hcju/proofwikiqa huggingface dataset:
Let $m, n \in \N: m \le n$. From natural number subtraction, $\exists p \in \N: m + p = n$ such that $n - m = p$. As $m, n, p \in \N$, it follows that $m, n, p \in \Z$ as well. However, as $\Z$ is the inverse completion of $\N$, it follows that $-m \in \Z$ as well, so it makes sense to express the following: {{begin-eqn}} {{eqn | l = \paren {n + \paren {-m} } + m       | r = n + \paren {\paren {-m} + m}       | c =  }} {{eqn | r = n       | c =  }} {{eqn | r = p + m       | c =  }} {{eqn | r = \paren {n - m} + m       | c =  }} {{end-eqn}} Thus, as all elements of $\Z$ are cancellable, it follows that $n + \paren {-m} = n - m$. So: : $\forall m, n \in \Z, m \le n: n + \paren {-m} = n - m = n -_\N m$ and the result follows. {{qed}} Category:Integers Category:Subtraction 7wy2x7ny7d5h653ngmzazw9eqd8qcub
Let $x y = x z$. There are two cases to investigate: $x > 0$ and $x < 0$. Let $x > 0$. From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$. By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication. {{qed|lemma} Let $x < 0$. We know that the Integers form Integral Domain and are thus a ring. Then $-x > 0$ and so: {{begin-eqn}} {{eqn | l = x y       | r = x z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {-\paren {-x} } y       | r = \paren {-\paren {-x} } z       | c = Negative of Ring Negative }} {{eqn | ll= \leadsto       | l = -\paren {\paren {-x} y}       | r = -\paren {\paren {-x} z}       | c = Product with Ring Negative }} {{eqn | ll= \leadsto       | l = -y       | r = -z       | c = as $-x$ is (strictly) positive, the above result holds }} {{eqn | ll= \leadsto       | l = y       | r = z       | c = $\struct {\Z, +}$ is a group: {{GroupAxiom|3}} }} {{end-eqn}} {{qed|lemma}} So whatever non-zero value $x$ takes, it is cancellable for multiplication. {{Qed}}
Let $y, z \in \Z: y \ne z$. {{begin-eqn}} {{eqn | l = y       | o = \ne       | r = z       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y - z       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \paren {y - z}       | o = \ne       | r = 0       | c = Ring of Integers has no Zero Divisors }} {{eqn | ll= \leadstoandfrom       | l = x y - x z       | o = \ne       | r = 0       | c = Integer Multiplication Distributes over Subtraction }} {{end-eqn}} The result follows by transposition. {{qed}}
Let $x y = x z$. There are two cases to investigate: $x > 0$ and $x < 0$. Let $x > 0$. From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$. By the Extension Theorem for Distributive Operations and Ordering on Natural Numbers is Compatible with Multiplication, $x$ is cancellable for multiplication. {{qed|lemma}} Let $x < 0$. We know that the Integers form Integral Domain and are thus a ring. Then $-x > 0$ and so: {{begin-eqn}} {{eqn | l = \paren {-x} y       | r = -\paren {x y}       | c = Product with Ring Negative }} {{eqn | r = -\paren {x z}       | c = $\struct {\Z, +}$ is a group: {{GroupAxiom|3}} }} {{eqn | r = \paren {-x} z       | c = Product with Ring Negative }} {{eqn | ll= \leadsto       | l = y       | r = z       | c = from above: case where $x > 0$ }} {{end-eqn}} {{qed|lemma}} So whatever non-zero value $x$ takes, it is cancellable for multiplication. {{Qed}}
Let $z > 0$. Let $M_z: \Z \to \Z$ be the mapping defined as: :$\forall x \in \Z: \map {M_z} x = z x$ All we need to do is show that $M_z$ is an order embedding from $\struct {\Z, +, \le}$ to itself. By Monomorphism from Total Ordering, we just need to show that: :$x < y \implies z x < z y$ If $x < y$, then $0 < y - x$, so $z \in \N$ and $y - x \in \N$ by Natural Numbers are Non-Negative Integers. Thus by Ordering on Natural Numbers is Compatible with Multiplication: :$z \paren {y - x} \in \N$ Therefore :$0 < z \paren {y - x} = z y - z x$ That is: :$z x < z y$ {{qed}}
