hcju/proofwikiqa huggingface dataset:
Proof by induction: From the definition of natural number multiplication, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m \times 0       | r = 0 }} {{eqn | l = m \times n^+       | r = \left({m \times n}\right) + m }} {{end-eqn}} For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$\forall m \in \N: m \times n = n \times m$ === Basis for the Induction === From Zero is Zero Element for Natural Number Multiplication, we have: :$\forall m \in \N: m \times 0 = 0 = 0 \times m$ Thus $P \left({0}\right)$ is seen to be true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true. So this is our induction hypothesis $P \left({k}\right)$: :$\forall m \in \N: m \times k = k \times m$ Then we need to show that $P \left({k^+}\right)$ follows from $P \left({k}\right)$: :$\forall m \in \N: m \times k^+ = k^+ \times m$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = m \times k^+       | r = \left({m \times k}\right) + m       | c = Definition of Natural Number Multiplication }} {{eqn | r = m + \left({m \times k}\right)       | c = Natural Number Addition is Commutative }} {{eqn | r = m + \left({k \times m}\right)       | c = Induction Hypothesis }} {{eqn | r = k^+ \times m       | c = Natural Number Multiplication Distributes over Addition }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \N: m \times n = n \times m$ {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\forall a \in \N_{> 0}: a \times n = n \times a$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = a \times 1       | r = a       | c = Axiom $\text A$ }} {{eqn | r = 1 \times a       | c = Axiom $\text A$ }} {{end-eqn}} and so $\map P 1$ 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: :$\forall a \in \N: a \times k = k \times a$ Then we need to show: :$\forall a \in \N: a \times \paren {k + 1} = \paren {k + 1} \times a$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = a \times \paren {k + 1}       | r = \paren {a \times k} + \paren {a \times 1}       | c = Left Distributive Law for Natural Numbers }} {{eqn | r = \paren {k \times a} + \paren {a \times 1}       | c = Induction hypothesis }} {{eqn | r = \paren {k \times a} + \paren {1 \times a}       | c = Basis for the Induction }} {{eqn | r = \paren {k + 1} \times a       | c = Right Distributive Law for Natural Numbers }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}
Natural number multiplication is recursively defined as: :$\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \left({n + 1}\right) & = m \times n + m \end{cases}$ From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition; that is, such that: :$\forall n \in \N: \begin{cases} f \left({0}\right) = 0 \\ f \left({n + 1}\right) = f \left({n}\right) + m \end{cases}$ Consider now $f'$ defined as $f' \left({n}\right) = n \times m$. Then by Zero is Zero Element for Natural Number Multiplication: :$f' \left({0}\right) = 0 \times m = 0$ Furthermore: {{begin-eqn}} {{eqn|l = f' \left({n + 1}\right)      |r = \left({n + 1}\right) \times m }} {{eqn|r = n \times m + m      |c = Natural Number Multiplication Distributes over Addition }} {{eqn|r = f' \left({n}\right) + m }} {{end-eqn}} showing that $f'$ also satisfies the definition of $m \times n$. By the Principle of Recursive Definition it follows that: :$m \times n = n \times m$ {{qed}}
We are to show that: :$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$ for all $x, y, n \in \N$. From the definition of natural number multiplication, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m \times 0       | r = 0       | c =  }} {{eqn | l = m \times \left({n + 1}\right)       | r = \left({m \times n}\right) + m       | c =  }} {{end-eqn}} Let $x, y \in \N$ be arbitrary. For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left({x \times y}\right) \times 0       | r = 0 }} {{eqn | r = x \times 0 }} {{eqn | r = x \times \left({y \times 0}\right) }} {{end-eqn}} and so $P \left({0}\right)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$\left({x \times y}\right) \times k = x \times \left({y \times k}\right)$ Then we need to show: :$\left({x \times y}\right) \times \left({k + 1 }\right) = x \times \left({y \times \left({k + 1}\right)}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({x \times y}\right) \times \left({k + 1 }\right)       | r = \left({\left({x \times y}\right) \times k}\right) + \left({x \times y}\right)       | c = Definition of Natural Number Multiplication }} {{eqn | r = \left({x \times \left({y \times k}\right)}\right) + \left({x \times y}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({x \times y}\right) + \left({x \times \left({y \times k}\right)}\right)       | c = Natural Number Addition is Commutative }} {{eqn | r = x \times \left({y + \left({y \times k}\right)}\right)       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | r = x \times \left({\left({y \times k}\right) + y}\right)       | c = Natural Number Addition is Commutative }} {{eqn | r = x \times \left({y \times \left({k + 1 }\right)}\right)       | c = Definition of Natural Number Multiplication }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. {{qed}}
From Index Laws for Semigroup: Product of Indices we have: :$+^{z \times y} x = +^z \left({+^y x}\right)$ By definition of multiplication, this amounts to: :$x \times \left({z \times y}\right) = \left({x \times y}\right) \times z$ From Natural Number Multiplication is Commutative, we have: :$x \times \left({z \times y}\right) = x \times \left({y \times z}\right)$ {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Let $x, y \in \N_{> 0}$ be arbitrary. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\paren {x \times y} \times n = x \times \paren {y \times n}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \paren {x \times y} \times 1       | r = x \times y       | c = Axiom $\text A$ }} {{eqn | r = x \times \paren {y \times 1}       | c = Axiom $\text A$ }} {{end-eqn}} and so $\map P 1$ 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 $\map P {k^+}$ is true. So this is our induction hypothesis: :$\paren {x \times y} \times k = x \times \paren {y \times k}$ Then we need to show: :$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {x \times y} \times \paren {k + 1}       | r = \paren {\paren {x \times y} \times k} + \paren {x \times y}       | c = Axiom $\text B$ }} {{eqn | r = \paren {x \times \paren {y \times k} } + \paren {x \times y}       | c = Induction Hypothesis }} {{eqn | r = \paren {x \times y} + \paren {x \times \paren {y \times k} }       | c = Natural Number Addition is Commutative }} {{eqn | r = x \times \paren {y + \paren {y \times k} }       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | r = x \times \paren {\paren {y \times k} + y}       | c = Natural Number Addition is Commutative }} {{eqn | r = x \times \paren {y \times \paren {k + 1} }       | c = {{Defof|Natural Number Multiplication}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
We are to show that: :$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$ for all $x, y, z \in \N$. From the definition of natural number multiplication, we have by definition that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m \times 0       | r = 0 }} {{eqn | l = m \times n^+       | r = \left({m \times n}\right) + m }} {{end-eqn}} Let $x, y \in \N$ be arbitrary. For all $z \in \N$, let $P \left({z}\right)$ be the proposition: :$\forall x, y \in \N: \left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left({x + y}\right) \times 0       | r = 0       | c = Definition of Natural Number Multiplication }} {{eqn | r = 0 + 0       | c = Definition of Natural Number Addition }} {{eqn | r = x \times 0 + y \times 0       | c = Definition of Natural Number Multiplication }} {{end-eqn}} and so $P \left({0}\right)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true. So this is our induction hypothesis: :$\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$ Then we need to show: :$\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({x + y}\right) \times k^+       | r = \left({x + y}\right) \times k + \left({x + y}\right)       | c = Definition of Natural Number Multiplication }} {{eqn | r = \left({x \times k}\right) + \left({y \times k}\right) + \left({x + y}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({\left({x \times k}\right) + x}\right) + \left({\left({y \times k}\right) + y}\right)       | c = Natural Number Addition is Commutative and Associative }} {{eqn | r = \left({x \times k^+}\right) + \left({y \times k^+}\right)       | c = Definition of Natural Number Multiplication }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction: :$\forall x, y, z \in \N: \left({x + y}\right) \times n = \left({x \times z}\right) + \left({y \times z}\right)$ {{qed|lemma}} Next we need to show that: :$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$ for all $x, y, z \in \N$. So: {{begin-eqn}} {{eqn | l = z \times \left({x + y}\right)       | r = \left({x + y}\right) \times z       | c = Natural Number Multiplication is Commutative }} {{eqn | r = \left({x \times z}\right) + \left({y \times z}\right)       | c = from above }} {{eqn | r = \left({z \times x}\right) + \left({z \times y}\right)       | c = Natural Number Multiplication is Commutative }} {{end-eqn}} Thus we have proved: :$\forall x, y, z \in \N: z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$ {{qed}}
{{begin-eqn}} {{eqn | l = \paren {x + y} \times z       | r = +^z \paren {x + y}       | c = {{Defof|Natural Number Multiplication}} }} {{eqn | r = \paren {+^z x} + \paren {+^z y}       | c = Power of Product of Commuting Elements in Semigroup equals Product of Powers }} {{eqn | r = x \times z + y \times z }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = z \times \paren {x + y}       | r = +^{x + y} z       | c = {{Defof|Natural Number Multiplication}} }} {{eqn | r = \paren {+^x z} + \paren {+^y z}       | c = Index Laws for Semigroup: Sum of Indices }} {{eqn | r = z \times x + z \times y }} {{end-eqn}} {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} === Left Distributive Law for Natural Numbers === First we show that: :$n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$ {{:Left Distributive Law for Natural Numbers}}{{qed|lemma}} === Right Distributive Law for Natural Numbers === Then we show that: :$\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$ {{:Right Distributive Law for Natural Numbers}}{{qed|lemma}} The result follows. {{qed}}
This is demonstrated in these pages: :Natural Number Multiplication Distributes over Addition :Integer Multiplication Distributes over Addition :Rational Multiplication Distributes over Addition :Real Multiplication Distributes over Addition :Complex Multiplication Distributes over Addition {{qed}}
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $\times$ is distributive over $+$ on $\Q$. {{qed}} Category:Rational Addition Category:Rational Multiplication Category:Distributive Operations bjfg58n5cz46tzt2g5hmag0hzq9g1cj
Let a first magnitude $A$ be the same multiple of a second $B$ that a third $C$ is of a fourth $D$. Let equimultiples $EF, GH$ be taken of $A, C$. We need to show that $EF$ is the same multiple of $B$ that $GH$ is of $D$. We have that $EF$ is the same multiple of $A$ that $GH$ is of $C$. Therefore as many magnitudes as there are in $EF$ equal to $A$, so many also are there in $GH$ equal to $C$. Let $EF$ be divided into the magnitudes $EK, KF$ equal to $A$, and $GH$ ito the magnitudes $GL, LH$ equal to $C$. Then the multitude of the magnitudes $EK, KF$ will be equal to the multitude of the magnitudes $GL, LH$. :310px We have that $A$ is the same multiple of $B$ that $C$ is of $D$, while $EK = A$ and $GL = C$. So $EK$ is the same multiple of $B$ that $GL$ is of $D$. For the same reason, $KF$ is the same multiple of $B$ that $LH$ is of $D$. So we have that: :a first magnitude $EK$ is the same multiple of a second $B$ that a third $GL$ is of a fourth $D$ :a fifth $KF$ is also of the same multiple of the second $B$ that a sixth $LH$ is of the fourth $D$. Therefore the sum of the first and fifth, $EF$, is also the same multiple of the second $B$ that the sum of the third and sixth, $GH$ is of the fourth $D$. {{qed}}
By definition of subtraction: {{begin-eqn}} {{eqn | l = a - \paren {b - c}       | r = a + \paren {-\paren {b + \paren {-c} } }       | c =  }} {{eqn | r = a + \paren {-b} + c       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {a - b} - c       | r = \paren {a + \paren {-b} } + \paren {-c}       | c =  }} {{eqn | r = a + \paren {-b} + \paren {-c}       | c =  }} {{end-eqn}} So we see that: :$a - \paren {b - c} = \paren {a - b} - c \iff c = 0$ and so in general: :$a - \paren {b - c} \ne \paren {a - b} - c$ {{qed}}
