hcju/proofwikiqa huggingface dataset:
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$. Proof by induction: For all $n \in \N^*$, let $\map P n$ be the proposition: :$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$. $\map P 1$ is true, as this just says $J_1$ is an ideal of $R$. === Basis for the Induction === $\map P 2$ is the case: :$J_1 + J_2$ is an ideal of $R$ which is proved in Sum of Ideals is Ideal. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$J_1 + J_2 + \cdots + J_k$ is an ideal of $R$. Then we need to show: :$J_1 + J_2 + \cdots + J_k + J_{k + 1}$ is an ideal of $R$. === Induction Step === This is our induction step: Let $J = J_1 + J_2 + \cdots + J_k$. From the induction hypothesis, $J$ is an ideal. From the base case, $J + J_{k + 1}$ is an ideal. That is: :$J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: J_1 + J_2 + \cdots + J_n$ is an ideal of $R$. {{qed}}
{{AimForCont}} this supposition is false. Let $m$ be the smallest integer which can not be expressed as the product of primes. As a prime number is trivially a product of primes, $m$ can not itself be prime. Hence: :$\exists r, s \in \Z: 1 < r < m, 1 < s < m: m = r s$ As $m$ is our least counterexample, both $r$ and $s$ can be expressed as the product of primes. Say $r = p_1 p_2 \cdots p_k$ and $s = q_1 q_2 \cdots q_l$, where all of $p_1, \ldots, p_k, q_1, \ldots, q_l$ are prime. Hence $m = r s = p_1 p_2 \cdots p_k q_1 q_2 \cdots q_l$, which is a product of primes. Hence there is no such counterexample. {{qed}}
If $n$ is prime, the result is immediate. Let $n$ be composite. Then by Composite Number has Two Divisors Less Than It: :$\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s < n$ This being the case, the set $S_1 = \set {d: d \divides n, 1 < d < n}$ is nonempty, and bounded below by $1$. By Set of Integers Bounded Below by Integer has Smallest Element, $S_1$ has a smallest element, which we will call $p_1$. {{AimForCont}} $p_1$ is composite. By Composite Number has Two Divisors Less Than It, there exist $a, b$ such that $a, b \divides p_1$ and $1 < a < p_1, 1 < b < p_1$. But by Divisor Relation on Positive Integers is Partial Ordering, it follows that $a, b \divides n$ and hence $a, b \in S$. This contradicts the assertion that $p_1$ is the smallest element of $S_1$. Thus, $p_1$ is necessarily prime. We may now write $n = p_1 n_1$, where $n > n_1 > 1$. If $n_1$ is prime, the proof is complete. Otherwise, the set $S_2 = \set {d: d \divides n_1, 1 < d < n_1}$ is nonempty, and bounded below by $1$. By the above argument, the smallest element $p_2$ of $S_2$ is prime. Thus we may write $n_1 = p_2 n_2$, where $1 < n_2 < n_1$. This gives us $n = p_1 p_2 n_2$. If $n_2$ is prime, we are done. Otherwise, we continue this process. Since $n > n_1 > n_2 > \cdots > 1$ is a (strictly) decreasing sequence of positive integers, there must be a finite number of $n_i$'s. That is, we will arrive at some prime number $n_{k - 1}$, which we will call $p_k$. This results in the prime decomposition $n = p_1 p_2 \cdots p_k$. {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 1}$, let $\map P n$ be the proposition: :$n$ can be expressed as a product of prime numbers. First note that if $n$ is prime, the result is immediate. === Basis for the Induction === $\map P 2$ is the case: :$n$ can be expressed as a product of prime numbers. As $2$ itself is a prime number, and the result is immediate. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :For all $j \in \N$ such that $2 \le j \le k$, $j$ can be expressed as a product of prime numbers. from which it is to be shown that: :$k + 1$ can be expressed as a product of prime numbers. === Induction Step === This is the induction step: If $k + 1$ is prime, then the result is immediate. Otherwise, $k + 1$ is composite and can be expressed as: :$k + 1 =  r s$ where $2 \le r < k + 1$ and $2 \le s < k + 1$ That is, $2 \le r \le k$ and $2 \le s \le k$. Thus by the induction hypothesis, both $r$ and $s$ can be expressed as a product of primes. So $k + 1 = r s$ can also be expressed as a product of primes. So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore, for all $n \in \N_{> 1}$: :$n$ can be expressed as a product of prime numbers.
