hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = p \implies q       | o = \dashv \vdash       | r = \neg p \lor q       | c = Rule of Material Implication }} {{eqn | o = \dashv \vdash       | r = \neg p \lor \neg \neg q       | c = Double Negation Introduction }} {{eqn | o = \dashv \vdash       | r = p \uparrow \neg q       | c = NAND as Disjunction of Negations }} {{eqn | o = \dashv \vdash       | r = p \uparrow \left({q \uparrow q}\right)       | c = NAND with Equal Arguments }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = p \iff q       | o = \dashv \vdash       | r = \neg \paren {p \oplus q}       | c = Exclusive Or is Negation of Biconditional }} {{eqn | o = \dashv \vdash       | r = \neg \paren {\paren {p \lor q} \land \neg \paren {p \land q} }       | c = {{Defof|Exclusive Or}} }} {{eqn | o = \dashv \vdash       | r = \neg \paren {\paren {p \lor q} \land \paren {p \uparrow q} }       | c = {{Defof|Logical NAND}} }} {{eqn | o = \dashv \vdash       | r = \neg \paren {\paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \land \paren {p \uparrow q} }       | c = Disjunction in terms of NAND }} {{eqn | o = \dashv \vdash       | r = \paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \uparrow \paren {p \uparrow q}       | c = {{Defof|Logical NAND}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === {{:Inverse of Injective and Surjective Mapping is Mapping}} === Sufficient Condition === {{:Inverse is Mapping implies Mapping is Injection and Surjection}}
=== Necessary Condition === {{:Inverse of Injective and Surjective Mapping is Mapping/Proof 2}}{{qed|lemma}} === Sufficient Condition === Let $f^{-1}: T \to S$ be a mapping. By Inverse Mapping is Bijection, both $f$ and $f^{-1}$ are bijections. Hence, in particular, $f$ is a bijection. {{qed}}
=== Definition 1 iff Definition 2 === {{:Equivalence of Definitions of Bijection/Definition 1 iff Definition 2}}{{qed|lemma}} === Definition 1 iff Definition 3 === {{:Equivalence of Definitions of Bijection/Definition 1 iff Definition 3}}{{qed|lemma}} === Definition 1 iff Definition 4 === {{:Equivalence of Definitions of Bijection/Definition 1 iff Definition 4}}{{qed|lemma}} === Definition 3 iff Definition 4 === {{:Equivalence of Definitions of Bijection/Definition 3 iff Definition 4}}{{qed|lemma}} === Definition 3 iff Definition 5 === {{:Equivalence of Definitions of Bijection/Definition 3 iff Definition 5}} Category:Bijections Category:Equivalence of Definitions of Bijection e2fhj7sz36otnfuxncqhz4qh2elwb91
From Identity Mapping is Bijection, $I_S$ is a bijection. From Identity Mapping is Injection, $I_S$ is an injection. From Injection if Composite is Injection it follows that $f$ is an injection. From Identity Mapping is Surjection, $I_S$ is a surjection. From Surjection if Composite is Surjection it follows that $g$ is a surjection. {{qed}}
=== Necessary Condition === Let $f$ be a bijection. Then for each $y \in T$ there exists one and only one $x \in S$ such that $\map f x = y$. That is, that there exists a mapping $g: T \to S$ with the property that: : $\forall y \in T: \exists x \in S: \map g y = x$ Let $y \in T$. Let $x = g \map g y$. Then: {{begin-eqn}} {{eqn | l = y       | r = \map f {\map g y}       | c =  }} {{eqn | r = \map {f \circ g} y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x       | r = \map g {\map f x}       | c =  }} {{eqn | r = \map {f \circ g} x       | c =  }} {{end-eqn}} demonstrating that $g$ has the property that: :$g \circ f = I_S$ :$f \circ g = I_T$ {{qed|lemma}} === Sufficient Condition === Suppose there exists a $g$ which satisfies the conditions on $f$. By Condition for Composite Mapping to be Identity: :$(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an injection. :$(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a surjection. Thus, by definition, $f$ is a bijection. {{qed}}
