hcju/proofwikiqa huggingface dataset:
Let $S = \set {n \in \Z: n \ge m \in \Z: \neg P \paren n}$. That is, $S$ is the set of all elements in $\Z$ not less than $m$ for which the condition is false. Since: :$\neg \paren {\forall n \ge m: P \paren n}$ it follows that: :$S \ne \O$ Also, $S \subseteq \Z$ and is bounded below (by $m$). Therefore $S$ has a smallest element, which proves the result.
From Division Theorem: Positive Divisor: :$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ That is, the result holds for positive $b$. {{qed|lemma}} It remains to show that the result also holds for negative values of $b$. Let $b < 0$. Consider: :$\size b = -b > 0$ where $\size b$ denotes the absolute value of $b$: by definition $\size b > 0$. From Division Theorem: Positive Divisor, we have the existence of $\tilde q, \tilde r \in \Z$ such that: :$a = \tilde q \size b + \tilde r, 0 \le \tilde r < \size b$ Since $\size b = -b$: :$a = \tilde q \paren {-b} + \paren {\tilde r} = \paren {-\tilde q} b + \tilde r$ Taking: :$q = -\tilde q, r = \tilde r$ the existence has been proved of integers $q$ and $r$ that satisfy the requirements. The proof that they are unique is the same as that for the proof for positive $b$, but with $\size b$ replacing $b$. {{qed}}
=== Existence === Consider the arithmetic sequence: :$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$ which extends in both directions. Then by the Well-Ordering Principle, there must exist a smallest non-negative element, denoted by $r$. So $r = a - q b$ for some $q \in \Z$. $r$ must be in the interval $\hointr 0 b$ because otherwise $r - b$ would be smaller than $r$ and a non-negative element in the sequence. {{qed|lemma}} === Uniqueness === Suppose we have another pair $q_0$ and $r_0$ such that $a = b q_0 + r_0$, with $0 \le r_0 < b$. Then: :$b q + r = b q_0 + r_0$ Factoring we see that: :$r - r_0 = b \paren {q_0 - q}$ and so: :$b \divides \paren {r - r_0}$ Since $0 \le r < b$ and $0 \le r_0 < b$, we have that: :$-b < r - r_0 < b$ Hence: :$r - r_0 = 0 \implies r = r_0$ So now: :$r - r_0 = 0 = b \paren {q_0 - q}$ which implies that: :$q = q_0$ Therefore the solution is unique. {{Qed}}
From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$. So, if there is a counterexample to be found, it will have a degree. {{AimForCont}} there exists at least one counterexample. By a version of the Well-Ordering Principle, we can assign a number $m$ to the lowest degree possessed by any counterexample. So, let $f$ denote a counterexample which has that minimum degree $m$. If $m < n$, the equation $f = 0_F \circ d + f$ would show that $f$ was not a counterexample. Therefore $m \ge n$. Suppose $d \divides f$ in $F \sqbrk X$. Then: :$\exists q \in F \sqbrk X: f = q \circ d + 0_F$ and $f$ would not be a counterexample. So $d \nmid f$ in $F \sqbrk X$. So, suppose that: {{begin-eqn}} {{eqn | l = f       | r = \sum_{k \mathop = 0}^m {a_k \circ X^k} }} {{eqn | l = d       | r = \sum_{k \mathop = 0}^n {b_k \circ X^k} }} {{eqn | l = m       | o = \ge       | r = n }} {{end-eqn}} We can create the polynomial $\paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ which has the same degree and leading coefficient as $f$. Thus $f_1 = f - \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ is a polynomial of degree less than $m$. Since $d \nmid f$, $f_1$ is a non-zero polynomial. There is no counterexample of degree less than $m$. Therefore: :$f_1 = q_1 \circ d + r$ for some $q_1, r \in F \sqbrk X$, where either: :$r = 0_F$ or: :$r$ is non-zero with degree strictly less than $n$. Hence: {{begin-eqn}} {{eqn | l = f       | r = f_1 + \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d       | c =  }} {{eqn | r = \paren {q_1 + a_m \circ b_n^{-1} \circ X^{m - n} } \circ d + r       | c =  }} {{end-eqn}} Thus $f$ is not a counterexample. From this contradiction follows the result. {{qed}}