{{AimForCont}} the supposition false. That is, suppose there is at least one positive integer that can be expressed in more than one way as a product of primes. Let the smallest of these be $m$. Thus: :$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$ where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime. By definition, $m$ is not itself prime. Therefore: :$r, s \ge 2$ Let us arrange that the primes which compose $m$ are in order of size: :$p_1 \le p_2 \le \dots \le p_r$ and: :$q_1 \le q_2 \le \dots \le q_s$ Let us arrange that $p_1 \le q_1$. Suppose $p_1 = q_1$. Then: :$\dfrac m {p_1} = p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s = \dfrac m {q_1}$ But then we have the positive integer $\dfrac m {p_1}$ being expressible in two different ways. This contradicts the fact that $m$ is the smallest positive integer that can be so expressed. Therefore: :$p_1 \ne q_1 \implies p_1 < q_1 \implies p_1 < q_2, q_3, \ldots, q_s$ as we arranged them in order. From Prime not Divisor implies Coprime: :$1 < p_1 < q_j: 1 < j < s \implies p_1 \nmid q_j$    But: :$p_1 \divides m \implies p_1 \divides q_1 q_2 \ldots q_s$ where $\divides$ denotes divisibility. Thus from Euclid's Lemma for Prime Divisors: :$\exists j: 1 \le j \le s: p_1 \divides q_j$ But $q_j$ was supposed to be a prime. This is a contradiction. Hence, by Proof by Contradiction, the supposition was false. {{Qed}}
{{AimForCont}} $n$ has two prime factorizations: :$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$ where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.  Since $p_1 \divides q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 = q_j$ for some $1 \le j \le s$. Thus: :$p_1 \ge q_1$ Similarly, since $q_1 \divides p_1 p_2 \dots p_r$, from Euclid's Lemma for Prime Divisors: :$q_1 \ge p_1$ Thus, $p_1 = q_1$, so we may cancel these common factors, which gives: :$p_2 p_3 \cdots p_r = q_2 q_3 \dots q_s$ This process is repeated to show that: :$p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r$ If $r < s$, we arrive at $1 = q_{r + 1} q_{r + 2} \cdots q_s$ after canceling all common factors. But by Divisors of One, the only divisors $1$ are $1$ and $-1$. Hence $q_{r + 1}, q_{r + 2}, \ldots, q_s$ cannot be prime numbers From that contradiction it follows that $r = s$. Thus: :$p_1 = q_1, p_2 = q_2, \ldots, p_r = q_s$ which means the two factorizations are identical. Therefore, the prime factorizations of $n$ is unique. {{Qed}}
The proof proceeds by strong induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :the prime decomposition for $n$ is unique up to order of presentation. Note that it has been established in Integer is Expressible as Product of Primes that $n$ does in fact have at least $1$ prime decomposition. === Basis for the Induction === $\map P 2$ is the case: :$n = 2$ which is trivially unique. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. This is the induction hypothesis: :the prime decomposition for all $j$ such that $0 \le j \le k$ is unique up to order of presentation. from which it is to be shown that: :the prime decomposition for $k + 1$ is unique up to order of presentation. === Induction Step === Either $k + 1$ is prime or it is composite. If $k + 1$ is prime, there is only one way to express it, that is, as the prime $k + 1$ itself. So suppose $k + 1$ is composite. {{AimForCont}} $k + 1$ has the two prime decompositions: $k + 1 = p_1 p_2 \dotsm p_r = q_1 q_2 \dotsm q s$ where: :$p_1 \le p_2 \le \dotsb \le p_r$ and: :$q_1 \le q_2 \le \dotsb \le q_s$ Because $q_1$ is a divisor of $k + 1$: :$q_1 \divides p_1 \le p_2 \le \dotsb \le p_r$ where $\divides$ indicates divisibility. Thus by Euclid's Lemma for Prime Divisors: :$\exists p_i \in \set {p_1, p_2, \ldots, p_r}: q_1 \divides p_i$ But as $q_1$ and $p_i$ are both primes, it follows by definition that $q_1 = p_i$. In a similar way it can be shown that: :$\exists q_j \in \set {q_1, q_2, \ldots, q_s}: p_1 = q_j$ So we have: :$p_1 = q_j \ge q_1$ and: :$q_1 = p_i \ge p_1$ Thus: :$p_1 \ge q_1 \ge p_1$ and so: :$p_1 = q_1$ Thus $\dfrac {k + 1} {p_1}$ is an integer such that: :$\dfrac {k + 1} {p_1} \le k$ and so: :$p_2 p_3 \dotsm p_r \dfrac {k + 1} {p_1} = q_2 q_3 \dotsm q s$ But by the induction hypothesis: :$p_2 = q_2, p_3 = q_3, \dotsc, p_r = q_s$ where furthermore $r = s$. Therefore the prime decomposition for $k + 1$ is unique. So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction: :$\forall n \in \Z_{\ge 2}$, the prime decomposition for $n$ is unique up to order of presentation. {{qed}}
