hcju/proofwikiqa huggingface dataset:
{{NotZFC}} Let $x \in B$. Let $x'$ be the transitive closure of $x$. Let $L = x' \cap B$. Then $x \in L$, so $L$ is not empty. Since $x'$ is a set, so is $L$, by the axiom of subset. Thus by the Axiom of Foundation, $L$ has an $\in$-minimal element $m$. By the definition of intersection, $m \in B$. Suppose for the sake of contradiction that there is an element $b \in B$ such that $b \in m$. Then since $m \in x'$ and $x'$ is transitive, $b \in x'$. Thus $b \in L$, contradicting the minimality of $m$. So $m$ is an $\in$-minimal element of $B$. {{qed}}
Let $\left({G, \circ}\right)$ be a group with identity $e_G$. By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy: :$\phi (e) = e_G$ Let us define the mapping $\phi$ in this way. By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniqueness. Now to verify that $\phi$ is actually a group homomorphism. Since $1$ has only the element $e$, this is verified by: {{begin-eqn}} {{eqn | l = \phi (e) \circ \phi (e)       | r = e_G \circ e_G       | c = Definition of $\phi$ }} {{eqn | r = e_G       | c = Definition of identity }} {{eqn | r = \phi (e) }} {{eqn | r = \phi \left({e * e}\right) }} {{end-eqn}} where we used $*$ to denote the group operation on $1$. Thus $\phi$ is a group homomorphism. We have thus established that there is a unique morphism $1 \to G$ in $\mathbf{Grp}$ for all $G$. That is, $1$ is an initial object in $\mathbf{Grp}$. {{qed}}
Let $\left({G, \circ}\right)$ be any group. By Singleton is Terminal Object of Category of Sets, there is precisely one mapping: :$!: G \to \left\{{e}\right\}$ defined by: :$\forall g \in G: ! (g) = e$ By definition, any group homomorphism is also a mapping. Hence, there is at most one morphism $\left({G, \circ}\right) \to \left\{{e}\right\}$ in $\mathbf{Grp}$. Now to verify that the mapping $!$ is a group homomorphism. For any $g, h \in G$, we have (using $*$ for the group operation on $\left\{{e}\right\}$): {{begin-eqn}} {{eqn | l = ! (g) * ! (h)       | r = e * e       | c = Definition of $!$ }} {{eqn | r = e       | c = Definition of $*$ }} {{eqn | r = ! \left({g \circ h}\right)       | c = Definition of $!$ }} {{end-eqn}} That is, $!$ is a group homomorphism. Thus for all groups $\left({G, \circ}\right)$, there is a unique group homomorphism $!: G \to \left\{{e}\right\}$. That is, $\left\{{e}\right\}$ is a terminal object of $\mathbf{Grp}$. {{qed}}
From the definition of orbit: :$\Orb H = \set {y \in G: \exists g \in G: y = g \circ H}$ The result follows from the definition of (left) coset space. {{qed}}
We have that: :$\Orb H = \set {g \circ H \circ g^{-1}: g \in G}$ from the definition. The result follows by definition of conjugate subgroup. {{Qed}}
Let $R \sqbrk x$ denote the image of $x$ under $R$. Set: :$\displaystyle \map G x = x \cup \bigcup_{i \mathop = 1}^n \mathcal R_i \sqbrk x \cup \bigcup_{i \mathop = 1}^m \SS_i \sqbrk x$ Using the Principle of Recursive Definition, construct the function $F$ as follows: :$\map F 0 = T$ :$\map F {n + 1} = \map G {\map F n}$ Define $X$ as follows: :$\displaystyle X = \bigcup_{n \mathop \in \omega} \map F n$ === Proof that $T \subseteq X$ === {{begin-eqn}} {{eqn | l = T       | r = \map F 0       | c = Definition of $F$ }} {{eqn | o = \subseteq       | r = \bigcup_{n \mathop \in \omega} \map F n       | c = Set is Subset of Union }} {{end-eqn}} {{qed|lemma}} === Proof of Closure === Take any $\RR_i$. Suppose $x \RR_i y$ and $x \in \map F n$ for some $n$. Then $y \in \map F {n + 1}$ by the definition of image. Thus $y \in X$ and $R_i$ is closed with respect to $X$. Similarly, take any $\SS_i$. Suppose $\paren {x S_i y} = z$ and that $x \in X$ and $y \in X$. It follows that $x, y \in \map F n$ for some $n$. So: :$z \in \map F {n + 1}$ Therefore: :$z \in X$ {{qed|lemma}} === Proof that $X$ is the Smallest Relational Closure === Suppose $T \subseteq Y$ and that $R_i$ and $S_i$ are closed with respect to $Y$. $\map F n \subseteq Y$ shall be proven by induction: For all $n \in \omega$, let $\map P n$ be the proposition: :$\map F n \subseteq Y$ === Basis for the Induction === $\map P 0$ is the case: :$\map F 0 \subseteq Y$ which has been proved above. 