Let $x > 0$ and $x y > 0$. Suppose $y \le 0$. Then from Multiplicative Ordering on Integers and Ring Product with Zero: :$x y \le x \, 0 = 0$ From this contradiction we deduce that $y > 0$. Now, if we have $x > 0$ and $x y = 1$, then $y > 0$ and hence $y \in \N$ by Natural Numbers are Non-Negative Integers. Hence by Invertible Elements under Natural Number Multiplication it follows that $x = 1$. Thus $1$ is the only element of $\N$ that is invertible for multiplication. Therefore by Natural Numbers are Non-Negative Integers and Product with Ring Negative, the result follows. {{qed}}
First we establish that $g$ is a homomorphism. Suppose $z \in \Z'$ such that $z \ne 0'$. Then by Ordering of Inverses in Ordered Monoid, either $z >' 0'$ or $-z >' 0'$. Thus either: :$z \in \N' \setminus \left\{{0'}\right\}$ or: :$-z \in \N' \setminus \left\{{0'}\right\}$ and thus $\N' \setminus \left\{{0'}\right\}$ is not empty. Therefore $\N' \setminus \left\{{0'}\right\}$ has a minimal element. Call this minimal element $1'$. It is clear that $\N'$ is an ordered semigroup satisfying $(NO 1)$, $(NO 2)$ and $(NO 4)$ of the naturally ordered semigroup axioms. Also: {{begin-eqn}} {{eqn | o =        | r = 0' \le' x \le' y       | c =  }} {{eqn | o = \implies       | r = 0' \le' y - x       | c =  }} {{eqn | o = \implies       | r = y - x \in \N' \land x +' \left({y - x}\right) = y       | c =  }} {{end-eqn}} Thus $\N'$ also satisfies $(NO 3)$ of the naturally ordered semigroup axioms. So $\left({\N', +', \le'}\right)$ is a naturally ordered semigroup. So, by Naturally Ordered Semigroup is Unique, the restriction to $\N$ of $g$ is an isomorphism from $\left({\N, +, \le}\right)$ to $\left({\N', +', \le'}\right)$. By Index Law for Sum of Indices, $g$ is a homomorphism from $\left({\Z, +}\right)$ into $\left({\Z', +'}\right)$. Next it is established that $g$ is surjective. Let $y \in \Z': y <' 0'$. {{begin-eqn}} {{eqn | o =        | r = y <' 0'       | c =  }} {{eqn | o = \implies       | r = -y >' 0'       | c =  }} {{eqn | o = \implies       | r = \exists n \in \N: -y = g \left({n}\right)       | c =  }} {{eqn | o = \implies       | r = y = - g \left({n}\right) = g \left({-n}\right)       | c = Homomorphism with Identity Preserves Inverses }} {{end-eqn}} Therefore $g$ is a surjection. Now to show that $g$ is a monomorphism, that is, it is injective. Let $n < m$. {{begin-eqn}} {{eqn | o =        | r = n < m       | c =  }} {{eqn | o = \implies       | r = m - n \in \N_{>0}       | c =  }} {{eqn | o = \implies       | r = g \left({m}\right) - g \left({n}\right) - g \left({m - n}\right) \in \N' \setminus \left\{ {0'}\right\}       | c =  }} {{eqn | o = \implies       | r = g \left({n}\right) <' \left({g \left({m}\right) - g \left({n}\right)}\right) +' g \left({n}\right) = g \left({m}\right)       | c = Strict Ordering Preserved under Product with Cancellable Element }} {{end-eqn}} Therefore it can be seen that $g$ is strictly increasing. It follows from Monomorphism from Total Ordering that $g$ is a monomorphism from $\left({\Z, +, \le}\right)$ to $\left({\Z', +', \le'}\right)$. A surjective monomorphism is an isomorphism, and the result follows. {{Qed}}
=== $\struct {\Z, +, \le}$ is an Ordered Structure === ==== $(1)$ ==== By Integer Addition is Closed, $\struct {\Z, +}$ is an algebraic structure. ==== $(2)$ ==== $\le$ is an ordering on $\Z$. Thus, $\struct {\Z, \le}$ is an ordered set. ==== $(3)$ ==== By Ordering is Preserved on Integers by Addition and Integer Addition is Commutative, $\le$ is compatible with $+$. Thus, $\struct {\Z, +, \le}$ is an ordered structure. {{qed|lemma}} === $\struct {\Z, +, \le}$ is a Totally Ordered Structure === By definition, the ordered structure $\struct {\Z, +, \le}$ is a totally ordered structure {{iff}} $\le$ is a total ordering. This follows from Ordering on Integers is Total Ordering. {{qed|lemma}} === $\struct {\Z, +, \le}$ is a Totally Ordered Group === By definition, the totally ordered structure $\struct {\Z, +, \le}$ is a totally ordered group {{iff}} $\struct {\Z, +}$ is a group. This follows from Integers under Addition form Abelian Group. {{qed}} Category:Integer Addition Category:Additive Group of Integers 15qysqr4r51uk7597b8j8vjkp15ng6w
