nyuuzyou/wikis huggingface dataset:
Exponential of Sum/Real Numbers/Proof 6

== Theorem ==
== Proof ==
Fix $a \in \R$ and define the function $f_a : \R \to \R$ by: 
:$\map {f_a} x = \map \exp {a - x} \exp x$
for all $x \in \R$.
We aim to establish that: 
:$\map {f_a} x = \map \exp {a - x} \exp x = \exp a$
for all $a, x \in \R$. 
Then, we can fix $x, y \in \R$ and set $a = x + y$ to obtain: 
:$\map {f_a} x = \exp y \exp x = \map \exp {x + y}$
which is the claim.
Note that $f_a$ is differentiable and we have: 
From Zero Derivative implies Constant Function, $f_a$ is constant.
That is, there exists $C \in \R$ such that: 
:$\map {f_a} x = C$
for all $x \in \R$.
We have: 
So: 
:$\map {f_a} x = \exp a$
for all $a, x \in \R$. 
That is: 
:$\map \exp {a - x} \map \exp x = \exp a$
for all $a, x \in \R$. 
Hence the result.
Category:Exponential of Sum
Exponential of Sum of Real Numbers/Corollary

== Corollary to Exponential of Sum of Real Numbers ==
Let $x, y \in \R$ be real numbers.
Let $\exp x$ be the exponential of $x$.
Then:
:$\map \exp {x - y} = \dfrac {\exp x} {\exp y}$
== Proof ==
By Exponential of Sum of Real Numbers:
:$\map \exp {x - y} = \exp x \, \map \exp {-y}$
By Reciprocal of Complex Exponential:
:$\dfrac 1 {\exp y} = \map \exp {-y}$
Combining these two, we obtain the result:
:$\map \exp {x - y} = \dfrac {\exp x} {\exp y}$
Category:Exponential of Sum
Exponential of Zero

== Theorem ==
Let $\exp x$ be the exponential of $x$.
Then:
: $\exp 0 = 1$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Sources ==
* : $\S 14.4$
Category:Exponential Function
Category:Exponential of Zero
Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x are Linearly Independent

== Theorem ==
Let $x, a, b \in \R$.
Then $e^{a x} \map \cos {b x}$ and $e^{b x} \map \sin {b x}$ are linearly independent.
== Proof ==
Let $\alpha_1, \alpha_2 \in \R$ be such that:
:$\alpha_1 e^{a x} \map \cos {b x} + \alpha_2 e^{a x} \map \sin {b x} = 0$
Suppose $x = 0$.
Then $\alpha_1 = 0$.
Suppose $\ds x = \frac \pi {2 b}$.
Then:
:$\ds \alpha_2 \map \exp {\frac {\pi a} {2 b}} = 0$
Hence, $\alpha_2 = 0$.
The conclusion follows from the definition of linearly independent real functions.
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Derivatives involving Exponential Function
Category:Derivatives involving Sine Function
Category:Derivatives involving Cosine Function
Category:Linear Independence
Exponential of x not less than 1+x

== Theorem ==
:$e^x \ge 1 + x$
for all $x \in \R$.
== Proof ==
For $x > - 1$:
as required.
For $x \le -1$, the inequality follows from the fact that $e^x$ is positive for all $x$.
Category:Exponential Function
Category:Inequalities
Exponential on Complex Plane is Group Homomorphism

== Theorem ==
Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ be the mapping:
:$x \mapsto \map \exp x$
where $\exp$ is the complex exponential function.
Then $\exp$ is a group homomorphism.
== Proof ==
If $z \in \C$, then by the definition of the complex exponential function, $\exp$ is a mapping $\C \to \C_{\ne 0}$.
Let $z_1, z_2 \in \C$.
By Exponential of Sum:
:$\map \exp {z_1 + z_2} = \map \exp {z_1} \, \map \exp {z_2}$
Therefore $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ is a group homomorphism.
== Sources ==
* : $\S 47$. Homomorphisms and their elementary properties: Illustration
Category:Examples of Group Homomorphisms
Category:Exponential Function
Exponential on Real Numbers is Group Isomorphism

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\R_{> 0}, \times}$ be the multiplicative group of positive real numbers.
Let $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:
:$x \mapsto \map \exp x$
where $\exp$ is the exponential function.
Then $\exp$ is a group isomorphism.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 7.3$. Isomorphism: Example $135$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.5$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \alpha$
* : $\S 3.2$: Groups; the axioms: Exercise $(8)$
Category:Examples of Group Isomorphisms
Category:Exponential Function
Category:Exponential on Real Numbers is Group Isomorphism
Exponential on Real Numbers is Group Isomorphism/Proof 1