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: :$\map F k \subseteq Y$ Then we need to show: :$\map F {k + 1} \subseteq Y$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map F k       | o = \subseteq       | r = Y       | c = Inductive Hypothesis }} {{eqn | ll= \leadsto       | l = R_i \sqbrk {\map F k} \subseteq Y        | o = \land       | r = S_i \sqbrk {\map F k \times \map F k} \subseteq Y       | c = Image Preserves Subsets and $R_i$ and $S_i$ are closed with respect to $Y$ }} {{eqn | ll= \leadsto       | l = \map F k \cup \bigcup_{i \mathop = 1}^n R_i “ \map F k \cup \bigcup_{i \mathop = 1}^m S_i “ \map F k       | o = \subseteq       | r = Y       | c = Indexed Union Subset }} {{eqn | ll= \leadsto       | l = \map G {\map F k}       | o = \subseteq       | r = Y       | c = Definition of $G$ }} {{eqn | ll= \leadsto       | l = \map F {k + 1}       | o = \subseteq       | r = Y       | c = Definition of $\map F {k + 1}$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \omega: \map F n \subseteq R$ Hence by Indexed Union Subset: :$T \subseteq Y$ {{qed}}
{{NotZFC}} The proof shall proceed by Transfinite Induction on $x$. === Basis for the Induction === We have that: :$V_0 = \O$ and $\O$ is supertransitive by the very fact that it has no elements. This proves the basis for the induction. {{qed|lemma}} === Induction Step === Let $\map V x$ be supertransitive. First, to prove transitivity. Let $\map V x$ be transitive. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \map V {x^+} }} {{eqn | ll= \leadsto       | l = y       | o = \subseteq       | r = \map V x       | c =  }} {{eqn | ll= \leadsto       | lo= \forall z:       | l = z \in y       | o = \implies       | r = z \subseteq \map V x       | c = Transitivity of $\map V x$ }} {{eqn | ll= \leadsto       | lo= \forall z:       | l = z \in y       | o = \implies       | r = z \in \map V {x^+}       | c = {{Defof|Von Neumann Hierarchy}} }} {{eqn | ll= \leadsto       | l = y       | o = \subseteq       | r = \map V {x^+}       | c = {{Defof|Subset}} }} {{end-eqn}} Next, to prove supertransitivity: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \powerset {\map V x}       | c = Hypothesis }} {{eqn | ll= \leadsto       | l = y       | o = \subseteq       | r = \map V x       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = \powerset y       | o = \subseteq       | r = \powerset {\map V x}       | c = Power Set Preserved Under Subset }} {{eqn | r = \map V {x^+}       | c = {{Defof|Von Neumann Hierarchy}} }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === Let $x$ be a limit ordinal. Furthermore, let $\map V y$ be transitive for all $y \in x$. Then: {{begin-eqn}} {{eqn | l = z       | o = \in       | r = \map V x       | c = Hypothesis }} {{eqn | ll= \leadsto       | lo= \exists y \in x:       | l = z       | o = \in       | r = \map V y       | c = {{Defof|Von Neumann Hierarchy}} }} {{eqn | ll= \leadsto       | lo= \exists y \in x:       | l = z       | o = \subseteq       | r = \map V y       | c = Transitivity of $V_y$ }} {{eqn | o = \subseteq       | r = \map V x       | c = Set is Subset of Union }} {{end-eqn}} This