hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$ We have that $\struct {R, \circ, *}$ is a ringoid, and so: :$\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$ :$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$ === Basis for the Induction === $\map P 1$ is the case: :$\displaystyle \forall m \in \Z_{>0}: \paren {\sum_{i \mathop = 1}^m a_i} * b_1 = \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_1}$ This is demonstrated in Lemma 1. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j}$ Then we need to show: :$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j}       | r = \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\paren {\sum_{j \mathop = 1}^k b_j} \circ b_{k + 1} }       | c = {{Defof|Summation}} }} {{eqn | r = \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \paren {\paren {\sum_{i \mathop = 1}^m a_i} * b_{k + 1} }       | c = as $\struct {R, \circ, *}$ is a ringoid }} {{eqn | r = \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j} \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}       | c = Associativity of $\circ$ in $\struct {R, \circ, *}$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$ {{qed}} The same result can be obtained by fixing $n$ and using induction on $m$, which requires Lemma 2 to be used for its base case.
For each $k \in \N_{>0}$, let $\left|{B_k}\right| = p_k$. Let: :$r_0 = 0$ :$\displaystyle \forall k \in \N_{>0}: r_k = \sum_{j \mathop = 1}^k {p_j}$ and: :$p = r_n$ Then: : $r_k - r_{k-1} = p_k$ So, by Isomorphism to Closed Interval, both $\left[{1 \,.\,.\, p_k}\right]$ and $\left[{r_{k - 1} + 1 \,.\,.\, r_k}\right]$ have $p_k$ elements. By Unique Isomorphism between Finite Totally Ordered Sets, there is a unique isomorphism $\tau_k: \left[{1 \,.\,.\, p_k}\right] \to \left[{r_{k - 1} + 1 \,.\,.\, r_k}\right]$ as both are totally ordered. The orderings on both of these are those induced by the ordering on $\N$. It is clear that $\tau_k$ is defined as: : $\forall j \in \left[{1 \,.\,.\, p_k}\right]: \tau_k \left({j}\right) = r_k + j$ For each $k \in \left[{1 \,.\,.\, n}\right]$, let $\rho_k: \left[{1 \,.\,.\, p_k}\right] \to B_k$ be a bijection. By Strictly Increasing Sequence induces Partition, the mapping $\sigma: \left[{1 \,.\,.\, p}\right] \to A$ defined by: : $\displaystyle \forall j \in \left[{r_{k-1}+1 \,.\,.\, r_k}\right]: \forall k \in \left[{1 \,.\,.\, n}\right]: \sigma \left({j}\right) = \rho_k \left({\tau_k^{-1} \left({j}\right)}\right)$ is a bijection. Let $\forall j \in \left[{1 \,.\,.\, p}\right]: y_j = x_{\sigma \left({j}\right)}$. By definition: : $\displaystyle \prod_{\alpha \mathop \in A} {x_\alpha} = \prod_{j \mathop = 1}^p {x_{\sigma \left({j}\right)}} = \prod_{j \mathop = 1}^p {y_j}$ Also: : $\displaystyle \forall k \in \left[{1 \,.\,.\, n}\right]: \prod_{\alpha \mathop \in B_k} {x_\alpha} = \prod_{i \mathop = 1}^{p^k} {x_{\rho_k \left({i}\right)}}$ Also by definition: : $\displaystyle \prod_{j \mathop = r_{k - 1} + 1}^{r_k} y_j = \prod_{i \mathop = 1}^{p_k} y_{\tau_k \left({i}\right)} = \prod_{i \mathop = 1}^{p_k} x_{\sigma \left({\tau_k \left({i}\right)}\right)} = \prod_{i \mathop = 1}^{p_k} x_{\rho_k \left({i}\right)}$ So by the General Associativity Theorem: {{begin-eqn}} {{eqn | l = \prod_{a \mathop \in A} {x_a}       | r = \prod_{j \mathop = 1}^p y_j       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} {y_j} }\right)       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{i \mathop = 1}^{p_k} x_{\rho_k \left({i}\right)} }\right)       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{a \mathop \in B_k} x_\alpha}\right)       | c =  }} {{end-eqn}} {{qed}}
Taking the group axioms in turn: === G0: Closure === From External Direct Product Closure it follows that $\struct {G_1 \times G_2, \circ}$ is closed. {{qed|lemma}} === G1: Associativity === From External Direct Product Associativity it follows that $\circ$ is associative on $G_1 \times G_2$. {{qed|lemma}} === G2: Identity === From External Direct Product Identity it follows that $\tuple {e_1, e_2}$ is the identity element of $\struct {G_1 \times G_2, \circ}$. {{qed|lemma}} === G3: Inverses === From External Direct Product Inverses it follows that $\tuple {g_1^{-1}, g_2^{-1} }$ is the inverse element of $\tuple {g_1, g_2}$ in $\struct {G_1 \times G_2, \circ}$. {{qed|lemma}} All group axioms are fulfilled, hence the result. {{qed}}
