hcju/proofwikips huggingface dataset:
Similar Polygons are composed of Similar Triangles/Porism
Similar Polygons are composed of Similar Triangles/Porism 0 35954 197167 2014-10-17T22:56:20Z Prime.mover 59 Created page with "== Porism to Similar Polygons are composed of Similar Triangles ==  {{:Euclid:Proposition/VI/20/Porism}}  == Proof ==  Follows immediately from..." wikitext text/x-wiki == Porism to Similar Polygons are composed of Similar Triangles == {{:Euclid:Proposition/VI/20/Porism}} == Proof == Follows immediately from Similar Polygons are composed of Similar Triangles.  {{Euclid Note|20|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Similar Polygons are composed of Similar Triangles|next=Similarity of Polygons is Equivalence Relation}}: Book $\text{VI}$. Propositions
Commensurability of Squares on Proportional Straight Lines/Lemma
== Lemma to Commensurability of Squares on Proportional Straight Lines == {{:Euclid:Proposition/X/14/Lemma}} == Proof == :400px Let $AB$ and $C$ be two unequal straight lines. Let $AB > C$. Let the semicircle $ADB$ be described with $AB$ as the diameter. Using Fitting Chord Into Circle, let $AD$ be fitted into $ADB$ equal to $C$. Let $DB$ be joined. From Relative Sizes of Angles in Segments, $\angle ADB$ is a right angle. From Pythagoras's Theorem: :$AB^2 = AD^2 + DB^2$ and so $AB^2$ is greater than $AD^2$, that is, $C^2$, by $DB^2$. Conversely, given two straight lines $A$ and $B$ the same technique can be used to find the straight line the square of whose length equals the sum of the squares on $A$ and $B$.  {{Euclid Note|14|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Commensurable Magnitudes are Incommensurable with Same Magnitude|next=Commensurability of Squares on Proportional Straight Lines}}: Book $\text{X}$. Propositions
Straight Line Commensurable with Medial Straight Line is Medial/Porism
== Porism to Straight Line Commensurable with Medial Straight Line is Medial == {{:Euclid:Proposition/X/23/Porism}} == Proof ==  :''And in the same way as was explained in the case of the rationals it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called '''''medial''''' and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.'' :''But, if any straight line be commensurable in square with a medial straight line , then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.'' {{Euclid Note|23|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Straight Line Commensurable with Medial Straight Line is Medial|next = Rectangle Contained by Medial Straight Lines Commensurable in Length is Medial}}: Book $\text{X}$. Propositions
Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 1
== Lemma to Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater == {{:Euclid:Proposition/X/29/Lemma 1}} == Proof == Let $a$ and $b$ be two natural numbers such that $a > b$. Let $a$ and $b$ be either both even or both odd. From: :Even Number minus Even Number is Even and: :Odd Number minus Odd Number is Even their difference $c = a - b$ is even. Let $d = \dfrac c 2$. Let $a$ and $b$ be similar plane numbers. By definition of similar plane numbers it is noted that $a$ and $b$ may both be square. Then by Square of Sum less Square: :$a b + d^2 = \left({a - d}\right)^2$ From Product of Similar Plane Numbers is Square, $a b$ is a square number. So two square numbers $a b$ and $d^2$ have been found whose sum is square. Similarly: :$a b = \left({a - d}\right)^2 - d^2$ from which it is noted that two square numbers $\left({a - d}\right)^2$ and $d^2$ have been found whose difference is square.  {{Euclid Note|29|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Construction of Components of Second Bimedial|next=Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 2}}: Book $\text{X}$. Propositions