Fix an arbitrary $g \in G$ and consider the corresponding vector $g \in K \left[{G}\right]$. Let $\lambda$ be an eigenvalue of $g$, that is $\lambda$ is an eigenvalue of the map in $\operatorname{Aut} \left({V}\right): \vec v \mapsto g \vec v$. Then by definition of an eigenvalue we have: :$\exists \vec v_\lambda \in V : g \vec v_\lambda = \lambda \vec v_\lambda$ Let $n$ be the order of $g$ in $G$. Then: {{begin-eqn}} {{eqn | l = \vec v_\lambda       | r = g^n \vec v_\lambda       | c = as $g^n = e$ and $e$ acts trivially on $V$ }} {{eqn | r = \lambda g^{n-1} \vec v_\lambda       | c = as $G$ acts linearly on $V$ }} {{eqn | r = \vdots       | c = Continue exchanging $g^k \vec v_\lambda$ with $\lambda g^{k-1} \vec v_\lambda$ }} {{eqn | r = \lambda^n \vec v_\lambda       | c = as $g^0 = e$ which acts trivially on $V$ }} {{end-eqn}} {{explain|"as $g^n {{=}} e$ and $e$ acts trivially on $V$" - this needs to be linked to and / or proved}} {{explain|"as $G$ acts linearly on $V$" - this needs to be linked to and / or proved}} {{explain|"as $g^0 = e$ which acts trivially on $V$" - this needs to be linked to and / or proved}} Thus: :$\vec v_\lambda = \lambda^n \vec v_\lambda$ which means: :$\lambda^n = 1$ and thus by definition, $\lambda$ is an $n$th root of unity. {{qed}} la77w1qnzjsm0z4avytxiwx9t7fuetp
By the definition of character: :$\chi \left({g}\right) = \operatorname{Tr} \left({\rho_g}\right)$ {{explain|$\operatorname{Tr} \left({\rho_g}\right)$, and indeed $\rho_g$ itself.}} where: : $\rho \in \hom \left({\C \left[{G}\right], \operatorname{Aut} \left({V}\right)}\right): \vec {e_g} \mapsto \rho_g$ by definition. {{explain|The above definition is not stated in that form on {{ProofWiki}}. Link to $\hom \left({\C \left[{G}\right]$ and $\operatorname{Aut} \left({V}\right)}\right)$, clarify specifically what $\C \left[{G}\right]$ is, and define $\vec {e_g}$.}} Fix an arbitrary $g \in G$. Let $\left\vert{g}\right\vert$ denote the order of $g$. The trace $\operatorname{Tr} \left({\rho_g}\right)$ of $\rho_g$ is defined as the sum of the eigenvalues of $\rho_g$. {{explain|The definition of Trace needs to be reviewed.}} From Eigenvalues of G-Representation are Roots of Unity, we have that any eigenvalue $\lambda$ of $\rho_g$ is a root of unity whose order is $\left\vert{g}\right\vert$. We have that $\lambda$ satisfies the monic polynomial $x^{\left\vert{g}\right\vert} - 1$ Hence we have that $\lambda$ is an algebraic integer. From Ring of Algebraic Integers, we have that the sum of the eigenvalues is also an algebraic integer. Thus $\chi \left({g}\right)$ is an algebraic integer. {{qed}} Category:Module Theory Category:Group Theory Category:Complex Analysis ncn9awjawffn7x5umv4yc1k7pls6q23
Let $a \in S$. We need to show that there exists $b \in S$ such that $a = \map f b$. Consider what happens when $f$ is applied repeatedly on $S$. Let $f^2$ denote $f \circ f$ and, generally, $f^n := f \circ f^{n-1}$. Consider the sequence of elements of $S$: :$a, \map f a, \map {f^2} a, \ldots$ Because $S$ is a finite set, there must be repetitions. That is, there must exist $r, s \in \N$ such that: :$\map {f^r} a = \map {f^s} a$ where $r \ne s$. {{WLOG}}, assume $r > s$. $f$ is an injection. Therefore by Composite of Injections is Injection, $f^s$ is an injection. By Injection iff Left Cancellable, $f^s$ is left cancellable. Thus: {{begin-eqn}} {{eqn | l = \map {f^r} a       | r = \map {f^s} a       | c =  }} {{eqn | ll= \leadsto       | l = \map {f^s \circ f^{r - s} } a       | r = \map {f^s} a }} {{eqn | ll= \leadsto       | l = \map {f^{r - s} } a       | r = a       | c = {{Defof|Left Cancellable Mapping}} }} {{end-eqn}} That is, $b = \map {f^{r - s - 1} } a$. {{qed}}