Suppose $\map \deg f < \map \deg d$. Then we take $\map q X = 0$ and $\map r X = \map a X$ and the result holds. Otherwise, $\map \deg f \ge \map \deg d$. Let: :$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$ :$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$ We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$: :$\map f X - \map d X \cdot \dfrac {a_m} {b_n} x^{m - n} = \map p X$ where $\map p X$ is some polynomial whose degree is less than that of $f$. If $\map p X$ still has degree higher than that of $d$, we do the same thing again. Eventually we reach: :$\map f X - \map d X \cdot \paren {\dfrac {a_m} {b_n} x^{m - n} + \dotsb} = \map r X$ where either $r = 0_F$ or $r$ has degree that is less than $n$. This approach can be formalised using the Principle of Complete Induction. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$ === Basis for the Induction === $\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$. This is shown trivially to be true by taking $q = r = 0$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k$ Then we need to show: :$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k + 1$ === Induction Step === This is our induction step: Let $f$ be such that $\map \deg f = n$. Let: :$f = a_0 + a_1 \circ x + a_2 \circ x^2 + \cdots + a_n \circ x^n$ where $a_n \ne 0$ Let: :$d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ where $b_j \ne 0$ If $n < l$ then take $q = 0, r = f$. If $n \ge l$, consider: :$c := f - a_n b_j^{-1} x^{n - j} d$ This has been carefully arranged so that the coefficient of $x^n$ in $c$ is zero. Thus $\map \deg c < n$. Therefore, by the induction hypothesis: :$c = d q_0 + r$ where $\map \deg r < \map \deg d$. Therefore: {{begin-eqn}} {{eqn | l = f       | r = d \paren {q_0 + a_n b_j^{-1} x^{n - j} } + r       | c =  }} {{eqn | r = d q + r       | c = where $\map \deg r < \map \deg d$ and $q = q_0 + a_n b_j^{-1} x^{n - j}$ }} {{end-eqn}} Thus the existence of $q$ and $r$ have been established. As for uniqueness, assume: :$d q + r = d q' + r'$ with $\map \deg r < \map \deg d, \map \deg {r'} < \map \deg d$ Then: :$d \paren {q - q'} = r' - r$ By Degree of Sum of Polynomials: :$\map \deg {r' - r} \le \max \set {\map \deg {r'}, \map \deg r} < \map \deg d$ and by Degree of Product of Polynomials over Integral Domain: :$\map \deg {d \paren {q - q'} } = \map \deg d + \map \deg {q - q'}$ That is: :$\map \deg d < \map \deg d + \map \deg {q - q'}$ and the only way for that to happen is for: :$\map \deg {q - q'} = -\infty$ that is, for $q - q'$ to be the null polynomial. That is, $q - q' = 0_F$ and by a similar argument $r' - r = 0_F$, demonstrating the uniqueness of $q$ and $r$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either: ::$(1): \quad r = 0_F$ :or: ::$(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$. {{qed}}
Consider the set of integer multiples $x \size b$ of $\size b$ less than or equal to $a$: :$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$ We have that: :$-\size a \size b \le -\size a \le a$ and so $M \ne \O$. From Set of Integers Bounded Above by Integer has Greatest Element, $M$ has a greatest element $h \size b$. Then $h \size b \le a$ and so: :$a = h \size b + r$ where $r \ge 0$. On the other hand: :$\paren {h + 1} \size b = h \size b + \size b > h \size b$ So: :$\paren {h + 1} \size b > a$ and: :$h \size b + \size b > h \size b + r$ Thus: :$r \le b$ Setting: :$q = h$ when $b > 0$ :$q = -h$ when $b < 0$ it follows that: :$h \size b = q b$ and so: :$a = q b + r$ as required. {{Qed}}
Follows directly from the Division Theorem. {{qed}} Category:Odd Integers mpi9q6vy7fvjpowo6jxu7u75i7r217j