From Integers form Commutative Ring, $\struct {\Z, +, \times}$ is a commutative ring. We need to show that $\le$ is a compatible ordering on $\Z$. That is, that: :$(1): \quad \le$ is compatible with $+$ :$(2): \quad \forall x, y \in \Z: 0 \le x, 0 \le y \implies 0 \le x \times y$ The first one follows from the fact that Integers under Addition form Totally Ordered Group. Now, from Multiplicative Ordering on Integers, we have: :$\forall z, x, y \in \Z, 0 < y: z \le x \iff y \times z \le y \times x$ Let $z = 0, 0 \le x$. Then: :$0 = x \times z \le x \times y = y \times x$ So if $y \ne 0$, the condition $0 \le x, 0 \le y \implies 0 \le x \times y$ holds. Now if $y = 0$, we have $x \times y = 0$ and thus $0 \le x \times y$. Thus $\le$ is compatible with $\Z$, and the result follows. {{qed}}
From: :Diagonal Relation is Universally Compatible and :Trivial Relation is Universally Congruent we know that both these relations are compatible with both addition and multiplication on $\Q$. Now we need to show that these are the ''only'' such relations. Let $\RR$ be a congruence on $\Q$, such that $\RR \ne \Delta_\Q$. {{begin-eqn}} {{eqn | l = \RR       | o = \ne       | r = \Delta_\Q       | c =  }} {{eqn | ll= \leadsto       | l = \exists r, s \in \Q: r       | o = \ne       | r = s \land r \RR s       | c =  }} {{eqn | ll= \leadsto       | l = \exists h \in \Q, h \ne 0: h       | r = r - s       | c =  }} {{eqn | ll= \leadsto       | l = h       | o = \RR       | r = 0       | c = as $\RR$ is a congruence relation compatible with $+$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \forall x \in \Q: \paren {x / h}       | o = \RR       | r = \paren {x / h}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {x / h} \times h}       | o = \RR       | r = \paren {\paren {x / h} \times 0}       | c = as $\RR$ is a congruence relation compatible with $\times$ }} {{eqn | ll= \leadsto       | l = x       | o = \RR       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 0 \RR       | r = \Q       | c =  }} {{eqn | ll= \leadsto       | l = \RR       | r = \Q \times \Q       | c =  }} {{end-eqn}} {{qed}}
Let the partition of $S$ be $S_1, S_2, \ldots, S_n$. Then: :$\forall k \in \set {1, 2, \ldots, n}: \card {S_k} = m$ By definition of multiplication: :$\displaystyle \sum_{k \mathop = 1}^n \card {S_k} = n m$ and the result follows from the Fundamental Principle of Counting. {{qed}}
The result is obvious for $S = T$ or $T = \O$. Otherwise, $\set {T, S \setminus T}$ is a partition of $S$. Let $\card {S \setminus T} = p$. Then by the Fundamental Principle of Counting: : $m + p = n$ and the result follows. {{qed}}
Let $\card S = n$ and $\card T = m$. If either $n = 0$ or $m = 0$, then from Cartesian Product is Empty iff Factor is Empty: :$S \times T = \O$ and the result holds, as $n m = 0 = \card \O$ from Cardinality of Empty Set. So, we assume that $n > 0$ and $m > 0$. For each $a \in S$, we define the mapping $g_a: T \to \set a \times T$ such that: :$\forall y \in T: \map {g_a} y = \tuple {a, y}$ The mapping $g_a$ is a bijection, so: :$\card {\set a \times T} = m$ Now let: :$\mathbb T = \set {\set a \times T: a \in S}$ We define the mapping $h: S \to \mathbb T$: :$\forall a \in S: \map h a = \set a \times T$ The mapping $h$ is a bijection, so $\card {\mathbb T} = n$. Thus $\mathbb T$ is a partition of $S \times T$ containing $n$ sets. Hence from Number of Elements in Partition: :$\card {S \times T} = n m$ and the result follows. {{qed}}