Follows directly from the definition of domain and codomain: :$\Dom f = S$ and $\Cdm f = T$ :$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$ That is: :$\Dom {f^{-1} } = T = \Cdm f$ {{qed}}
Follows directly from the definition of domain and codomain: :$\Dom f = S$ and $\Cdm f = T$ :$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$ That is: :$\Cdm {f^{-1} } = S = \Dom f$ {{qed}}
If $f$ is a bijection, then by definition $f$ is also a surjection. Then: :$T \setminus \Img f = \O$ :and we have that $g = f^{-1}$. As $f^{-1}$ is uniquely defined $g$ is itself unique. {{qed|lemma}} If $S$ is a singleton then there can only be one mapping $g: T \to S$: :$\forall t \in T: \map g t = s$ {{qed|lemma}} If $f$ is not a bijection, then as it is an injection it can not be a surjection. Then: :$T \setminus \Img f \ne \O$ Let $t \in T \setminus \Img f$. We can now choose any $x_0 \in S$ such that $\map g t = x_0$. If $S$ is not a singleton, such an $x_0$ is not unique. Hence the result. {{qed}}
If $f$ is not an injection then: :$\exists y \in T: \exists x_1, x_2 \in S: \map f {x_1} = y = \map f {x_2}$ Hence we have more than one choice in $\map {f^{-1} } {\set y}$ for how to map $\map g y$. That is, $\map g y$ is not unique. This does not happen {{iff}} $f$ is an injection. Hence the result. {{qed}}
=== Reflexivity === {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S       | c =  }} {{eqn | ll= \leadsto       | l = \map f x       | o = \in       | r = T       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = \map f x       | o = \precsim       | r = \map f x       | c = as $\precsim$ is a preordering on $T$ }} {{eqn | ll= \leadsto       | l = x       | o = \RR       | r = x       | c = Definition of $\RR$ }} {{end-eqn}} Thus $\RR$ is reflexive. {{qed|lemma}} === Transitivity === {{begin-eqn}} {{eqn | l = x       | o = \RR       | r = y       | c =  }} {{eqn | lo= \land       | l = y       | o = \RR       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = \map f x       | o = \precsim       | r = \map f y       | c =  }} {{eqn | lo= \land       | l = \map f y       | o = \precsim       | r = \map f z       | c = Definition of $\RR$ }} {{eqn | ll= \leadsto       | l = \map f x       | o = \precsim       | r = \map f z       | c = as $\precsim$ is a preordering on $T$ }} {{eqn | ll= \leadsto       | l = x       | o = \RR       | r = z       | c = Definition of $\RR$ }} {{end-eqn}} Thus $\RR$ is transitive. {{qed|lemma}} So, by definition, $\RR$ is a preordering. {{qed}}
Proof by induction: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\paren {1 + x}^n \ge 1 + nx$ === Basis for the Induction === $\map P 0$ is the case: :$\paren {1 + x}^0 \ge 1$ so $\map P 0$ holds. 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: :$\left({1 + x}\right)^k \ge 1 + kx$ We need to show that: :$\paren {1 + x}^{k + 1} \ge 1 + \paren {k + 1} x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {1 + x}^{k + 1}       | r = \paren {1 + x}^k \paren {1 + x} }} {{eqn | o = \ge       | r = \paren {1 + k x} \paren {1 + x}       | c = Induction Hypothesis }} {{eqn | r = 1 + \paren {k + 1} x + k x^2 }} {{eqn | o = \ge       | r = 1 + \paren {k + 1} x }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Let $y = 1 + x$. Then $y \ge 0$, and: :$\paren {1 + x}^n = 1 + \paren {y^n - 1}$ If $y \ge 1$, then by Sum of Geometric Sequence: :$\displaystyle y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$ If $y < 1$, then by Sum of Geometric Sequence: :$\displaystyle y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0}^{n - 1} y^k \ge -n \paren {1 - y} = n x$ Hence the result. {{qed}}
By definition, a proper element is a non-zero element which has no product inverse. Also by definition, a unit is an element which does have a product inverse. Because $0 \circ x = 0$ there can be no $x \in R$ such that $0 \times x = 1$, and so $0$ is not a unit. Hence the result. {{qed}} Category:Ring Theory 76cxzcd0t4nheq7khu1x1x3rp3cc4sw