Let $\tuple {s_1, s_2} \in S_1 \times S_2$. So, $s_1 \in S_1$ and $s_2 \in S_2$. From the definition of the canonical injection, we have: :$\map {\inj_1} {s_1} = \tuple {s_1, e_2}$ :$\map {\inj_2} {s_2} = \tuple {e_1, s_2}$ From the definition of the first projection: :$\map {\pr_1} {s_1, e_2} = s_1$ and similarly from the definition of the second projection: :$\map {\pr_2} {e_1, s_2} = s_2$ Thus: :$\map {\pr_1 \circ \inj_1} {s_1} = s_1$ :$\map {\pr_2 \circ \inj_2} {s_2} = s_2$ and the result follows from the definition of the identity mapping. {{qed}}
The set of natural numbers under addition can be denoted $\left ({\N, +}\right)$. From Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\left ({\N, +}\right)$ is a commutative monoid. Therefore by definition of commutative monoid, $\left ({\N, +}\right)$ is a commutative semigroup. From Natural Number Addition is Cancellable, all of the elements of $\left ({\N, +}\right)$ are cancellable. The result follows from the Inverse Completion Theorem. {{qed}} Category:Inverse Completions Category:Natural Numbers q7fkui5qs8edlrc74tqw2t2cnac9r61
Let $l = \displaystyle \lim_{x \mathop \to a^+} \frac{f' \left({x}\right)}{g' \left({x}\right)}$. Let $\epsilon > 0$. By the definition of limit, we ought to find a $\delta > 0$ such that: :$\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l}\right\vert < \epsilon$ Fix $\delta$ such that: :$\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f' \left({x}\right)} {g' \left({x}\right)} - l}\right\vert < \epsilon$ which is possible by the definition of limit. Let $x$ be such that $\left\vert{x - a}\right\vert < \delta$. By the Cauchy Mean Value Theorem with $b = x$: : $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$ Since $f \left({a}\right) = g \left({a}\right) = 0$, we have: : $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right)} {g \left({x}\right)}$ Now, as $a < \xi < x$, it follows that $\left\vert{\xi - a}\right\vert < \delta$ as well. Therefore: :$\left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l }\right\vert = \left\vert{ \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} - l}\right\vert < \epsilon$ which leads us to the desired conclusion that: :$\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$ {{qed}}
Take the Cauchy Mean Value Theorem with $b = x$: :$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$ Then if $\map f a = \map g a = 0$ we have: :$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map g x}$ Note that $\xi$ depends on $x$; that is, different values of $x$ may require different values of $\xi$ to make the above statement valid. It follows from Limit of Function in Interval that $\xi \to a$ as $x \to a$. Also, $\xi \ne a$ when $x > a$. So from '''Hypothesis $2$''' of Limit of Composite Function, it follows that: :$\displaystyle \lim_{x \mathop \to a^+} \dfrac {\map {f'} \xi} {\map {g'} \xi} = \lim_{x \mathop \to a^+} \dfrac {\map {f'} x} {\map {g'} x}$ Hence the result. {{qed}}
Consider the natural numbers $\N$ defined as the naturally ordered semigroup. From the definition of the naturally ordered semigroup, it follows that $\struct {\N, +}$ is a commutative semigroup. From the definition of zero, $\struct {\N, +}$ has $0 \in \N$ as the identity, hence is a monoid. {{qed}}
The algebraic structure $\struct {\N, +}$ is a commutative monoid from Natural Numbers under Addition form Commutative Monoid. Then we have: :$(1): \quad$ $\times$ is closed :$(2): \quad$ Natural Number Multiplication is Associative :$(3): \quad$ Natural Number Multiplication is Commutative :$(4): \quad$ Natural Number Multiplication Distributes over Addition Thus $\struct {\N, \times}$ forms an algebraic structure which is closed such that $\times$ is associative and commutative. So by definition, $\struct {\N, \times}$ is a commutative semigroup. The result follows from definition of commutative semiring and the distributivity of $\times$ over $+$. {{qed}} Category:Natural Numbers Category:Semirings 3dw345e0hzrnc5uqio3ha4k7kqjsxnm
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as: :$\forall u \in \N_{>0}: \map \phi u = u'$ where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$. Let $v' = \eqclass {c + v, c} {}$. Then: {{begin-eqn}} {{eqn | l = u' v'       | r = \eqclass {b + u, b} {} \times \eqclass {c + v, c} {}       | c =  }} {{eqn | r = \eqclass {\paren {b + u} \paren {c + v} + b c, \paren {b + u} c + b \paren {c + v} } {}       | c =  }} {{eqn | r = \eqclass {b c + b v + c u + u v + b c, b c + u c + b c + b v} {}       | c =  }} {{eqn | r = \eqclass {b c + u v, b c} {}       | c =  }} {{eqn | r = \eqclass {b + u v, b} {}       | c =  }} {{eqn | r = \paren {u v}'       | c =  }} {{end-eqn}} {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. Integer multiplication is defined as: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$ We have that: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$ Also: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$ But: :$a c + b d \in \N, a d + b c \in \N$ So: :$\forall a, b, c, d \in \N: \eqclass {a c + b d, a d + b c} {} \in \Z$ Therefore integer multiplication is closed. {{qed}}