Let $T = \set {0, 1}$. For each $A \in \powerset S$, we consider the characteristic function $\chi_A: S \to T$ defined as: :$\forall x \in S: \map {\chi_A} x = \begin{cases} 1 & : x \in A \\ 0 & : x \notin A \end{cases}$ Now consider the mapping $f: \powerset S \to T^S$: :$\forall A \in \powerset S: \map f A = \chi_A$ where $T^S$ is the set of all mappings from $S$ to $T$. Also, consider the mapping $g: T^S \to \powerset S$: :$\forall \phi \in T^S: \map g \phi = \phi^{-1} \sqbrk {\set 1}$ where $\phi^{-1} \sqbrk {\set 1}$ is the preimage of $\set 1$ under $\phi$. Consider the characteristic function of $\phi^{-1} \sqbrk {\set 1}$, denoted $\chi_{\phi^{-1} \sqbrk {\set 1} }$. We have: {{begin-eqn}} {{eqn | l = \forall x \in S: \map {\chi_{\phi^{-1} \sqbrk {\set 1} } } x       | r = \begin{cases} 1 & : x \in \phi^{-1} \sqbrk {\set 1} \\ 0 & : x \notin \phi^{-1} \sqbrk {\set 1} \end{cases}       | c = }} {{eqn | r = \begin{cases} 1 & : \map \phi x = 1 \\ 0 & : \map \phi x = 0 \end{cases}       | c = }} {{eqn | r = \map \phi x       | c = }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \forall \phi \in T^S: \map {\paren {f \circ g} } \phi       | r = \map f {\phi^{-1} } {\sqbrk {\set 1} }       | c =  }} {{eqn | r = \chi_{\phi^{-1} \sqbrk {\set 1} }       | c =  }} {{eqn | r = \phi       | c =  }} {{end-eqn}} So $f \circ g = I_{T^S}$, that is, the identity mapping on $T^S$. So far so good. Now we consider the preimage of $\set 1$ under $\chi_A$: : $\chi_A^{-1} \sqbrk {\set 1} = A$ from the definition of the characteristic function $\chi_A$ above. Thus: {{begin-eqn}} {{eqn | l = \forall A \in \powerset S: \map {\paren {g \circ f} } A       | r = \map g {\map f A}       | c =  }} {{eqn | r = \map g {\chi_A}       | c =  }} {{eqn | r = \chi_A^{-1} \sqbrk {\set 1}       | c =  }} {{eqn | r = A       | c =  }} {{end-eqn}} So $g \circ f = I_{\powerset S}$, that is, the identity mapping on $\powerset S$. It follows from Bijection iff Left and Right Inverse that $f$ and $g$ are bijections. Thus by Cardinality of Set of All Mappings the result follows. {{Qed}}
Enumerating the subsets of $S$ is equivalent to counting all of the ways of selecting $k$ out of the $n$ elements of $S$ with $k = 0, 1, \ldots, n$. So, from Cardinality of Set of Subsets, the number we are looking for is: :$\displaystyle \card {\powerset S} = \sum_{k \mathop = 0}^n \binom n k$ But from the binomial theorem: :$\displaystyle \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ It follows that: :$2^n = \displaystyle \paren {1 + 1}^n = \sum_{k \mathop = 0}^n \binom n k \paren 1^{n - k} \paren 1^k = \sum_{k \mathop = 0}^n \binom n k = \card {\powerset S}$ {{Qed}}
Given an element $x$ of $S$, each subset of $S$ either includes $x$ or does not include $x$ (this follows directly from the definition of a set), which gives us two possibilities. The same reasoning holds for any element of $S$. One can intuitively see that this means that there are $\displaystyle \underbrace {2 \times 2 \times \ldots \times 2}_{\card S} = 2^{\card S}$ total possible combinations of elements of $S$. This is exactly $\card {\powerset S}$. {{qed}}
Proof by induction: For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$\left|{S}\right| = n \implies \left|{\mathcal P \left({S}\right)}\right| = 2^n$ Do not confuse $P \left({n}\right)$, which is a propositional function on $\N$, with $\mathcal P \left({S}\right)$, the power set of $S$. === Basis for the Induction === From Cardinality of Empty Set, we have: : $S = \varnothing \iff \left|{S}\right| = 0$ Then: : $\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$ has one element, that is, $\varnothing$. So: : $\left|{\mathcal P \left({\varnothing}\right)}\right| = \left|{\left\{{\varnothing}\right\}}\right| = 1 = 2^0$ thus confirming that $P \left({0}\right)$ holds. 