proves transitivity. Now, to prove supertransitivity: {{begin-eqn}} {{eqn | l = z       | o = \in       | r = \map V x       | c = Hypothesis }} {{eqn | ll= \leadsto       | lo= \exists y \in x:       | l = z       | o = \in       | r = \map V y       | c = {{Defof|Von Neumann Hierarchy}} }} {{eqn | ll= \leadsto       | lo= \exists y \in x:       | l = z       | o = \subseteq       | r = \map V y       | c = Transitivity of $V_y$ }} {{eqn | ll= \leadsto       | lo= \exists y \in x:       | l = \powerset z       | o = \subseteq       | r = \map V {y + 1}       | c = Power Set Preserved Under Subset }} {{eqn | o = \subseteq       | r = \map V x       | c = Set is Subset of Union }} {{end-eqn}} This proves the limit case. {{qed}}
{{NotZFC}} The proof shall proceed by Transfinite Induction on $y$. === Basis for the Induction === If $y = 0$, then $x \not < y$. This proves the basis for the induction. {{qed|lemma}} === Induction Step === Let $x < y \implies \map V x \in \map V y$. Then: {{begin-eqn}} {{eqn | l = x       | o = <       | r = y^+       | c =  }} {{eqn | ll= \leadsto       | l = x < y       | o = \lor       | r = x = y       | c = {{Defof|Successor Set}} }} {{eqn | l = x = y       | o = \implies       | r = \map V x \in \powerset {\map V x}       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = \map V x       | o = \in       | r = \map V {y^+}       | c = {{Defof|Von Neumann Hierarchy}} }} {{eqn | l = x < y       | o = \implies       | r = \map V x \in \map V y       | c = Inductive Hypothesis }} {{eqn | ll= \leadsto       | l = \map V x       | o = \in       | r = \map V {y^+}       | c = Von Neumann Hierarchy is Supertransitive }} {{end-eqn}} In either case: :$\map V x \in \map V {y^+}$ This proves the induction step. {{qed|lemma}} === Limit Case === Let $y$ be a limit ordinal. Let: :$\map V x \in \map V z$ for all $z \in y$ such that $x < z$. Since $x < y$, it follows that $x < z$ for some $z \in y$ by Union of Limit Ordinal. By the inductive hypothesis: :$\map V x \in \map V z$ But by Set is Subset of Union: Family of Sets: :$\map V z \subseteq \map V y$ Therefore: :$\map V x \in \map V y$ This proves the limit case. {{qed|lemma}} :$\map V x \subset \map V y$ follows by Von Neumann Hierarchy is Supertransitive. {{qed}}
{{NotZFC}} The proof shall proceed by Epsilon Induction on $S$. Suppose that all the elements $a \in S$ are well-founded. That is, $a \in V \left({x}\right)$ for some $x$. Let: :$\displaystyle F \left({a}\right) = \bigcap \left\{{x \in \operatorname{On} : a \in V \left({x}\right)}\right\}$ Take $\displaystyle \bigcup_{a \mathop \in S} F \left({a}\right)$. {{explain|... and do what with it?}} Take any $a \in S$. {{begin-eqn}} {{eqn | l = a       | o = \in       | r = V \left({F \left({a}\right)}\right)       | c = Definition of $F$ }} {{eqn | ll= \implies       | l = a       | o = \in       | r = V \left({\bigcup_{x \mathop \in S} F \left({x}\right) }\right)       | c = Set is Subset of Union: Family of Sets and Von Neumann Hierarchy Comparison }} {{eqn | ll= \implies       | l = S       | o = \subseteq       | r = V \left({\bigcup_{x \mathop \in S} F \left({x}\right) }\right)       | c = Definition of Subset }} {{eqn | ll= \implies       | l = S       | o = \in       | r = \mathcal P \left({ V\left({ \bigcup_{x \mathop \in S} F\left({x}\right) }\right) }\right)       | c = Definition of Power Set }} {{end-eqn}} Therefore: :$\displaystyle S \in V \left({ \bigcup_{x \mathop \in S} F \left({x}\right) + 1}\right)$ and $S \in V \left({x}\right)$ for some ordinal $x$. {{qed}}