Axiom:Peano's Axioms/Formulation 2
== Axioms == '''Peano's Axioms''' are intended to reflect the intuition behind $\N$, the mapping $s: \N \to \N: \map s n = n + 1$ and $0$ as an element of $\N$. Let there be given a set $P$, a mapping $s: P \to P$, and a distinguished element $0$. Historically, the existence of $s$ and the existence of $0$ were considered the first two of '''Peano's Axioms''': {{begin-axiom}} {{axiom | n = \text P 1         | q =          | m = 0 \in P         | t = $0$ is an element of $P$ }} {{axiom | n = \text P 2         | q = \forall n \in P         | m = \map s n \in P         | t = For all $n \in P$, its successor $\map s n$ is also in $P$ }} {{end-axiom}} The other three are as follows:  {{begin-axiom}} {{axiom | n = \text P 3         | q = \forall m, n \in P         | m = \map s m = \map s n \implies m = n         | t = $s$ is injective }} {{axiom | n = \text P 4         | m = \Img s \ne P         | t = $s$ is not surjective }} {{axiom | n = \text P 5         | q = \forall A \subseteq P         | m = \paren {\paren {\exists x \in A: \neg \exists y \in P: x = \map s y} \land \paren {\forall z \in A: \map s z \in A} }         | t = Principle of Mathematical Induction: }} {{axiom | m = \implies A = P         | t = Any subset $A$ of $P$, containing an element not }} {{axiom | t = in the image of $s$ and closed under $s$, }} {{axiom | t = is equal to $P$ }} {{end-axiom}}  == Also defined as == {{:Axiom:Peano's Axioms/Also defined as}} == Also see == * Equivalence of Formulations of Peano's Axioms {{LinkToCategory|Peano's Axioms|Peano's axioms}} {{NamedforAxiom|Giuseppe Peano|name2 = Richard Dedekind}} == Historical Note == {{:Axiom:Peano's Axioms/Historical Note}} == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Definition:Natural Numbers|next = Axiom:Peano's Axioms/Also defined as}}: Introduction $\S 4$: The natural numbers
Greatest Power of Two not Divisor
== Lemma == Let $S = \set {1, 2, 3, 4, \ldots, n}$ be a set of integers from $1$ through $n$. Let $2^t$ be the greatest power of $2$ in $S$. Then $2^t$ does not divide any other integer in $S$. That is, no other member of $S$ is a multiple of $2^t$. == Proof == Let $k$ be a multiple of $2^t$ in $S$. Then $k = 2^t \times \ell$ for some $\ell \in \Z, \ell \ge 1$. If $\ell = 2$, then $k = 2 \times 2^t = 2^{t + 1}$, which would contradict $2^t$ being the highest power of $2$ in $S$. Otherwise, if $\ell > 2$, then we would have $k = \ell \times 2^t > 2\times 2^t = 2^{t+1}$. As $S$ contains all integers up to $k$, we would have $2^{t+1} \in S$, contradicting $2^t$ being the highest power of $2$ in $S$. So $\ell = 1$, that is, $2^t$ is the only multiple of $2^t$ in $S$.   == Also see == * Harmonic Number is not Integer
Bertrand-Chebyshev Theorem/Lemma 1
== Lemma ==  For all $n \in \N$: :$\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n + 1}$ where $\dbinom {2 n} n$ denotes a binomial coefficient.  == Proof == From Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index, $\dbinom n k$ increases for $k < \dfrac n 2$. From Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index, $\dbinom n k$ decreases for $k > \dfrac n 2$. Therefore, $\dbinom {2 n} n$ is the largest term in the sequence $\dbinom {2 n} 0, \dbinom {2 n} 1, \ldots, \dbinom {2 n} {2 n}$. Finally, observe that the mean of the terms in the sequence is $\dfrac {2^{2 n}} {2 n + 1}$, by Sum of Binomial Coefficients over Lower Index. {{qed|lemma}}
Bertrand-Chebyshev Theorem/Lemma 2