== Theorem ==
== Proof ==
From Exponential of Sum we have:
:$\forall x, y \in \R: \map \exp {x + y} = \exp x \cdot \exp y$
That is, $\exp$ is a group homomorphism.
Then we have that Exponential is Strictly Increasing.
From Strictly Monotone Real Function is Bijective, it follows that $\exp$ is a bijection.
So $\exp$ is a bijective group homomorphism, and so a group isomorphism.
== Sources ==
* : $\S 7.1$. Homomorphisms: Example $128$
* : $\S 1.5$: Example $12$
Category:Exponential on Real Numbers is Group Isomorphism
Exponential on Real Numbers is Group Isomorphism/Proof 2

== Theorem ==
== Proof ==
From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {\R_{> 0}, \times}$ is a group.
We have that for all $y \in R_{> 0}$ there exists $x = \map \ln y \in R$ which satisfies $\map \exp x = y$.
Thus $\exp$ is a surjection.
Then from Exponential on Real Numbers is Injection:
:$\exp$ is an injection.
Therefore, $\exp$ is a bijection.
Let $x, y \in R$.
From Exponential of Sum:
:$\map \exp {x + y} = \map \exp x \, \map \exp y$
So $\exp$ is a homomorphism and a bijection.
It follows by definition that $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ is an isomorphism.
Category:Exponential on Real Numbers is Group Isomorphism
Exponential on Real Numbers is Injection

== Theorem ==
Let $\exp: \R \to \R$ be the exponential function:
:$\map \exp x = e^x$
Then $\exp$ is an injection.
== Proof ==
<onlyinclude>
From Exponential is Strictly Increasing:
:$\exp$ is strictly increasing on $\R$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
:$\exp$ is an injection.
== Sources ==
* : $\S 3.3$. Injective, surjective, bijective; inverse mappings: Example $48$
* : Chapter $2$: Maps and relations on sets: Example $2.5$
Category:Exponential Function
Category:Injections
Exponentiation Functor is Functor

== Theorem ==
Let $\mathbf C$ be a Cartesian closed metacategory.
Let $A$ be an object of $\mathbf C$.
Let $\left({-}\right)^A: \mathbf C \to \mathbf C$ be the exponentiation functor.
Then $\left({-}\right)^A$ is a functor.
== Proof ==
Let $B$ be an object of $\mathbf C$.
Let $\epsilon_B: B^A \times A \to B$ be the evaluation morphism at $B$.
Then, since:
it follows that $\left({\operatorname{id}_B}\right)^A = \operatorname{id}_{B^A}$.
Next, let $f: B \to C, g: C \to D$ be composable morphisms.
Then:
Thus, $g^A f^A$ is the morphism satisfying the UMP for $\left({g f}\right)^A$.
That is to say:
:$\left({g f}\right)^A = g^A f^A$
Hence, $\left({-}\right)^A$ is a functor.
== Sources ==
* : $\S 6.2$: Proposition $6.7$
Category:Category Theory
Exponentiation is Primitive Recursive

== Theorem ==
The function $\exp: \N^2 \to \N$, defined as:
:$\map \exp {n, m} = n^m$
is primitive recursive‎.
== Proof ==
We observe that:
:$\map \exp {n, 0} = n^0 = 1$
and that:
:$\map \exp {n, m + 1} = n^\paren {m + 1} = \paren {n^m} \times n = \map {\mathrm {mult} } {\map \exp {n, m}, n}$.
Thus $\exp$ is defined by primitive recursion from the primitive recursive function ‎$\mathrm {mult}$.
Hence the result.
Category:Primitive Recursive Functions
Exponents of Primes in Prime Decomposition are Less iff Divisor

== Theorem ==
Let $a, b \in \Z_{>0}$.
Then $a \divides b$ :
:$(1): \quad$ every prime $p_i$ in the prime decomposition of $a$ appears in the prime decomposition of $b$
and:
:$(2): \quad$ the exponent of each $p_i$ in $a$ is less than or equal to its exponent in $b$.
== Proof ==
Let $a, b \in \Z_{>0}$.
Let their prime decomposition be:
:$a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$
:$b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$
=== Necessary Condition ===
Let:
:$(1): \quad$ prime in the prime decomposition of $a$ appear in the prime decomposition of $b$
and:
:$(2): \quad$ its exponent in $a$ be less than or equal to its exponent in $b$.
Then:
where:
:$k_1 \le l_1, k_2 \le l_2, \dotsc, k_r \le l_r, r \le s$
Thus:
:$d = p_1^{l_1 - k_1} p_2^{l_2 - k_2} \dotsm p_r^{l_r - k_r} \in \Z$
and so:
:$b = a d$
So $a \divides b$.
=== Sufficient Condition ===
Let $a \divides b$.
Let $a = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ be the prime decomposition of $a$.
Then:
:$\forall i \in \N_r: p_i^{k_i} \divides a$
Hence by Divisor Relation on Positive Integers is Partial Ordering each $p_i^{k_i}$ also divides $b$.
Thus:
:$\exists c \in \Z: b = p_i^{k_i} c$
The prime decomposition of $b$ is therefore:
:$b = p_i^{k_i} \times \paren {\text {prime decomposition of $c$} }$
which may need to be rearranged.
So $p_i$ must occur in the prime decomposition of $b$ with an exponent at least as big as $k_i$.
The result follows.
Category:Prime Numbers
Exportation and Self-Conditional