To show that $F$ is a primitive of $f$ on $\closedint a b$, we need to establish the following: :$F$ is continuous on $\closedint a b$ :$F$ is differentiable on the open interval $\openint a b$ :$\forall x \in \closedint a b: \map {F'} x = \map f x$. === Proof that $F$ is Continuous === We have that $f$ is continuous on $\closedint a b$. It follows from Continuous Image of Closed Interval is Closed Interval that $f$ is bounded on $\closedint a b$. Suppose that: :$\forall t \in \closedint a b: \size {\map f t} < \kappa$ Let $x, \xi \in \closedint a b$. From Sum of Integrals on Adjacent Intervals for Continuous Functions, we have that: :$\ds \int_a^x \map f t \rd t + \int_x^\xi \map f t \rd t = \int_a^\xi \map f t \rd t$ That is: :$\ds \map F x + \int_x^\xi \map f t \rd t = \map F \xi$ So: :$\ds \map F x - \map F \xi = -\int_x^\xi \map f t \rd t = \int_\xi^x \map f t \rd t$ From the corollary to Upper and Lower Bounds of Integral: :$\size {\map F x - \map F \xi} < \kappa \size {x - \xi}$ Thus it follows that $F$ is continuous on $\closedint a b$. {{qed|lemma}} === Proof that $F$ is Differentiable and $f$ is its Derivative === It is now to be shown that that $F$ is differentiable on $\openint a b$ and that: :$\forall x \in \closedint a b: \map {F'} x = \map f x$ Let $x, \xi \in \closedint a b$ such that $x \ne \xi$. Then: {{begin-eqn}} {{eqn | l = \frac {\map F x - \map F \xi} {x - \xi} - \map f \xi       | r = \frac 1 {x - \xi} \paren {\map F x - \map F \xi - \paren {x - \xi} \map f \xi}       | c =  }} {{eqn | r = \frac 1 {x - \xi} \paren {\int_\xi^x \map f t \rd t - \paren {x - \xi} \map f \xi}       | c =  }} {{eqn | r = \frac 1 {x - \xi} \int_\xi^x \paren {\map f t - \map f \xi} \rd t       | c = Definite Integral of Function plus Constant, putting $c = \map f \xi$ }} {{end-eqn}} Now, let $\epsilon > 0$. If $\xi \in \openint a b$, then $f$ is continuous at $\xi$. So for some $\delta > 0$: :$\size {\map f t - \map f \xi} < \epsilon$ provided $\size {t - \xi} < \delta$. So provided $\size {x - \xi} < \delta$ it follows that: :$\size {\map f t - \map f \xi} < \epsilon$ for any $t$ in an interval whose endpoints are $x$ and $\xi$. So from the corollary to Upper and Lower Bounds of Integral, we have: {{begin-eqn}} {{eqn | l = \size {\frac {\map F x - \map F \xi} {x - \xi} - \map f \xi}       | r = \frac 1 {\size {x - \xi} } \size {\int_\xi^x \paren {\map f t - \map f \xi} \rd t}       | c =  }} {{eqn | o = <       | r = \frac 1 {\size {x - \xi} } \epsilon \size {x - \xi}       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} provided $0 < \size {x - \xi} < \delta$. But that is what this means: :$\dfrac {\map F x - \map F \xi} {x - \xi} \to \map f \xi$ as $x \to \xi$ So $F$ is differentiable on $\openint a b$, and: :$\forall x \in \closedint a b: \map {F'} x = \map f x$ {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \map F x       | r = \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_a^{x + \Delta x} \map f t \rd t - \int_a^x \map f t \rd t}       | c = {{Defof|Derivative of Real Function at Point}} }} {{eqn | r = \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^a \map f t \rd t + \int_a^{x + \Delta x} \map f t \rd t}       | c = because $\displaystyle \int_a^b \map f x \rd x = - \int_b^a \map f x \rd x$ }} {{eqn | r = \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}       | c = Sum of Integrals on Adjacent Intervals for Continuous Functions }} {{end-eqn}} Suppose $\Delta x > 0$ and both $x$ and $x + \Delta x$ are in the closed interval $\closedint a b$. By hypothesis, $f$ is continuous on the closed interval $\closedint a b$ Thus it is also continuous on the closed interval $\closedint x {x + \Delta x}$. Thus the conditions of the Mean Value Theorem for Integrals are fulfilled. So, by the Mean Value Theorem for Integrals, there exists some $k \in \closedint x {x + \Delta x}$ such that: :$\displaystyle \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$ Then: {{begin-eqn}} {{eqn | l = \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}       | r = \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \map f k \Delta x }} {{eqn | r = \lim_{\Delta x \mathop \to 0} \map f k }} {{end-eqn}} By the definition of $k$: :$x \le k \le x + \Delta x$ which means that $k \to x$ as $\Delta x \to 0$. So: {{begin-eqn}} {{eqn | l = \lim_{\Delta x \mathop \to 0} \map f k       | r = \lim_{k \mathop \to x} \map f k }} {{eqn | r = \map f x       | c = because $f$ is continuous }} {{end-eqn}} For $\Delta x < 0$, consider $k \in \closedint {x + \Delta x} x$, and the argument is similar. Hence the result, by the definition of primitive. {{qed}}
Let $G$ be defined on $\closedint a b$ by: :$\displaystyle \map G x = \int_a^x \map f t \rd t$ We have: :$\displaystyle \map G a = \int_a^a \map f t \rd t = 0$ from Integral on Zero Interval :$\displaystyle \map G b = \int_a^b \map f t \rd t$ from the definition of $G$ above. Therefore, we can compute: {{begin-eqn}} {{eqn | l = \int_a^b \map f t \rd t       | r = \int_a^a \map f t \rd t + \int_a^b \map f t \rd t       | c = Sum of Integrals on Adjacent Intervals for Continuous Functions }} {{eqn | r = \map G a + \map G b       | c = From the definition of $G$ }} {{eqn | r = \map G b - \map G a       | c = $\map G a = 0$ }} {{end-eqn}} By the first part of the theorem, $G$ is a primitive of $f$ on $\closedint a b$. By Primitives which Differ by Constant, we have that any primitive $F$ of $f$ on $\closedint a b$ satisfies $\map F x = \map G x + c$, where $c$ is a constant. Thus we conclude: {{begin-eqn}} {{eqn | l = \int_a^b \map f t \rd t       | r = \paren {\map G b + c} - \paren {\map G a + c} }} {{eqn | r = \map F b - \map F a }} {{end-eqn}} {{qed}}
From the definition of addition modulo $m$, we have: :$\eqclass x m +_m \eqclass y m = \eqclass {x + y} m$ By the Division Theorem: :$x + y = q m + r$ where $0 \le r < m$ Therefore for all $0 \le r < m$: :$\eqclass {x + y} m = \eqclass r m$. Therefore from the definition of integers modulo $m$: :$\eqclass {x + y} m \in \Z_m$ {{Qed}}
From the definition of addition modulo $z$, we have: :$\eqclass x z +_z \eqclass y z = \eqclass {x + y} z$ As $x, y \in R$, we have that $x + y \in \R$ as Real Addition is Closed. Hence by definition of congruence, $\eqclass {x + y} z \in \R_z$. {{qed}}
First note that if $\phi$ is the trivial homomorphism then $\phi \sqbrk K = 0_{F_2}$ and so is not a field. Since $K$ is a field, we have that: :$0_{F_1} \in K$ :$1_{F_1} \in K$ and so by Ring Homomorphism Preserves Zero and Field Homomorphism Preserves Unity: :$\map \phi {0_{F_1} } = 0_{F_2} \in \phi \sqbrk K$ :$\map \phi {1_{F_1} } = 1_{F_2} \in \phi \sqbrk K$ So $\phi \sqbrk K$ contains at least the zero and unity. From Group Homomorphism Preserves Subgroups: :$\struct {\phi \sqbrk K, +_2}$ is a subgroup of $\struct {F_2, +_2}$ :$\struct {\phi \sqbrk K, \circ_2}$ is a subgroup of $\struct {F_2, \circ_2}$. Thus, as $\struct {F_2, +_2}$ and $\struct {F_2, \circ_2}$ are both groups, the result follows. {{qed}} Category:Field Homomorphisms Category:Subfields o840aywktlkuylc1hi0khw54ojop6ai
=== Intersection Subset of Union === Let $\displaystyle x \in S \cap \bigcup \mathbb T$. We need to show that $\displaystyle x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of subset we will have shown that $\displaystyle S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$. So, we have that $\displaystyle x \in S \cap \bigcup \mathbb T$. By definition of set intersection, $x \in S$ and $\displaystyle x \in \bigcup \mathbb T$. From $\displaystyle x \in \bigcup \mathbb T$ we know that: :$\exists X \in \mathbb T: x \in X$ and so: :$\displaystyle \exists X \in \mathbb T: x \in S \cap X$ So by definition of set union: :$\displaystyle x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ So: :$\displaystyle S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ {{qed|lemma}} === Union Subset of Intersection === Let $\displaystyle x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$. We need to show that $\displaystyle x \in