== Lemma ==  For all $m \in \N$: :$\displaystyle \prod_{p \mathop \le m} p \le 2^{2 m}$ where the product is taken over all prime numbers $p \le m$.  == Proof == It is plainly true for $m \le 2$; we will proceed by strong induction. If $m > 2$ is even then: {{begin-eqn}} {{eqn | l = \prod_{p \mathop \le m} p       | r = \prod_{p \mathop \le m - 1} p       | c =  }} {{eqn | r = 2^{2 \left({m - 1}\right)}       | o = \le        | c = by the induction hypothesis }} {{eqn | r = 2^{2 m}       | o = <       | c =  }} {{end-eqn}} By Sum of Binomial Coefficients over Lower Index: :$\displaystyle \sum_{r \mathop = 0}^{2 k + 1} \binom {2 k + 1} r = 2^{2 k + 1}$ Therefore: {{begin-eqn}} {{eqn | l = \binom {2 k + 1} k + \binom {2 k + 1} {k + 1}        | r = 2^{2 k + 1}       | o = \le       | c =  }} {{eqn | ll = \implies       | n = 1       | l = \binom {2 k + 1} k        | r = 2^{2 k}       | o = \le       | c = since $\displaystyle \binom {2 k + 1} k = \binom {2 k + 1} {k + 1}$ }} {{end-eqn}} If $m = 2 k + 1$ is odd, then all the prime numbers $k + 2 \le p \le 2 k + 1$ divide: :$\dbinom {2 k + 1} k = \dfrac {\left({2 k + 1}\right)!} {k! \left({k + 1}\right)!}$ Thus: {{begin-eqn}} {{eqn | l = \prod_{p \mathop \le m} p       | r = \prod_{p \mathop \le k + 1} p \prod_{k + 2 \mathop \le p \mathop \le 2 k + 1} p       | c =  }} {{eqn | r = \binom {2 k + 1} k \prod_{p \mathop \le k + 1} p        | o = \le       | c =  }} {{eqn | r = 2^{2 k} \prod_{p \mathop \le k + 1} p       | o = \le       | c = by $(1)$ }} {{eqn | r = 2^{2 k} 2^{2 \left({k + 1}\right)}       | o = \le       | c = by the induction hypothesis }} {{eqn | r = 2^{2 m}       | c =  }} {{end-eqn}} {{qed|lemma}}
Bertrand-Chebyshev Theorem/Lemma 3
== Lemma ==  If $p$ is a prime number and $p^k \mathrel \backslash \dbinom {2 n} n$, then $p^k \le 2 n$.  == Proof  == Let $l$ be the largest power of $p$ with $p^l \le 2 n$. By De Polignac's Formula, the largest power of $p$ dividing $n!$ is $\displaystyle \sum_{i \mathop \ge 1} \left \lfloor{\frac n {p^i}}\right \rfloor$. So: {{begin-eqn}} {{eqn | l = k       | r = \sum_{i \mathop \ge 1} \left \lfloor{\frac {2 n} {p^i} }\right \rfloor - 2 \sum_{i \mathop \ge 1} \left \lfloor{\frac n {p^i} }\right \rfloor       | o = \le       | c = Largest power of $p$ dividing $\dbinom {2 n} n$ }} {{eqn | r = \sum_{i \mathop = 1}^l \left({\left \lfloor{\frac {2 n} {p^i} }\right \rfloor - 2 \left \lfloor{\frac n {p^i} }\right \rfloor}\right)       | c = as terms with $i > l$ are zero }} {{eqn | r = \sum_{i \mathop = 1}^l 1       | o = \le       | c = since $\left \lfloor{2 x}\right \rfloor - 2 \left \lfloor{x}\right \rfloor \le 1$ }} {{eqn | r = l       | c =  }} {{end-eqn}} {{qed|lemma}}
Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index
Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index 0 37642 204674 2014-12-20T15:04:02Z Prime.mover 59 Created page with "== Corollary to Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index ==  Let $n \in \Z_{>0}$ be a Definition:Strictly Positive Integ..." wikitext text/x-wiki == Corollary to Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index ==  Let $n \in \Z_{>0}$ be a strictly positive integer. Let $\dbinom n k$ be the binomial coefficient of $n$ over $k$ for a positive integer $k \in \Z_{\ge 0}$. Let $S_n = \left\langle{x_k}\right\rangle$ be the sequence defined as: :$x_k = \dbinom n k$ Then $S_n$ is strictly decreasing exactly where $\dfrac n 2 < k \le n$.  == Proof == If $k > \dfrac n 2$ then it follows that $n - k < \dfrac n 2$. Thus: {{begin-eqn}} {{eqn | l = \binom n {k + 1}       | r = \binom n {n - \left({k + 1}\right)}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \binom n {n - k - 1}       | c =  }} {{eqn | o = <       | r = \binom n {n - k}       | c = Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index }} {{eqn | r = \binom n k       | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} Hence the result.