By definition: :$\displaystyle n \cdot x := \sum_{j \mathop = 1}^n x$ Thus: {{begin-eqn}} {{eqn | l = \left({n \cdot x} \right) \circ y       | r = \left({\sum_{j \mathop = 1}^n x} \right) \circ y       | c = by definition of integral multiple }} {{eqn | n = 1       | r = \sum_{j \mathop = 1}^n \left({x \circ y} \right)       | c = General Distributivity Theorem }} {{eqn | r = n \cdot \left({x \circ y}\right)       | c = by definition of integral multiple }} {{eqn | r = x \circ \left({\sum_{j \mathop = 1}^n y} \right)       | c = General Distributivity Theorem from $(1)$ }} {{eqn | r = x \circ \left({n \cdot y}\right)       | c = by definition of integral multiple }} {{end-eqn}} {{qed}} Category:Ring Theory g02rj9281x192zcs329rxqfssoa0b8q
Recall that Integers form Ordered Integral Domain. Then from Relation Induced by Strict Positivity Property is Compatible with Addition: :$\forall x, y, z \in \Z: x \le y \implies \paren {x + z} \le \paren {y + z}$ :$\forall x, y, z \in \Z: x \le y \implies \paren {z + x} \le \paren {z + y}$ So: {{begin-eqn}} {{eqn | l = a       | o = \le       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a + c       | o = \le       | r = b + c       | c = Relation Induced by Strict Positivity Property is Compatible with Addition }} {{end-eqn}} {{begin-eqn}} {{eqn | l = c       | o = \le       | r = d       | c =  }} {{eqn | ll= \leadsto       | l = b + c       | o = \le       | r = b + d       | c = Relation Induced by Strict Positivity Property is Compatible with Addition }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = a + c       | o = \le       | r = b + c }} {{eqn | l = b + c       | o = \le       | r = b + d }} {{eqn | ll= \leadsto       | l = a + c       | o = \le       | r = b + d       | c = {{Defof|Ordering}} }} {{end-eqn}} {{qed}}
Let $S$ be a finite set. Then by definition: :$\exists n \in \N: \card S = n$ where $\card S$ denotes the cardinality of $S$. From Cardinality of Power Set of Finite Set: :$\card {\powerset S} = 2^n$ where $\powerset S$ denotes the power set of $S$. As $n \in \N$ it follows that $2^n \in \N$ and so $\powerset S$ is also by definition a finite set. {{qed}} Category:Power Set tikj18ioh81t6cn77m6ng7119sxrh4b
As the set of integers form an integral domain, the concept '''divides''' is fully applicable to the integers. Therefore this result follows directly from Element of Integral Domain is Divisor of Itself. {{qed}}
From Integer Multiplication Identity is One: :$\forall n \in \Z: 1 \cdot n = n = n \cdot 1$ thus demonstrating that $n$ is a divisor of itself. {{qed}}
Let $c \divides a \land c \divides b$. From Common Divisor Divides Integer Combination: :$\forall p, q \in \Z: c \divides \paren {p a + q b}$ Putting $p = 1$ and $q = -1$: :$c \divides \paren {a - b}$ {{qed}}
{{begin-eqn}} {{eqn | l = c       | o = \divides       | r = a }} {{eqn | ll= \leadsto       | lo= \exists x \in \Z:       | l = a       | r = x c       | c = {{Defof|Divisor of Integer}} }} {{eqn | l = c       | o = \divides       | r = b }} {{eqn | ll= \leadsto       | lo= \exists y \in \Z:       | l = b       | r = y c       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = a - b       | r = x c - y c       | c = substituting for $a$ and $b$ }} {{eqn | r = \paren {x - y} c       | c = Integer Multiplication Distributes over Addition }} {{eqn | ll= \leadsto       | lo= \exists z \in \Z:       | l = a - b       | r = z c       | c = where $z = x - y$ }} {{eqn | ll= \leadsto       | l = c       | o = \divides       | r = \paren {a - b}       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{qed}}
Let $a \divides b$. From Integer Divides Zero: :$a \divides 0$ Thus $a$ is a common divisor of $b$ and $0$. From Common Divisor Divides Integer Combination: :$\forall p, q \in \Z: a \divides \paren {p \cdot b + q \cdot 0}$ Putting $p = c$ and $q = 1$ (for example): :$a \divides \paren {c b + 0}$ Hence the result. {{qed}}