S \cap \bigcup \mathbb T$ and then by definition of subset we will have shown that $\displaystyle \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$. So, we have that $\displaystyle x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$. By definition of set union: :$\exists X \in \mathbb T: x \in S \cap X$ By definition of set intersection, we have that $x \in S$ and $x \in X$. By definition of set union: :$\displaystyle x \in \bigcup \mathbb T$ So by definition of set intersection, we have that: :$\displaystyle x \in S \cap \bigcup \mathbb T$ So: :$\displaystyle \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$ {{qed|lemma}} So we have that: :$\displaystyle S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and: :$\displaystyle \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$ and so by definition of set equality: :$\displaystyle S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ {{qed}}
Let $\Img f$ denote the image of $f$. Let $\Img {\lambda f}$ denote the image of $\lambda f$. Consider the surjection $l_\lambda: \Img f \to \Img {\lambda f}$ defined by: :$\map {l_\lambda} {\map f x} := \lambda \map f x$ By Measurable Function is Simple Function iff Finite Image Set, $\card {\Img f}$ is finite. Hence Cardinality of Surjection yields that $\size {\Img {\lambda f} }$ is finite as well. The result follows from a second application of Measurable Function is Simple Function iff Finite Image Set. {{qed}} Category:Simple Functions rcbby2ajni37qx8w446glba8whhxttf
By definition of image, we have: :$\operatorname{Im} \left({f \circ g}\right) = \left\{{t \in T: \exists r \in R: f \circ g \left({r}\right) = t}\right\}$ and by definition of the image of a subset: :$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in \operatorname{Im} \left({g}\right): f \left({s}\right) = t}\right\}$ which, expanding what it means that $s \in \operatorname{Im} \left({g}\right)$, equals: :$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in S: \exists r \in R: g \left({r}\right) = s \land f \left({s}\right) = t}\right\}$ Now substituting $g \left({r}\right) = s$ in $f \left({s}\right) = t$, we obtain: :$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in S: \exists r \in R: f \left({g \left({r}\right)}\right) = t}\right\}$ which is seen to equal the expression for $\operatorname{Im} \left({f \circ g}\right)$ as soon as $S$ is non-empty. The remaining case to be checked is thus if $S = \varnothing$. From Null Relation is Mapping iff Domain is Empty Set, also $R = \varnothing$, so that $f$ and $g$ are empty mappings. By Image of Empty Set is Empty Set, we conclude: :$\operatorname{Im} \left({f \circ g}\right) = \operatorname{Im} \left({g}\right) = \varnothing$ and also: :$f \left({\operatorname{Im} \left({g}\right)}\right) = \varnothing$ which together yield the desired equality. Hence the result. {{qed}}
The integers $\struct {\Z, +, \times}$ form an integral domain. By definition of integral domain, therefore, the integers form a ring with no (proper) zero divisors. The result follows from Idempotent Elements of Ring with No Proper Zero Divisors. {{qed}}
Let us cast the proposition in the form: :$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$ For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = a \times \paren {b + 1}       | r = \paren {a \times b} + a       | c = Axiom $\text B$ }} {{eqn | r = \paren {a \times b} + \paren {a \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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$ Then we need to show: :$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = a \times \paren {b + \paren {k + 1} }       | r = a \times \paren {\paren {b + k} + 1}       | c = Axiom $\text C$ }} {{eqn | r = \paren {a \times \paren {b + k} } + a       | c = Axiom $\text B$ }} {{eqn | r = \paren {\paren {a \times b} + \paren {a \times k} } + a       | c = Induction hypothesis }} {{eqn | r = \paren {a \times b} + \paren {\paren {a \times k} + a}       | c = Natural Number Addition is Associative }} {{eqn | r = \paren {a \times b} + \paren {a \times \paren {k + 1} }       | c = Axiom $\text B$ }} {{end-eqn}} The result follows by the Principle of Mathematical Induction.