Isometry between Metric Spaces is Continuous/Corollary
== Corollary to Isometry between Metric Spaces is Continuous ==  Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces. Let $\phi: M_1 \to M_2$ be an isometry. Then its inverse $\phi^{-1}: M_2 \to M_1$ is a continuous mapping.  == Proof == From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry. The result follows from Isometry between Metric Spaces is Continuous.  == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev = Isometry between Metric Spaces is Continuous|next = Definition:Homeomorphism/Metric Spaces/Definition 1}}: $\S 2.7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.5$
Axiom:Neighborhood Space Axioms
== Axioms == A neighborhood space is a set $S$ such that, for each $x \in S$, there exists a set of subsets $\mathcal N_x$ of $S$ satisfying the following conditions:  {{begin-axiom}} {{axiom|n = N1        |lc= There exists at least one element in $\mathcal N_x$        |q = \forall x \in S        |m = \mathcal N_x \ne \varnothing }} {{axiom|n = N2        |lc= Each element $x$ is in its own $\mathcal N_x$        |q = \forall x \in S        |m = \forall N \in \mathcal N_x: x \in N }} {{axiom|n = N3        |lc= Each superset of $N \in \mathcal N_x$ is also in $\mathcal N_x$        |q = \forall x \in S: \forall N \in \mathcal N_x        |m = N' \supseteq N \implies N' \in \mathcal N_x }} {{axiom|n = N4        |lc= Intersection of $2$ elements of $\mathcal N_x$ is also in $\mathcal N_x$        |q = \forall x \in S: \forall M, N \in \mathcal N_x        |m = M \cap N \in N_x }} {{axiom|n = N5        |lc= Exists $N' \subseteq N \in \mathcal N_x$ which is $\mathcal N_y$ of each $y \in N'$        |q = \forall x \in S: \forall N \in \mathcal N_x        |m = \exists N' \in \mathcal N_x, N' \subseteq N: \forall y \in N': N' \in \mathcal N_y }} {{end-axiom}}  These stipulations are called the '''neighborhood space Axioms'''. Each element of $\mathcal N_x$ is called a '''neighborhood''' of $x$. == Also see == * Basic Properties of Neighborhood in Topological Space == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev=Definition:Neighborhood Space|next=Definition:Neighborhood (Neighborhood Space)}}: $\S 3.3$: Neighborhoods and Neighborhood Spaces: Definition $3.4$
Finite Intersection of Open Sets of Neighborhood Space is Open
== Corollary to Intersection of two Open Sets of Neighborhood Space is Open ==  Let $\left({S, \mathcal N}\right)$ be a neighborhood space. Let $n \in \N_{>0}$ be a natural number. Let $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\left({S, \mathcal N}\right)$. Then $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$.  == Proof == Proof by induction: Let $U_1, U_2, \ldots$ be open sets of $\left({S, \mathcal N}\right)$. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$. $P \left({1}\right)$ is true, as this just says: :$U_1$ is an open set of $\left({S, \mathcal N}\right)$. === Basis for the Induction === $P \left({2}\right)$ is the case: :$U_1 \cap U_2$ is an open set of $\left({S, \mathcal N}\right)$ which has been proved above. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$. Then we need to show: :$\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i$ is an open set of $\left({S, \mathcal N}\right)$. === Induction Step === This is our induction step: We have that: :$\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i = \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ From the induction hypothesis: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$. From the basis for the induction: :$\displaystyle \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ is an open set of $\left({S, \mathcal N}\right)$. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$ \displaystyle\forall n \in \N_{>0}: \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$.  == Sources == * {{BookReference|Introduction to Topology|1962|Bert Mendelson|prev=Intersection of two Open Sets of Neighborhood Space is Open|next=Union of Open Sets of Neighborhood Space is Open}}: $\S 3.3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$
Form of Geometric Sequence of Integers/Corollary
== Corollary to Form of Geometric Sequence of Integers ==  Let $p$ and $q$ be integers. Then the finite sequence $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ defined as: :$a_j = p^j q^{n - j}$ is a geometric sequence whose common ratio is $\dfrac p q$.  == Proof == Let the greatest common divisor of $p$ and $q$ be $d$. Then by Integers Divided by GCD are Coprime: :$p = d r$ :$q = d s$ where $r$ and $s$ are coprime integers. Thus: :$a_j = p^j q^{n - j}$ {{begin-eqn}} {{eqn | l = a_j       | r = p^j q^{n - j}       | c =  }} {{eqn | r = \paren {d r}^j \paren {d s}^{n - j}       | c =  }} {{eqn | r = d^n r^j s^{n - j}       | c =  }} {{end-eqn}} and so by Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence whose common ratio is $\dfrac r s$. Then: {{begin-eqn}} {{eqn | l = \dfrac r s       | r = \paren {\dfrac p d} / \paren {\dfrac q d}       | c =  }} {{eqn | r = \dfrac p d \dfrac d q       | c =  }} {{eqn | r = \dfrac p q       | c =  }} {{end-eqn}} Hence the result.