Let $a \in \Z_{\ne 0}$. By Integer Divides its Absolute Value: :$a \divides \size a$ and $\size a \divides a$ Hence $\size a$ is an associate of $a$. Now we prove its uniqueness. Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$. Let $a \sim b$ and $a \sim c$ where $\sim$ denotes the relation of associatehood. By definition of associatehood: :$a \divides b$ and $b \divides a$ and: :$a \divides c$ and $c \divides a$ From Divisor Relation is Antisymmetric/Corollary/Proof 2: :$a = \pm b$ and :$a = \pm c$ That is: :$\pm b = \pm c$ which means: :$b = c$ or $b = -c$ But as both $b > 0$ and $c > 0$: :$b = c$ Hence the result. {{qed}}
By definition, a prime number is a positive integer which has exactly $2$ divisors which are themselves positive integers. From Divisors of One, the only divisors of $1$ are $1$ and $-1$. So the only divisor of $1$ which is a positive integer is $1$. As $1$ has only one such divisor, it is not classified as a prime number. {{qed}}
From Sigma Function of Prime Number, the sum $\map \sigma p$ of all the positive integer divisors of a prime number $p$ is $p + 1$. But from Sigma Function of 1, $\map \sigma 1 = 1$. If $1$ were to be classified as prime, then $\map \sigma 1$ would be an exception to the rule that $\map \sigma p = p + 1$. {{qed}}
By definition of substring, there exists a string $T'$ such that: :$S = TT'$ Hence $S$ is the concatenation of the null string, $T$, and $T'$. Thus by definition of substring, $T$ is a substring of $S$. {{qed}} Category:Formal Systems brdmwx4fr7kw0ci47xvs1zgz98lmxep
By Ring Product with Zero (applied to $0_R$): :$0_R \circ 0_R = 0_R$ which was to be proven.  {{qed}} Category:Ring Theory ss3g8gf6liesnsozzglyn9w1zk3odym
=== Definition 1 iff Definition 2 === This is proved in Prime Number has 4 Integral Divisors: {{:Prime Number has 4 Integral Divisors}} === Definition 1 iff Definition 3 === This is proved in Tau of Prime Number: {{:Tau of Prime Number}} === Definition 1 iff Definition 4 === From these two results: : $1$ Divides all Integers : Integer Divides Itself it follows that if $p$ has exactly two positive integer divisors then those are $1$ and $p$. By the same coin, if the only positive integer divisors of $p$ are $1$ and $p$, then $p$ has exactly two positive integer divisors. {{qed}} === Definition 4 iff Definition 5 === Let the only two positive integer divisors of $p$ be $1$ and $p$. Then the only divisor of $p$ strictly less than $p$ is $1$. Conversely, let the only divisor of $p$ strictly less than $p$ be $1$ From Integer Divides Itself we also have that $p$ is a divisor of $p$. From Absolute Value of Integer is not less than Divisors it follows that any positive integer greater than $p$ is not a divisor of $p$. Thus the only positive integer divisors of $p$ are $1$ and $p$. {{qed}} === Definition 2 iff Definition 6 === This is proved in Prime iff Equal to Product. {{qed}} === Definition 4 iff Definition 7 === Let the only two positive integer divisors of $p$ be $1$ and $p$. Then the only positive integer less than $p$ which divisors of $p$ is $1$ So if $a b = p$ then either $a$ or $b$ is $p$ and so it is not the case that both $a$ and $b$ are less than $p$. Now suppose that there do not exist $a$ and $b$ less than $p$ such that $a b = p$. Suppose $a \divides p$ such that $a < p$. That means that $\exists b \in \Z_{>0}: a b = p$. That means $b \divides p$. But $b \not < p$ by hypothesis. It follows that $b = p$ and so $a = 1$ Hence $1$ and $p$ are the only positive integer divisors of $p$. {{qed}}