{{begin-eqn}} {{eqn | l = 0       | o = \divides       | r = n       | c =  }} {{eqn | ll= \leadsto       | l = \exists q \in \Z: n       | r = q \times 0       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = n       | r = 0       | c = Integers have no zero divisors, as Integers form Integral Domain. }} {{end-eqn}} {{Qed}}
Suppose $a \divides c$ for some $c \ne 0$. From Negative of Absolute Value: : $a \le \size a$ Then: {{begin-eqn}} {{eqn | l = a       | o = \divides       | r = c       | c =  }} {{eqn | lll=\leadsto       | ll= \exists q \in \Z:       | l = c       | r = a q       | c = {{Defof|Divisor of Integer}} }} {{eqn | lll=\leadsto       | l = \size c       | r = \size a \size q       | c =  }} {{eqn | lll=\leadsto       | l = \size a \size q \ge \size a \times 1       | r = \size a       | c =  }} {{eqn | lll=\leadsto       | l = a \le \size a       | o = \le       | r = \size c       | c =  }} {{end-eqn}} {{qed}}
Checking in turn each of the criteria for an ordering: === Divisor Relation is Reflexive === {{:Integer Divisor Results/Integer Divides Itself/Proof 1}} === Divisor Relation is Transitive === :$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$: {{:Divisor Relation is Transitive/Proof 2}} === Divisor Relation is Antisymmetric === {{:Divisor Relation is Antisymmetric}} === Divisor Ordering is Partial === Let $a = 2$ and $b = 3$. Then neither $a \divides b$ nor $b \divides a$. Thus, while the divisor relation is an ordering, it is specifically a partial ordering {{qed}}
{{begin-eqn}} {{eqn | l = c       | o = \divides       | r = a }} {{eqn | ll= \leadsto       | lo= \exists x \in D:       | l = a       | r = x \times c       | c = {{Defof|Divisor of Ring Element}} }} {{eqn | l = c       | o = \divides       | r = b }} {{eqn | ll= \leadsto       | lo= \exists y \in D:       | l = b       | r = y \times c       | c = {{Defof|Divisor of Ring Element}} }} {{eqn | ll= \leadsto       | lo= \forall p, q \in D:       | l = p \times a + q \times b       | r = p \times x \times c + q \times y \times c       | c = substituting for $a$ and $b$ }} {{eqn | r = \paren {p \times x + q \times y} c       | c = as $\times$ is distributive over $+$ }} {{eqn | ll= \leadsto       | lo= \exists z \in D:       | l = p \times a + q \times b       | r = z \times c       | c = where $z = p \times x + q \times y$ }} {{eqn | ll= \leadsto       | l = c       | o = \divides       | r = \paren {p \times a + q \times b}       | c = {{Defof|Divisor of Ring Element}} }} {{end-eqn}} {{qed}}
=== Proof of Existence === We have that: :$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$ so $1$ is always a common divisor of any two integers. {{qed|lemma}} === Proof of there being a Largest === As the definition of $\gcd$ shows that it is symmetric, we can assume without loss of generality that $a \ne 0$. First we note that from Absolute Value of Integer is not less than Divisors: :$\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \size c \le \size a$ The same applies for $c \divides b$. Now we have three different results depending on $a$ and $b$: {{begin-eqn}} {{eqn | l = a \ne 0 \land b \ne 0       | o = \implies       | r = \gcd \set {a, b} \le \min \set {\size a, \size b}       | c =  }} {{eqn | l = a = 0 \lor b = 0       | o = \implies       | r = \gcd \set {a, b} = \max \set {\size a, \size b}       | c =  }} {{eqn | l = a = b = 0       | o = \implies       | r = \forall x \in \Z: x \divides a \land x \divides b       | c =  }} {{end-eqn}} So if $a$ and $b$ are ''both'' zero, then ''any'' $n \in \Z$ divides both, and there is no greatest common divisor. This is why the proviso that $a \ne 0 \lor b \ne 0$. So we have proved that common divisors exist and are bounded above. Therefore, from Set of Integers Bounded Above by Integer has Greatest Element there is always a '''greatest''' common divisor. {{qed}}
