "text", "text": "An **ordered ring** is a [https://arbital.com/p/-3gq](https://arbital.com/p/-3gq) $R=(X,\\oplus,\\otimes)$ with a [total order](https://arbital.com/p/540) $\\leq$ compatible with the ring structure. Specifically, it must satisfy these axioms for any $a,b,c \\in X$:\n\n- If $a \\leq b$, then $a \\oplus c \\leq b \\oplus c$.\n- If $0 \\leq a$ and $0 \\leq b$, then $0 \\leq a \\otimes b$\n\nAn element $a$ of the ring is called \"[https://arbital.com/p/-positive](https://arbital.com/p/-positive)\" if $0| Step | \n\tDistance Covered | \n\tDistance Remaining | \n
|---|---|---|
| 0 | \n\t0 | \n\t1 | \n
| 1 | \n\t\\(\\frac{1}{2}\\) | \n\t\\(\\frac{1}{2}\\) | \n
| 2 | \n\t\\(\\frac{1}{2}+\\frac{1}{4}\\) | \n\t\\(\\frac{3}{4}\\) | \n
| 3 | \n\t\\(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}\\) | \n\t\\(\\frac{7}{8}\\) | \n
| 4 | \n\t\\(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}\\) | \n\t\\(\\frac{15}{16}\\) | \n
| 5 | \n\t\\(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\frac{1}{32}\\) | \n\t\\(\\frac{31}{32}\\) | \n