This is demonstrated in these pages: * Identity Element of Natural Number Multiplication is One * Integer Multiplication Identity is One * Rational Multiplication Identity is One * Real Multiplication Identity is One * Complex Multiplication Identity is One {{qed}}
This is demonstrated in these pages: * Identity Element of Natural Number Addition is Zero * Integer Addition Identity is Zero * Rational Addition Identity is Zero * Real Addition Identity is Zero * Complex Addition Identity is Zero {{qed}}
Suppose $z_1$ and $z_2$ are both zeroes of $\struct {S, \circ}$. Then by the definition of zero element: : $z_2 \circ z_1 = z_1$ by dint of $z_1$ being a zero : $z_2 \circ z_1 = z_2$ by dint of $z_2$ being a zero. So $z_1 = z_2 \circ z_1 = z_2$. So $z_1 = z_2$ and there is only one zero after all. {{qed}}
Let $e \in G$ be the identity element of $G$. Let $z \in G$ be a zero element. Let $x \in G$ be any arbitrary element of $\struct {G, \circ}$. Then: {{begin-eqn}} {{eqn | l = x       | r = x \circ e       | c = {{GroupAxiom|2}} }} {{eqn | r = x \circ \paren {z \circ z^{-1} }       | c = {{GroupAxiom|3}} }} {{eqn | r = \paren {x \circ z} \circ z^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = z \circ z^{-1}       | c = {{Defof|Zero Element}}: $x \circ z = z$ }} {{eqn | r = e       | c = {{GroupAxiom|3}} }} {{end-eqn}} So whatever $x \in G$ is, it has to be the identity element of $G$. So $G$ can contain only that one element, and is therefore the trivial group. {{qed}}
A binary operation on $S$ is by definition a mapping from the cartesian product $S \times S$ to the set $S$. Thus we are looking to evaluate: :$N = \card {\set {f: S \times S \to S} }$  The domain of $f$ has $n^2$ elements, from Cardinality of Cartesian Product. The result follows from Cardinality of Set of All Mappings. {{qed}}
Let $\struct {S, \circ}$ be a magma. From Cardinality of Cartesian Product, there are $n^2$ elements in $S \times S$. The binary operations $\circ$ is commutative {{iff}}: :$\forall x, y \in S: x \circ y = y \circ x$ Thus for every pair of elements $\tuple {x, y} \in S \times S$, it is required that $\tuple {y, x} \in S \times S$. So the question boils down to establishing how many different ''unordered'' pairs there are in $S$. That is, how many doubleton subsets there are in $S$. From Cardinality of Set of Subsets, this is given by: :$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$ To that set of doubleton subsets, we also need to add those ordered pairs where $x = y$. There are clearly $n$ of these. So the total number of pairs in question is: :$\dfrac {n \paren {n - 1} } 2 + n = \dfrac {n \paren {n + 1} } 2$ The result follows from Cardinality of Set of All Mappings. {{qed}}
Let $S$ be a set such that $\left|{S}\right| = n$. Let $x \in S$ be an identity element. From Count of Binary Operations on Set, there are $n^{\left({n^2}\right)}$ binary operations in total. We also know that $a \in S \implies a \circ x = a = x \circ a$, so all operations on $x$ are already specified. It remains to count all possible combinations of the remaining $n-1$ elements. This is effectively counting the mappings $\left({S - \left\{{x}\right\}}\right) \times \left({S - \left\{{x}\right\}}\right) \to S$. From Count of Binary Operations on Set, this is $n^{\left({\left({n-1}\right)^2}\right)}$ structures with $x$ as the identity. {{qed}}
This follows by the arguments of Count of Binary Operations with Fixed Identity and Count of Commutative Binary Operations on Set. From Count of Binary Operations on Set, there are $n^{\left({n^2}\right)}$ binary operations in total. We also know that: :$a \in S \implies a \circ x = a = x \circ a$ so all binary operations on $x$ are already specified. It remains to count all possible combinations of the remaining $n-1$ elements. This is effectively counting the mappings: : $\left({S - \left\{{x}\right\}}\right) \times \left({S - \left\{{x}\right\}}\right) \to S$ So the question boils down to establishing how many different ''unordered'' pairs there are in $\left({S - \left\{{x}\right\}}\right)$. That is, how many doubleton subsets there are in $\left({S - \left\{{x}\right\}}\right)$. From Cardinality of Set of Subsets, this is given by: :$\dbinom {n - 1} 2 = \dfrac {\left({n - 1}\right) \left({n - 2}\right)} 2$ To that set of doubleton subsets, we also need to add those ordered pairs where $x = y$. There are clearly $n - 1$ of these. So the total number of pairs in question is: : $\dfrac {\left({n - 1}\right) \left({n - 2}\right)} 2 + n - 1 = \dfrac {n \left({n-1}\right)} 2$ The result follows from Cardinality of Set of All Mappings. {{qed}}