From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$. Then: {{begin-eqn}} {{eqn | l = x \times y       | r = \eqclass {a, b} {}\times \eqclass {c, d} {}       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{eqn | r = \eqclass {a c + b d, a d + b c} {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = \eqclass {c a + d b, d a + c b} {}       | c = Natural Number Multiplication is Commutative }} {{eqn | r = \eqclass {c a + d b, c b + d a} {}       | c = Natural Number Addition is Commutative }} {{eqn | r = \eqclass {c, d} {} \times \eqclass {a, b} {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = y \times x       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{end-eqn}} {{qed}}
From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$ for some $x, y, z \in \Z$. Then: {{begin-eqn}} {{eqn | l = x \times \paren {y \times z}       | r = \eqclass {a, b} {} \times \paren {\eqclass {c, d} {} \times \eqclass {e, f} {} }       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{eqn | r = \eqclass {a, b} {} \times \eqclass {c e + d f, c f + d e} {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = \eqclass {a \paren {c e + d f} + b \paren {c f + d e}, a \paren {c f + d e} + b \paren {c e + d f} } {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = \eqclass {a c e + a d f + b c f + b d e, a c f + a d e + b c e + b d f} {}       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | r = \eqclass {a c e + b d e + a d f + b c f, a c f + b d f + a d e + b c e} {}       | c = Natural Number Addition is Commutative and Associative }} {{eqn | r = \eqclass {\paren {a c + b d} e + \paren {a d + b c} f, \paren {a c + b d} f + \paren {a d + b c} e} {}       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | r = \eqclass {a c + b d, a d + b c} {} \times \eqclass {e, f} {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = \paren {\eqclass {a, b} {} \times \eqclass {c, d} {} } \times \eqclass {e, f} {}       | c = {{Defof|Integer Multiplication}} }} {{eqn | r = \paren {x \times y} \times z       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{end-eqn}} {{qed}}
Let us define $\Z$ as in the formal definition of integers. That is, $\Z$ is an inverse completion of $\N$. From Natural Numbers form Commutative Semiring, we have that: :All elements of $\N$ are cancellable for addition :Addition and multiplication are commutative and associative on the natural numbers $\N$ :Natural number multiplication is distributive over natural number addition. The result follows from the Extension Theorem for Distributive Operations. {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$. $\boxtimes$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$. In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested. From the method of construction, $\eqclass {c + 1, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $1 \in \N$. To ease the algebra, we will take $\eqclass {1, 0} {}$ as a canonical instance of this equivalence class. Thus it is to be shown that: :$\forall a, b \in \N: \eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$ From Natural Numbers form Commutative Semiring, we take it for granted that: :addition and multiplication are commutative and associative on the natural numbers $\N$ :natural number multiplication is distributive over natural number addition. So: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {} \times \eqclass {1, 0} {}       | r = \eqclass {1 \times a + 0 \times b, 0 \times a + 1 \times b} {}       | c =  }} {{eqn | r = \eqclass {a, b} {}       | c = from Construction of Inverse Completion: Members of Equivalence Classes }} {{end-eqn}} So: :$\eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {}$ The identity $\eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$ is demonstrated similarly. {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$. In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$. To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class. We need to show that: $\forall a, b, c \in \N: \eqclass {a, b} {} \times \eqclass {0, 0} {} = \eqclass {0, 0} {} = \eqclass {0, 0} {} \times \eqclass {a, b} {}$. From Natural Numbers form Commutative Semiring, we can take it for granted that addition and multiplication are commutative on the natural numbers $\N$. {{begin-eqn}} {{eqn | l = \eqclass {a, b} {} \times \eqclass {0, 0} {}       | r = \eqclass {a \times 0 + b \times 0, a \times 0 + b \times 0} {}       | c =  }} {{eqn | r = \eqclass {0, 0} {}       | c = Construction of Inverse Completion: Equivalence Class of Equal Elements }} {{eqn | r = \eqclass {0 \times a + 0 \times b, 0 \times a + 0 \times b} {}       | c =  }} {{eqn | r = \eqclass {0, 0} {} \times \eqclass {a, b} {}       | c =  }} {{end-eqn}} {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$. In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$. To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class. We need to show that: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {} \lor \eqclass {c, d} {} = \eqclass {0, 0} {}$ From Natural Numbers form Commutative Semiring, we can take it for granted that: :addition and multiplication are commutative and associative on the natural numbers $\N$ :natural number multiplication is distributive over natural number addition. So: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {} \times \eqclass {c, d} {}       | r = \eqclass {0, 0} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {a c + b d, a d + b c} {}       | r = \eqclass {0, 0} {}       | c =  }} {{eqn | ll= \leadsto       | l = a c + b d + 0       | r = a d + b c + 0       | c =  }} {{eqn | ll= \leadsto       | l = a c + b d       | r = a d + b c       | c =  }} {{end-eqn}} We have to be careful here, and bear in mind that $a, b, c, d$ are natural numbers, and we have not defined (and, at this stage, will not define) subtraction on such entities. {{WLOG}}, suppose that $\eqclass {c, d} {} \ne \eqclass {0, 0} {}$. Then $c \ne d$. {{WLOG}}, suppose also that $c > d$. {{wtd|Ordering in terms of Addition should have a version for $\N$}} From Ordering in terms of Addition, $\exists p \in \N: d + p = c$ where $p > 0$. Then: {{begin-eqn}} {{eqn | l = a c + b d       | r = a d + b c       | c =  }} {{eqn | ll= \leadsto       | l = a \paren {d + p} + b d       | r = a d + b \paren {d + p}       | c =  }} {{eqn | ll= \leadsto       | l = a d + a p + b d       | r = a d + b d + b p       | c =  }} {{eqn | ll= \leadsto       | l = a p       | r = b p       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | r = \eqclass {0, 0} {}       | c = Construction of Inverse Completion: Equivalence Class of Equal Elements }} {{end-eqn}} Similarly for when $c < d$. Thus: :$\eqclass {c, d} {} \ne \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {}$ A similar argument shows that: :$\eqclass {a, b} {} \ne \eqclass {0, 0} {} \implies \eqclass {c, d} {} = \eqclass {0, 0} {}$ The equivalence between the two forms of the statement of this theorem follows from De Morgan's Laws: Conjunction of Negations and the Rule of Transposition. {{qed}}
First we note that the integers form a commutative ring with unity whose zero is $0$ and whose unity is $1$. Next we see that the $\struct {\Z, +, \times}$ has no divisors of zero. So, by definition, the algebraic structure $\struct {\Z, +, \times}$ is an integral domain whose zero is $0$ and whose unity is $1$. {{Qed}}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$. In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. Thus, what we are trying to prove is: :$\eqclass {a, b} {} = \eqclass {c, d} {} \land b < a \implies d < c$ By definition: :$\eqclass {a, b} {} = \eqclass {c, d} {} \iff a + d = b + c$ So: {{begin-eqn}} {{eqn | l = b       | o = <       | r = a }} {{eqn | ll= \leadsto       | l = \exists p \in \N: a       | r = b + p }} {{eqn | ll= \leadsto       | l = b + p + d       | r = b + c }} {{eqn | ll= \leadsto       | l = p + d       | r = c }} {{eqn | ll= \leadsto       | l = d       | o = <       | r = c }} {{end-eqn}} {{qed}} Category:Integers tfoqxrhyoxppc1d4o3kos1sbvkcow8n
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. Let $x \in \Z_{>0}$ be a (strictly) positive integer. Thus by definition: :$x > 0$ This is equivalent to the condition that $a > b$. Hence: :$x = \eqclass {b + u, b} {}$ for some $u \in \N_{>0}$. It is immediate that: :$\forall c \in \N: \eqclass {b + u, b} {} = \eqclass {c + u, c} {}$ Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as: :$\forall u \in \N_{>0}: \map \phi u = u'$ where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$. From the above we have that $\phi$ is well-defined. Let $\eqclass {b + u, b} {} = \eqclass {c + v, c} {}$. Then: :$b + u + c = b + c + v$ and so: :$u = v$ Hence $\phi$ is an injection. Next note that all $u' \in \Z_{>0}$ can be expressed in the form: :$u' = \eqclass {b + u, b} {}$ for arbitrary $b$. Hence $\phi$ is a surjection. Hence $\phi$ is an isomorphism. By defining $\map \phi 0 = \eqclass {b, b} {}$ we see that $\N$ and $\Z_{\ge 0}$ are isomorphic. {{qed}}