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: :$\left|{S}\right| = k \implies \left|{\mathcal P \left({S}\right)}\right| = 2^k$ Then we need to show: :$\left|{S}\right| = k + 1 \implies \left|{\mathcal P \left({S}\right)}\right| = 2^{k + 1}$ === Induction Step === This is our induction step: Let $\left|{S}\right| = k + 1$. Let $x \in S$. Consider the set $S' = S \setminus \left\{{x}\right\}$ where $x$ is any element of $S$. We see that $\left|{S'}\right| = k$. Now adjoin $x$ to all the subsets of $S'$. Counting the subsets of $S$, we have: : all the subsets of $S'$ and: : all the subsets of $S'$ with $x$ adjoined to them. From the induction hypothesis, there are $2^k$ subsets of $S'$. Adding $x$ to each of these does not change their number, so there are another $2^k$ subsets of $S$ consisting of all the subsets of $S'$ with $x$ adjoined to them. In total that makes $2^k + 2^k = 2 \times 2^k = 2^{k + 1}$ subsets of $S$. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \left|{S}\right| = n \implies \left|{\mathcal P \left({S}\right)}\right| = 2^n$ {{qed}}
Follows directly from Cardinality of Set of Injections and Equivalence of Mappings between Sets of Same Cardinality. {{Qed}}
For each $X \subseteq \N_n$ and $Y \subseteq S$, let $B \left({X, Y}\right)$ be the set of all bijections from $X$ onto $Y$. Let $\Bbb S$ be the set of all subsets of $S$ with $m$ elements. By Cardinality of Power Set of Finite Set and Cardinality of Subset of Finite Set, $\Bbb S$ is finite, so let $s = \left|{\Bbb S}\right|$. Let $\beta: B \left({\N_n, S}\right) \to \Bbb S$ be the mapping defined as $\forall f \in B \left({\N_n, S}\right): \beta \left({f}\right) = f \left({\N_m}\right)$. For each $Y \in \Bbb S$, the mapping: :$\Phi_Y: \beta^{-1} \left({Y}\right) \to B \left({\N_m, Y}\right) \times B \left({\N_n - \N_m, S - Y}\right)$ defined as: :$\Phi_Y \left({f}\right) = \left({f_{\N_m}, f_{\N_n - \N_m}}\right)$ is also (clearly) a bijection. By Cardinality of Set of Bijections: :$\left|{B \left({\N_m, Y}\right)}\right| = m!$ and: :$\left|{B \left({\N_n - \N_m, S - Y}\right)}\right| = \left({n - m}\right)!$ So by Cardinality of Cartesian Product: :$\left|{\beta^{-1} \left({Y}\right)}\right| = m! \left({n - m}\right)!$ It is clear that $\left\{{\beta^{-1} \left({Y}\right): Y \in \Bbb S}\right\}$ is a partition of $B \left({\N_n, S}\right)$. Therefore by Number of Elements in Partition: :$\left|{B \left({\N_n, S}\right)}\right| = m! \left({n - m}\right)! s$ Consequently, as $\left|{B \left({\N_n, S}\right)}\right| = n!$ by Cardinality of Set of Bijections, it follows that: :$m! \left({n - m}\right)! s = n!$ and the result follows. {{qed}}
Let ${}^m C_n$ be the number of subsets of $m$ elements of $S$. From Number of Permutations, the number of $m$-permutations of $S$ is: :${}^m P_n = \dfrac {n!} {\paren {n - m}!}$ Consider the way ${}^m P_n$ can be calculated. First one makes the selection of which $m$ elements of $S$ are to be arranged. This number is ${}^m C_n$. Then for each selection, the number of different arrangements of these is $m!$, from Number of Permutations. So: {{begin-eqn}} {{eqn | l = m! \cdot {}^m C_n       | r = {}^m P_n       | c = Product Rule for Counting }} {{eqn | r = \frac {n!} {\paren {n - m}!}       | c = Number of Permutations }} {{eqn | ll= \leadsto       | l = {}^m C_n       | r = \frac {n!} {m! \paren {n - m}!}       | c = Product Rule for Counting }} {{end-eqn}} {{qed}}
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$. Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$. Let $r \in \N: 0 < r \le n$. Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$: :$B_r  := \set {S \subseteq \N_n: \card S = r}$ Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as: :$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$ where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$. From Group Action of Symmetric Group on Subset it is established that $*$ is a group action. The stabilizer of any $U \in B_r$ is the set of permutations on $\N_n$ that fix $U$. Let $U = \set {1, 2, \ldots, r}$. So: :$\tuple {a_1, a_2, \ldots, a_r}$ can be any one of the $r!