{{NotZFC}} The rank of $S$ is an intersection of a set of ordinals $B$. $B$ is nonempty by the fact that Every Set in Von Neumann Universe. Thus, $B$ has a minimal element, which is the rank of $S$ plus $1$. Therefore, the rank is itself an ordinal. {{qed}}
{{NotZFC}} === Sufficient Condition === Suppose that $x$ is equal to the rank of $S$. Then $S \in V \left({x+1}\right)$ by the definition of rank. Suppose $S \in V \left({x}\right)$.  Then $x \ne 0$, so $x = y^+$ for some ordinal $y$ or $x$ is a limit ordinal. Suppose $x = y^+$. Then, $S \in V \left({y+1}\right)$. This contradicts the fact that the rank is the smallest set $x$ such that $S \in V\left({x+1}\right)$. Suppose $x$ is a limit ordinal. Then, $S \in V \left({y}\right)$ for some $y$, so $S \in V \left({y+1}\right)$ and $y+1 < x$. This contradicts the fact that the rank is the smallest set $x$ such that $S \in V \left({x+1}\right)$. Therefore, $S \notin V \left({x}\right)$. {{qed|lemma}} === Necessary Condition === Suppose $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$. Then, if $y < x$, then $S \notin V \left({y+1}\right)$ by Von Neumann Hierarchy Comparison. If $x < y$, then $S \in V\left({y}\right)$. Therefore, $x$ is the unique ordinal that satisfies $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$. Moreover, the rank of $S$ also satisfies $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$. Therefore, $x = \operatorname{rank} \left({S}\right)$. {{qed}}
When $s = 1$ the result is trivial. Assume the result holds for $s = k$. Then $H = G_1 \times G_2 \times \ldots \times G_k$ is cyclic of order $n_1 n_2 \ldots n_k$. Applying the main result to $H \times G_{k + 1}$ gives us the result for $s = k + 1$. The result follows by induction. {{qed}}
=== Necessary Condition === Let $p$ be a prime number. Then, by definition, the only positive divisors of $p$ are $1$ and $p$. Hence by definition of the tau function: :$\map \tau p = 2$ {{qed|lemma}} === Sufficient Condition === Suppose $\map \tau p = 2$. Then by One Divides all Integers we have: :$1 \divides p$ Also, by Integer Divides Itself we have: :$p \divides p$ So if $p > 1$ it follows that $\map \tau p \ge 2$. Now for $\map \tau p = 2$ it must follow that the only divisors of $p$ are $1$ and $p$. That is, that $p$ is a prime number. {{qed}}
Let $a, b \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn|l = a \vee b      |r = \sup \left\{ {a, b}\right\}      |c = Definition of join }} {{eqn|r = \sup \left\{ {b, a}\right\}      |c = Axiom of Extension }} {{eqn|r = b \vee a      |c =  Definition of join }} {{end-eqn}} Hence the result. {{qed}}
Let $a, b \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn|l = a \wedge b      |r = \inf \left\{ {a, b}\right\}      |c = Definition of Meet }} {{eqn|r = \inf \left\{ {b, a}\right\}      |c = Axiom of Extension }} {{eqn|r = b \wedge a      |c =  Definition of Meet }} {{end-eqn}} Hence the result. {{qed}}
By definition of join: :$a \vee b = \sup \left\{{a, b}\right\}$ where $\sup$ denotes supremum. Since a supremum is a fortiori an upper bound: :$a \preceq \sup \left\{{a, b}\right\}$ :$b \preceq \sup \left\{{a, b}\right\}$ as desired. {{qed}}
By definition of meet: :$a \wedge b = \inf \left\{{a, b}\right\}$ where $\inf$ denotes infimum. Since an infimum is a fortiori a lower bound: :$\inf \left\{{a, b}\right\} \preceq a$ :$\inf \left\{{a, b}\right\} \preceq b$ as desired. {{qed}} Category:Join and Meet 6uisdb7i4lgfu5gmx3se5uxb9bzcd84