Greatest Common Measure of Commensurable Magnitudes/Porism
== Porism to Greatest Common Measure of Commensurable Magnitudes == {{:Euclid:Proposition/X/3/Porism}} == Proof == Apparent from the construction.  {{Euclid Note|3|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Greatest Common Measure of Commensurable Magnitudes|next=Greatest Common Measure of Three Commensurable Magnitudes}}: Book $\text{X}$. Propositions
Magnitudes with Rational Ratio are Commensurable/Porism
== Porism to Magnitudes with Rational Ratio are Commensurable == {{:Euclid:Proposition/X/6/Porism}} == Proof == Apparent from the construction.  {{Euclid Note|6|X}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Magnitudes with Rational Ratio are Commensurable|next = Incommensurable Magnitudes have Irrational Ratio}}: Book $\text{X}$. Propositions
Ratios of Equal Magnitudes/Porism
== Porism to Ratios of Equal Magnitudes == {{:Euclid:Proposition/V/7/Porism}} == Proof == Apparent from the construction.  {{Euclid Note|7|V}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Ratios of Equal Magnitudes|next=Relative Sizes of Ratios on Unequal Magnitudes}}: Book $\text{V}$. Propositions
Construction of Incommensurable Lines/Lemma
== Lemma to Construction of Incommensurable Lines == {{:Euclid:Proposition/X/10/Lemma}} == Proof == The propositions referred to are: :{{EuclidPropLink|book = VIII|prop = 26|title = Similar Plane Numbers have Same Ratio as between Two Squares}} and its converse, which {{AuthorRef|Euclid}} does not explicitly prove.  :''For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.'' :''Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.''  {{Euclid Note|10|X|It was suggested by {{AuthorRef|Johan Ludvig Heiberg|Heiberg}} that this lemma is a later interpolation, and not part of the original work by {{AuthorRef|Euclid}}. The theorem Construction of Incommensurable Lines that it supports is itself also questionable.<br/><br/>Moreover, there is no reason why numbers which are not similar plane numbers should be introduced here.}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 3|1926|ed = 2nd|edpage =Second Edition|Sir Thomas L. Heath|prev = Commensurability of Squares/Porism|next = Construction of Incommensurable Lines}}: Book $\text{X}$. Propositions
Sum of Möbius Function over Divisors/Lemma
== Lemma to Sum of Möbius Function over Divisors == Let $n \in \Z_{>0}$, i.e. let $n$ be a strictly positive integer. Let $\displaystyle \sum_{d \mathop \divides n}$ denote the sum over all of the divisors of $n$. Let $\map \mu d$ be the Möbius function. Then:  :$\displaystyle \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$ That is: :$\mu * u = \iota$ where $u$ and $\iota$ are the unit arithmetic function and identity arithmetic function respectively.  == Proof == The lemma is clearly true if $n=1$. Assume, then, that $n > 1$ and write, by the Fundamental Theorem of Arithmetic: :$n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ In the sum $\displaystyle \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the divisors of $n$ which are products of distinct primes. Thus: {{begin-eqn}} {{eqn | l = \sum_{d \mathop \divides n} \map \mu d       | r = \map \mu 1 + \map \mu {p_1} + \dotsb + \map \mu {p_k} + \map \mu {p_1 p_2} + \dotsb + \map \mu {p_{k - 1} p_k} + \dotsb + \map \mu {p_1 p_2 \dotsm p_k}       | c =  }} {{eqn | r = \dbinom k 0 + \dbinom k 1 \paren {-1} + \dbinom k 2 \paren {-1}^2 + \cdots + \dbinom k k \paren {-1}^k       | c =  }} {{eqn | r = 0       | c = Alternating Sum and Difference of Binomial Coefficients for Given n }} {{end-eqn}} {{Handwaving|The jump from the expression in $\mu$ to the one in binomial coefficients is far too big. Intermediate steps need to be included.}} Hence, the sum is $1$ for $n = 1$, and $0$ for $n > 1$, which are precisely the values of $\floor {\dfrac 1 n}$.