$ permutations of $1, 2, \ldots, r$ :$\tuple {a_{r + 1}, a_{r + 2}, \ldots, _n}$ can be any one of the $\paren {n - r}!$ permutations of $r + 1, r + 2, \ldots, n$. Thus: :$\order {\Stab U} = r! \paren {n - r}!$ From Group Action of Symmetric Group on Subset is Transitive: :$B_r = \Orb U$ and so: :$\card {B_r} = \card {\Orb U}$ From the Orbit-Stabilizer Theorem: :$\card {\Orb U} = \dfrac {\order {S_n} } {\order {\Stab U} } = \dfrac {n!} {r! \paren {n - r}!}$ But $\card {B_r}$ is the number of subsets of $\N_n$ of cardinality $r$. Hence the result. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$ === Basis for the Induction === $\map P 1$ is the case: :${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$ {{begin-eqn}} {{eqn | l = {}^m C_1       | r = \begin {cases} 1 & : m = 0 \text { or } m = 1 \\ 0 & : \text {otherwise} \end {cases}       | c =  }} {{eqn | r = \dfrac {1!} {0! \paren {1 - 0}!}       | c = for $m = 0$ }} {{eqn | r = \dfrac {1!} {1! \paren {1 - 1}!}       | c = for $m = 1$ }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :${}^m C_k = \dfrac {k!} {m! \paren {k - m}!}$ from which it is to be shown that: :${}^m C_{k + 1} = \dfrac {\paren {k + 1}!} {m! \paren {k + 1 - m}!}$ === Induction Step === This is the induction step: The number of ways to choose $m$ elements from $k + 1$ elements is: :the number of ways to choose $m$ elements elements from $k$ elements (deciding not to select the $k + 1$th element) added to: :the number of ways to choose $m - 1$ elements elements from $k$ elements (after having selected the $k + 1$th element for the $n$th selection) {{begin-eqn}} {{eqn | l = {}^m C_{k + 1}       | r = {}^m C_k + {}^{m - 1} C_k       | c =  }} {{eqn | r = \binom k m + \binom k {m - 1}       | c = Induction Hypothesis }} {{eqn | r = \binom {k + 1} m       | c = Pascal's Rule }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: {}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
We have that $S$ is bounded below in $\Z$. So: : $\exists m \in \Z: \forall s \in S: m \le s$ Hence: : $\forall s \in S: 0 \le s - m$ Thus: :$T = \set {s - m: s \in S} \subseteq \N$ The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$. Hence: :$\paren {\forall s \in S: b_T \le s - m} \land \paren {\exists b_S \in S: b_T = b_S - m}$ So: {{begin-eqn}} {{eqn | o =        | r = s \in S: b_S - m \le s - m       | c =  }} {{eqn | o = \leadsto       | r = \forall s \in S: b_S \le s       | c = Cancellability of elements of $\Z$ }} {{eqn | o = \leadsto       | r = b_S \in S \land \paren {\forall s \in S: b_S \le s}       | c = {{Defof|Smallest Element}} }} {{end-eqn}} So $b_S$ is the smallest element of $S$. {{Qed}}
$S$ is bounded above, so $\exists M \in \Z: \forall s \in S: s \le M$. Hence $\forall s \in S: 0 \le M - s$. Thus the set $T = \set {M - s: s \in S} \subseteq \N$. The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$. Hence: :$\paren {\forall s \in S: b_T \le M - s} \land \paren {\exists g_S \in S: b_T = M - g_S}$ So: {{begin-eqn}} {{eqn | o =        | r = \forall s \in S: M - g_S \le M - s       | c =  }} {{eqn | o = \leadsto       | r = \forall s \in S: -g_S \le -s       | c = Cancellability of elements of $\Z$ }} {{eqn | o = \leadsto       | r = \forall s \in S: g_S \ge s       | c =  }} {{eqn | o = \leadsto       | r = g_S \in S \land \paren {\forall s \in S: g_S \ge s}       | c = which is how the greatest element is defined. }} {{end-eqn}} 		  So $g_S$ is the greatest element of $S$. {{qed}}
Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$. Hence we can define the set $S' = \set {-s: s \in S}$. $S'$ is bounded below by $-M$. So from Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a smallest element, $-g_S$, say, where $\forall s \in S: -g_S \le -s$. Therefore $g_S \in S$ (by definition of $S'$) and $\forall s \in S: s \le g_S$. So $g_S$ is the greatest element of $S$. {{qed}}