Let $a, b, c \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn|l = a \wedge \left({b \wedge c}\right)      |r = \inf \left\{ {a, b \wedge c} \right\}      |c = Definition of meet }} {{eqn|r = \inf \left\{ {\inf \left\{ {a} \right\}, \inf \left\{ {b, c} \right\} } \right\}      |c = Infimum of Singleton }} {{eqn|r = \inf \left\{ {a, b, c} \right\}      |c = Infimum of Infima }} {{eqn|r = \inf \left\{ {\inf \left\{ {a, b} \right\}, \inf \left\{ {c} \right\} } \right\}      |c = Infimum of Infima }} {{eqn|r = \inf \left\{ {a, b} \right\} \wedge c      |c = Infimum of Singleton }} {{eqn|r = \left({a \wedge b}\right) \wedge c      |c = Definition of meet }} {{end-eqn}} Hence the result. {{qed}}
=== Necessary Condition === If $s = t$, then trivially also: :$s^\preccurlyeq = t^\preccurlyeq$ {{qed|lemma}} === Sufficient Condition === Suppose that: :$s^\preccurlyeq = t^\preccurlyeq$ By definition of weak lower closure, we have: :$s \in s^\preccurlyeq$ :$t \in t^\preccurlyeq$ and hence: :$s \in t^\preccurlyeq$ :$t \in s^\preccurlyeq$ which by definition of weak lower closure means: :$s \preccurlyeq t$ and $t \preccurlyeq s$ Since $\preccurlyeq$ is antisymmetric it follows that $s = t$. {{qed}} Category:Lower Closures r29q5hrjf5ddxur17xbil4vsqzm2i35
=== Necessary Condition === If $s = t$, then trivially also: :$s^\succcurlyeq = t^\succcurlyeq$ {{qed|lemma}} === Sufficient Condition === Suppose that: :$s^\succcurlyeq = t^\succcurlyeq$ By definition of weak upper closure, we have: :$s \in s^\succcurlyeq$ :$t \in t^\succcurlyeq$ and hence: :$s \in s^\succcurlyeq$ :$t \in s^\succcurlyeq$ which by definition of weak upper closure means: :$t \preccurlyeq s$ and $s \preccurlyeq t$ Since $\preccurlyeq$ is antisymmetric it follows that $s = t$. {{qed}} Category:Upper Closures cduy6671bisz78p983f2cvsa6e68q6k
Let $a, b, c \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = a \vee \paren {b \vee c}       | r = a \vee \sup \set {b, c}       | c = {{Defof|Join (Order Theory)}} }} {{eqn | r = \sup \set {\sup \set a, \sup \set {b, c} }       | c = Supremum of Singleton }} {{eqn | r = \sup \set {a, b, c}       | c = Supremum of Suprema }} {{eqn | r = \sup \set {\sup \set {a, b}, \sup \set c}       | c = Supremum of Suprema }} {{eqn | r = \sup \set {a, b} \vee c       | c = Supremum of Singleton }} {{eqn | r = \paren {a \vee b} \vee c       | c = {{Defof|Join (Order Theory)}} }} {{end-eqn}} Hence the result. {{qed}}
=== Necessary Condition === Let $p$ be a prime number. Then by definition, the only divisors of $p$ are $\pm 1$ and $\pm p$. Thus, if $p = a b$ then either $a = \pm 1$ and $b = \pm p$ or $a = \pm p$ and $b = \pm 1$. {{qed|lemma}} === Sufficient Condition === Suppose that: :$\forall a, b \in \Z: p = a b \implies p = \pm a \lor p = \pm b$ This means that the only divisors of $p$ are $\pm 1$ and $\pm p$. That is, that $p$ is a prime number. {{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       | lo= \forall p, q \in \Z:       | l = p a + q b       | r = p x c + q y c       | c = Substitution for $a$ and $b$ }} {{eqn | r = \paren {p x + q y} c       | c = Integer Multiplication Distributes over Addition }} {{eqn | ll= \leadsto       | lo= \exists z \in \Z:       | l = p a + q b       | r = z c       | c = where $z = p x + q y$ }} {{eqn | ll= \leadsto       | l = c       | o = \divides       | r = \paren {p a + q b}       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{qed}}
We have that Integers form Integral Domain. The result then follows from Multiple of Divisor in Integral Domain Divides Multiple. {{qed}}
By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$. Then: :$\paren {a d} c = b c$ that is: :$\paren {a c} d = b c$ which follows because Integer Multiplication is Commutative and Integer Multiplication is Associative. Hence the result. {{qed}}