By definition of greatest common divisor, we aim to show that there exists $c \in \Z_{>0}$ such that: {{begin-eqn}} {{eqn | l = c       | o = \divides       | r = a }} {{eqn | l = c       | o = \divides       | r = b }} {{end-eqn}} and: :$d \divides a, d \divides b \implies d \divides c$ Consider the set $S$: :$S = \set {s \in \Z_0: \exists x, y \in \Z: s = a x + b y}$ $S$ is not empty, because by setting $x = 1$ and $y = 0$ we have at least that $a \in S$. From the Well-Ordering Principle, there exists a smallest $c \in S$. So, by definition, we have $c > 0$ is the smallest such that $c = a x + b y$ for some $x, y \in \Z$. Let $d$ be such that $d \divides a$ and $d \divides b$. Then from Common Divisor Divides Integer Combination: :$d \divides a x + b y$ That is: :$d \divides c$ We have that: {{begin-eqn}} {{eqn | lo= \exists t, u \in \Z:       | l = a       | r = c t + u:       | rr= 0 \le u < c       | c = Division Theorem }} {{eqn | ll= \leadsto       | l = a       | r = a x t + b y t + u       | c = Definition of $c$ }} {{eqn | ll= \leadsto       | l = u       | r = r \paren {1 - x t} + b \paren {-y t}        | c = rearranging }} {{eqn | ll= \leadsto       | l = u       | r = 0       | c = as $u < c$ and the definition of $c$ }} {{eqn | ll= \leadsto       | l = c       | o = \divides       | r = a       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} Mutatis mutandis: :$c \divides b$ Now suppose $c'$ is such that: {{begin-eqn}} {{eqn | l = c'       | o = \divides       | r = a }} {{eqn | l = c'       | o = \divides       | r = b }} {{end-eqn}} and: :$d \divides a, d \divides b \implies d \divides c'$ Then we have immediately that: :$c' \divides c$ and by the same coin: $c \divides c'$ and so: :$c = c'$ demonstrating that $c$ is unique. {{qed}}
From One Divides all Integers: :$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$ and so: :$1 \le \gcd \set {a, b}$ as required. {{qed}} Category:Greatest Common Divisor 2nrji97dz25wb80mslueng8xuvr7oza
$a \divides b$ by hypothesis, $a \divides a$ from Integer Divides Itself. Thus $a$ is a common divisor of $a$ and $b$. Note that from Absolute Value of Integer is not less than Divisors: :$\forall x \in \Z: x \divides a \implies x \le \size a$. As $a$ and $b$ are both positive, the result follows. {{qed}}
Note that $\size a = \pm a$. Suppose that: :$u \divides a$ where $\divides$ denotes divisibility. Then: :$\exists q \in \Z: a = q u$ Then: :$\size a = \pm q u = \paren {\pm q} u \implies u \divides \size a$ So every divisor of $a$ is a divisor of $\size a$. Similarly, note that: :$a = \pm \size a$ so every divisor of $\size a$ is a divisor of $a$. So it follows that the common divisors of $a$ and $b$ are the same as those of $a$ and $\size b$, and so on. In particular: :$\gcd \set {a, b} = \gcd \set {a, \size b}$ and so on. {{Qed}} Category:Greatest Common Divisor le9bxzi7pir8rzelwcgd7h3vqlvza11
Follows from: : Integer Divides Zero : GCD for Negative Integers. {{qed}}
=== Necessary Condition === Let $d = \gcd \set {a, b}$. Then: :$d \divides c \implies \exists m \in \Z: c = m d$ So: {{begin-eqn}} {{eqn | l = \exists p, q \in \Z: d       | r = p a + q b       | c = Bézout's Lemma }} {{eqn | ll= \leadsto       | l = c       | r = m d       | c =  }} {{eqn | r = m p a + m q b       | c =  }} {{eqn | r = \paren {m p} a + \paren {m q} b       | c =  }} {{eqn | ll= \leadsto       | l = \exists x, y \in \Z: c       | r = x a + y b       | c =  }} {{end-eqn}} Thus: :$\gcd \set {a, b} \divides c \implies \exists x, y \in \Z: c = x a + y b$ {{qed|lemma}} === Sufficient Condition === Suppose $\exists x, y \in \Z: c = x a + y b$. From Common Divisor Divides Integer Combination, we have: :$\gcd \set {a, b} \divides \paren {x a + y b}$ It follows directly that $\gcd \set {a, b} \divides c$ and the proof is finished. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \gcd \set {a, b} \divides a \land \gcd \set {a, b} \divides b       | c = {{Defof|Greatest Common Divisor of Integers}} }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {a, b} \divides \paren {a - q b}       | c = Common Divisor Divides Integer Combination }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {a, b} \divides r       | c = as $r = a - q b$ }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {a, b} \le \gcd \set {b, r}       | c = {{Defof|Greatest Common Divisor of Integers}} }} {{end-eqn}} The argument works the other way about: {{begin-eqn}} {{eqn | o =        | r = \gcd \set {b, r} \divides b \land \gcd \set {b, r} \divides r       | c = {{Defof|Greatest Common Divisor of Integers}} }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {b, r} \divides \paren {q b + r}       | c = Common Divisor Divides Integer Combination }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {b, r} \divides a       | c = as $a = q b + r$ }} {{eqn | ll= \leadsto       | o =        | r = \gcd \set {b, r} \le \gcd \set {a, b}       | c = {{Defof|Greatest Common Divisor of Integers}} }} {{end-eqn}} Thus: :$\gcd \set {a, b} = \gcd \set {b, r}$ {{qed}} Category:Greatest Common Divisor 52xifv9uor3uhps3z23ov9u5wtec8hn
{{begin-eqn}} {{eqn | l = a       | o = \perp       | r = b }} {{eqn | ll= \leadsto       | l = \gcd \set {a, b}       | r = 1       | c = {{Defof|Coprime Integers}} }} {{eqn | ll= \leadsto       | l = \exists m, n \in \Z: m a + n b       | r = 1       | c = Bézout's Lemma }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \exists m, n \in \Z: m a + n b       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {a, b}       | o = \divides       | r = 1       | c = Set of Integer Combinations equals Set of Multiples of GCD }} {{eqn | ll= \leadsto       | l = \gcd \set {a, b}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \perp       | r = b       | c = {{Defof|Coprime Integers}} }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$. Let $d$ be a divisor of both $a$ and $b$. Then: :$d \divides m a + n b$ and so: :$d \divides 1$ Thus: :$d = \pm 1$ and so: :$\gcd \set {a, b} = 1$ Thus, by definition, $a$ and $b$ are coprime. {{qed|lemma}} === Necessary Condition === Let $a \perp b$. Thus they are not both $0$. Let $S$ be defined as: :$S = \set {a m + b n: m, n \in \Z}$ $S$ contains at least one strictly positive integer, because for example $a^2 + b^2 \in S$. By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive. Let $d = a x + b y$. It remains to be shown that $d = 1$. By the Division Theorem: :$a = d q + r$ where $0 \le r < d$ Then: {{begin-eqn}} {{eqn | l = r       | r = a - d q       | c =  }} {{eqn | r = a - \paren {a x + b y} q       | c =  }} {{eqn | r = a \paren {1 - x q} + b \paren {- y q}       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} But we have that $0 \le r < d$. We have defined $d$ as the smallest element of $S$ which is strictly positive Hence it follows that $r$ cannot therefore be strictly positive itself. Hence $r = 0$ and so $a = d q$. That is: :$d \divides a$ By a similar argument: :$d \divides b$ and so $d$ is a common divisor of both $a$ and $b$. But the GCD of $a$ and $b$ is $1$. Thus it follows that, as $d \in S$: :$\exists m, n \in \Z: m a + n b = 1$ {{qed}}