From Count of Binary Operations with Fixed Identity, there are $n^{\paren {n - 1}^2}$ such binary operations for each individual element of $S$. As Identity is Unique, if $x$ is the identity, no other element can also be an identity. As there are $n$ different ways of choosing such an identity, there are $n \times n^{\left({n-1}\right)^2}$ different magmas with an identity. These are guaranteed not to overlap by the uniqueness of the identity. Hence the result. {{qed}}
From Count of Commutative Binary Operations with Fixed Identity, there are $n^{\frac {n \left({n-1}\right)}2}$ such binary operations for each individual element of $S$. As Identity is Unique, if $x$ is the identity, no other element can also be an identity. As there are $n$ different ways of choosing such an identity, there are $n \times n^{\frac {n \left({n-1}\right)}2}$ different algebraic structures with an identity. These are guaranteed not to overlap by the uniqueness of the identity. Hence the result. {{qed}}
From Count of Binary Operations on Set, the total number of operations is $n^{\left({n^2}\right)}$. From Count of Binary Operations with Identity, the total number of operations with an identity is $n^{\left({n-1}\right)^2 + 1}$. So the total number of operations without an identity is: :$n^{\left({n^2}\right)} - n^{\left({n-1}\right)^2 + 1} = n^{\left({\left({n-1}\right)^2 + 1}\right)} \left({n^{2 \left({n-1}\right)} - 1}\right)$ Hence the result. {{qed}}
Proof by induction: For all $n \in \N^*$, let $\map P n$ be the compound proposition: :$x^n \in H \text{ and } x^{-n} \in H$. $\map P 0$ is true, as this just says $x^0 \in H$. By Powers of Group Elements, $x^0 = e$. This follows by Identity of Subgroup. === Basis for the Induction === $\map P 1$ is true, as this just says $x^1 = x \in H$. By the Two-Step Subgroup Test, we also have that $x^{-1} \in H$. This is our basis for the induction. === Induction Hypothesis === Now we need to show 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 our induction hypothesis: :$x^k \in H \text{ and } x^{-k} \in H$. Then we need to show: :$x^{k + 1} \in H \text{ and } x^{-\paren {k + 1} } \in H$. === Induction Step === This is our induction step: :By Powers of Group Elements, $x^{k + 1} = x \circ x^k$. :By the base case, $x \in H$. :By the induction hypothesis, $x^k \in H$. :By the closure axiom, $x \circ x^k \in H$. :By the Two-Step Subgroup Test, it follows that $x^{-\paren {k + 1} } \in H$. So $\map P k \implies \map P {k + 1}$. The result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: x^n \in H$ :$\forall n \in \N: x^{-n} \in H$ Hence the result. {{qed}}
Let $n \in \Z$. We have: {{begin-eqn}} {{eqn | l = n       | r = q k + r       | c =  }} {{eqn | ll= \leadstoandfrom       | l = n - r       | r = q k       | c =  }} {{eqn | ll= \leadstoandfrom       | l = k       | o = \divides       | r = \paren {n - r}       | c =  }} {{end-eqn}} The result follows from Equal Powers of Finite Order Element. {{qed}}
Let $k \in \N$ be the smallest such that $a^k = e$ as per the hypothesis. === Necessary Condition === Let $a^n = e$. Let $n = q k + r, 0 \le r < k$. By Element to Power of Remainder: : $a^r = a^n = e$ But $0 \le r < k$. Since $k$ is the smallest such that $a^k = e$: :$1 \le s < k \implies a^s \ne e$ Thus $r = 0$. That is: : $k \divides n$ === Sufficient Condition === Suppose $k \divides n$. Then: : $\exists s \in \Z: n = s k$ So: {{begin-eqn}} {{eqn | l = a^n       | r = a^{s k}       | c =  }} {{eqn | r = \paren {a^k}^s       | c =  }} {{eqn | r = e^s       | c =  }} {{eqn | r = e       | c =  }} {{end-eqn}} {{qed}}
By Element to Power of Remainder, every power of $a$ is equal to one appearing in the list $a^0, a^1, a^2, \ldots, a^{k - 1}$. This list has to be repetition free, otherwise it would contain $a^m = a^n$ with $0 \le m < n < k$ which contradicts the hypothesis. {{qed}}