We have that Integers form Integral Domain. The result then follows directly from Divisor Relation in Integral Domain is Transitive. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \divides       | r = y       | c =  }} {{eqn | ll= \leadsto       | lo= \exists q_1 \in \Z:       | l = q_1 x       | r = y       | c = {{Defof|Divisor of Integer}} }} {{eqn | l = y       | o = \divides       | r = z       | c =  }} {{eqn | ll= \leadsto       | lo= \exists q_2 \in \Z:       | l = q_2 y       | r = z       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = q_2 \paren {q_1 x}       | r = z       | c = substituting for $y$ }} {{eqn | ll= \leadsto       | l = \paren {q_2 q_1} x       | r = z       | c = Integer Multiplication is Associative }} {{eqn | ll= \leadsto       | lo= \exists q \in \Z:       | l = q x       | r = z       | c = where $q = q_1 q_2$ }} {{eqn | ll= \leadsto       | l = x       | o = \divides       | r = z       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{Qed}}
We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a ring. Suppose $f$ and $g$ are polynomials in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$. If $\map \deg f = \map \deg g = 0$ then $f$ and $g$ are elements of $F$. As $F$ is a field and a field is an integral domain, $f g \ne 0_f$. Otherwise from Degree of Product of Polynomials over Integral Domain: :$\map \deg {f g} = \map \deg f + \map \deg g$ and so: :$\map \deg {f g} > 0$ which means: :$f g \ne 0_F$ Hence the result, by definition of integral domain. {{qed}}
We have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a commutative ring with unity. The result follows from Ring of Polynomial Forms is Integral Domain. {{qed}}
Proof by induction on $m$: For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition: :$\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$ $\map P 0$ is true, as this just says: :$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$ === Basis for the Induction === $\map P 1$ is true, by definition of power to an integer: :$r^{n + 1} = r^n \times r = r^n \times r^1$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall n \in \Z_{\ge 0}: r^{n + k} = r^n \times r^k$ Then we need to show: :$\forall n \in \Z_{\ge 0}: r^{n + k + 1} = r^n \times r^{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = r^n \times r^{k + 1}       | r = r^n \times \paren {r^k \times r}       | c = {{Defof|Integer Power}} }} {{eqn | r = \paren {r^n \times r^k} \times r       | c = Real Multiplication is Associative }} {{eqn | r = r^{n + k} \times r       | c = Induction Hypothesis }} {{eqn | r = r^{n + k + 1}       | c = {{Defof|Integer Power}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n, m \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$ {{qed}}
Proof by induction on $m$: For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition: :$\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$ $\map P 0$ is true, as this just says: :$\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$ === Basis for the Induction === $\map P 1$ is true, by definition of power to an integer: :$\paren {r^n}^1 = r^n = r^{n \times 1}$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall n \in \Z_{\ge 0}: \paren {r^n}^k = r^{n k}$ Then we need to show: :$\forall n \in \Z_{\ge 0}: \paren {r^n}^{\paren {k + 1} } = r^{n \paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {r^n}^{\paren {k + 1} }       | r = \paren {\paren {r^n}^k} \times r^n       | c = {{Defof|Integer Power}} }} {{eqn | r = r^{n k} \times r^n       | c = Induction Hypothesis }} {{eqn | r = r^{n k + n}       | c = Sum of Indices }} {{eqn | r = r^{n \paren {k + 1} }       | c = Integer Multiplication Distributes over Addition }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n, m \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$ {{qed}}