=== Sufficient Condition === Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$. Let $d$ be a divisor of both $a$ and $b$. Then: :$d \mathrel \backslash m a + n b$ and so: :$d \mathrel \backslash 1$ Thus: : $d = \pm 1$ and so: : $\gcd \left\{ {a, b}\right\} = 1$ Thus, by definition, $a$ and $b$ are coprime. {{qed|lemma}} === Necessary Condition === Let $a \perp b$. Thus they are not both $0$. Let $S$ be defined as: :$S = \left\{ {\lambda a + \mu b: \lambda, \mu \in \Z}\right\}$ $S$ contains at least one strictly positive integer, because for example: : $a \in S$ (setting $\lambda = 1$ and $\mu = 0$) : $b \in S$ (setting $\lambda = 0$ and $\mu = 1$) By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive. Let $d = \alpha a + \beta b$. Let $c \in S$, such that $\lambda_0 a + \mu_0 b = c$ for some $\lambda_0, \mu_0 \in \Z$. By the Division Algorithm: :$\exists \gamma, \delta \in \Z: c = \gamma d + \delta$ where $0 \le \delta < d$ Then: {{begin-eqn}} {{eqn | l = \delta       | r = c - \gamma d       | c =  }} {{eqn | r = \left({\lambda_0 a + \mu_0 b}\right) - \gamma \left({\alpha a + \beta b}\right)       | c =  }} {{eqn | r = \left({\lambda_0 - \gamma \alpha}\right) a + \left({\mu_0 - \gamma \beta}\right) b       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} But we have that $0 \le \delta < d$. We have defined $d$ as the smallest element of $S$ which is strictly positive Hence it follows that $\delta$ cannot therefore be strictly positive itself. Hence $\delta = 0$ and so $c = \gamma d$. That is: :$d \mathrel \backslash c$ and so the smallest element of $S$ which is strictly positive is a divisor of both $a$ and $b$. But $a$ and $b$ are coprime. Thus it follows that, as $d \mathrel \backslash 1$: :$d = 1$ and so, by definition of $S$: :$\exists m, n \in \Z: m a + n b = 1$ {{qed}}
Let $d \in \Z_{>0}: d \divides c \land d \divides a$. Then: {{begin-eqn}} {{eqn | l = d       | o = \divides       | r = \paren {a + b}       | c = as $c \divides \paren {a + b}$ }} {{eqn | ll= \leadsto       | l = d       | o = \divides       | r = \paren {a + b - a}       | c =  }} {{eqn | ll= \leadsto       | l = d       | o = \divides       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = d       | r = 1       | c = as $d \divides a$ and $d \divides b$ which are coprime }} {{end-eqn}} A similar argument shows that if $d \divides c \land d \divides b$ then $d \divides a$. It follows that: : $\gcd \set {a, c} = \gcd \set {b, c} = 1$ Hence the result. {{qed}}
Let $d = \gcd \set {a, b}$. We have: :$d \divides a \iff \exists s \in \Z: a = d s$ :$d \divides b \iff \exists t \in \Z: b = d t$ So: {{begin-eqn}} {{eqn | l = \exists m, n \in \Z: d       | r = m a + n b       | c = Bézout's Identity }} {{eqn | ll= \leadstoandfrom       | l = d       | r = m d s + n d t       | c = Definition of $s$ and $t$ }} {{eqn | ll= \leadstoandfrom       | l = 1       | r = m s + n t       | c = dividing through by $d$ }} {{eqn | ll= \leadstoandfrom       | l = \gcd \set {s, t}       | r = 1       | c = Bézout's Identity }} {{eqn | ll= \leadstoandfrom       | l = \gcd \set {\frac a d, \frac b d}       | r = 1       | c = Definition of $s$ and $t$ }} {{end-eqn}} {{qed}}
We have: :$a \divides c \implies \exists r \in \Z: c = a r$ :$b \divides c \implies \exists s \in \Z: c = b s$ So: {{begin-eqn}} {{eqn | l = a       | o = \perp       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = \exists m, n \in \Z: m a + n b       | r = 1       | c = Integer Combination of Coprime Integers }} {{eqn | ll= \leadsto       | l = c m a + c n b       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = b s m a + a r n b       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = a b \paren {s m + r n}       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = a b       | o = \divides       | r = c       | c =  }} {{end-eqn}} {{qed}}