From Prime not Divisor implies Coprime, if $p$ does not divide an integer $x$, it is relatively prime to $x$. From Absolute Value of Integer is not less than Divisors: Corollary, $p$ does not divide an integer smaller than $p$. It follows that $p$ is relatively prime to all smaller (strictly) positive integers. The special case when $x = 0$ is excluded as from Integers Coprime to Zero, $p$ is not relatively prime to $0$. {{qed}}
This result can be split into two parts: === Proof of Existence === {{:Division Theorem/Positive Divisor/Positive Dividend/Existence}} === Proof of Uniqueness === {{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness}} Category:Division Theorem 5mooiyv6ltjxhds9a5ijyjvoog14knh
This result can be split into two parts: === Proof of Existence === {{:Division Theorem/Positive Divisor/Existence}} === Proof of Uniqueness === {{:Division Theorem/Positive Divisor/Uniqueness}}
Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given. Let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer: :$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$ By setting $z = 0$ we have that $a \in S$. Thus $S \ne \O$. We have that $S$ is bounded below by $0$. From Set of Integers Bounded Below by Integer has Smallest Element it follows that $S$ has a smallest element $r$. Thus: :$\exists q \in \Z: a - q b = r$ and so: :$a = q b + r$ So we have proved the existence of $q$ and $r$ such that $a = q b + r$. It remains to be shown that $0 \le r < b$. We have that $r \in S$ which is bounded below by $0$. Therefore $0 \le r$. {{AimForCont}} $b \le r$. So: {{begin-eqn}} {{eqn | l = b       | o = \le       | r = r       | c =  }} {{eqn | o = <       | r = r + b       | c = as $b > 0$ }} {{eqn | ll= \leadsto       | l = 0       | o = \le       | r = r - b       | c =  }} {{eqn | o = <       | r = r       | c = subtracting $b$ throughout }} {{end-eqn}} But then: {{begin-eqn}} {{eqn | l = a       | r = q b + r       | c = from above }} {{eqn | ll= \leadsto       | l = r - b       | r = \paren {a - q b} - b       | c =  }} {{eqn | r = a - b \paren {q + 1}       | c =  }} {{eqn | ll= \leadsto       | l = r - b       | o = \in       | r = S       | c = as $r - b$ is of the correct form }} {{end-eqn}} But $r - b < r$ contradicts the choice of $r$ as the least element of $S$. Hence $r < b$ as required. Thus the existence of $q$ and $r$ satisfying $a = q b + r, 0 \le r < b$ has been demonstrated. {{qed}}
Let $a = 0$. It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$. {{qed|lemma}} Let $a > 0$ and $b = 1$. Then from the condition $0 \le r < b$ it follows that $r = 0$ and hence $q = a$. {{qed|lemma}} Let $a > 0$ and $b > 1$. By the Basis Representation Theorem, $a$ has a unique representation to the base $b$: {{begin-eqn}} {{eqn | l = a       | r = \sum_{k \mathop = 0}^s r_k b^k       | c =  }} {{eqn | r = b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0       | c =  }} {{eqn | r = b q + r       | c = where $0 \le r = r_0 < b$ }} {{end-eqn}} {{qed}}
Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$. That is: {{begin-eqn}} {{eqn | l = a       | r = q_1 b + r_1, 0 \le r_1 < b       | c =  }} {{eqn | l = a       | r = q_2 b + r_2, 0 \le r_2 < b       | c =  }} {{end-eqn}} This gives: :$0 = b \paren {q_1 - q_2} + \paren {r_1 - r_2}$ {{AimForCont}} that $q_1 \ne q_2$. {{WLOG}}, suppose that $q_1 > q_2$. Then: {{begin-eqn}} {{eqn | l = q_1 - q_2       | o = \ge       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = r_2 - r_1       | r = b \left({q_1 - q_2}\right)       | c =  }} {{eqn | o = \ge       | r = b \times 1       | c = as $b > 0$ }} {{eqn | r = b       | c =  }} {{eqn | ll= \leadsto       | l = r_2       | o = \ge       | r = r_1 + b       | c =  }} {{eqn | o = \ge       | r = b       | c =  }} {{end-eqn}} This contradicts the assumption that $r_2 < b$. A similar contradiction follows from the assumption that $q_1 < q_2$. Therefore $q_1 = q_2$ and so $r_1 = r_2$. Thus it follows that $q$ and $r$ are unique. {{qed}}