== Theorem ==
:$p \land q \implies r \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
== Proof ==
From the Rule of Exportation:
:$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$
Then by Self-Distributive Law for Conditional:
:$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
Category:Conditional
Category:Conjunction
Expression for Closure of Set in Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$. 
Let $\mathcal V$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Then: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$
where $A^-$ is the closure of $A$.
=== Corollary ===
== Proof ==
From Condition for Point being in Closure: Topological Vector Space, we have $x \in A^-$ :
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. 
We have that $\paren {x + V} \cap A \ne \O$ there exists some $u \in X$ with $u \in A$ and $u \in x + V$.
This is equivalent to, $u \in A$ and there exists $v \in V$ such that $u = x + v$. 
This is equivalent to, $x = u - v$ for $u \in A$ and $v \in V$.
That is, equivalent to $x \in A - V$.
So, we have that $x \in A^-$ $x \in A - V$ for all open neighborhoods $V$ of ${\mathbf 0}_X$.
So, we have: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A - V}$
From Dilation of Open Set in Topological Vector Space is Open, we have that $-V$ is open $V$ is open.
Writing $A - V = A + \paren {-V}$, we therefore see that: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$
== Sources ==
* : $1.13$: Theorem
Category:Topological Vector Spaces
Category:Set Closures
Category:Expression for Closure of Set in Topological Vector Space
Expression for Closure of Set in Topological Vector Space/Corollary

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$. 
Let $A \subseteq X$. 
Let $\BB$ be a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$. 
Then:
:$\ds A^- = \bigcap_{V \in \BB} \paren {A + V}$
where $A^-$ is the closure of $A$.
== Proof ==
Let $\VV$ be the set of open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$.
From Intersection is Decreasing, we have: 
:$\ds \bigcap_{V \in \mathcal V} \paren {A + V} \subseteq \bigcap_{V \in \BB} \paren {A + V}$
Conversely, suppose that:
:$\ds x \in \bigcap_{V \in \BB} \paren {A + V}$
Let $O \in \VV$.
Since $\BB$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, there exists $U \in \BB$ such that $U \subseteq O$.
Since $x \in A + V$ for each $V \in \BB$, we in particular have $x \in A + U$. 
Since $U \subseteq O$, we have that $x \in A + O$.
Since $O \in \VV$ was arbitrary, we have:
:$\ds x \in \bigcap_{V \in \BB} \paren {A + V}$
Category:Expression for Closure of Set in Topological Vector Space
Expression for Integer as Product of Primes is Unique

== Theorem ==
Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Integer is Expressible as Product of Primes
== Sources ==
* : Book $\text{IX}$. Propositions
* : Chapter $\text {B}.16$: The Sequence of Primes: Problem $1 \ \text{(b)}$
* : Chapter $7$: Patterns in Numbers: Euclid
Category:Expression for Integer as Product of Primes is Unique
Category:Fundamental Theorem of Arithmetic
Category:Prime Decompositions
Category:Prime Numbers
Category:Factorization
Expression for Integer as Product of Primes is Unique/Proof 1

== Theorem ==
== Proof ==
 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.
== Sources ==
* : $\S 13.2$: The fundamental theorem of arithmetic
* : $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic
Category:Expression for Integer as Product of Primes is Unique
Expression for Integer as Product of Primes is Unique/Proof 2

== Theorem ==
== Proof ==
 $n$ 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.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Theorem $4$
Category:Expression for Integer as Product of Primes is Unique
Expression for Integer as Product of Primes is Unique/Proof 3

== Theorem ==
== Proof ==
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.
 $k + 1$ 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.
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Theorem $\text {2-5}$
Category:Expression for Integer as Product of Primes is Unique
Expression for Integers as Powers of Same Primes

== Theorem ==
Let $a, b \in \Z$ be integers.
Let their prime decompositions be given by:
Then there exist prime numbers:
:$t_1 < t_2 < \dotsb < t_v$
such that:
=== General Result ===
== Proof ==
In the prime decompositions $(1)$ and $(2)$, we have that:
:$q_1 < q_2 < \dotsb < q_r$
and:
:$s_1 < s_2 < \dotsb < s_u$
Hence we can define:
as all the $q_1, q_2, \dotsc, q_r$ are distinct, and all the $s_1, s_2, \dotsc, s_u$ are distinct.
Then let:
:$T = E \cup F$
and let the elements of $T$ be renamed as:
:$T = \set {t_1, t_2, \ldots, t_v}$
where all the $t_1, t_2, \dotsc, t_v$ are distinct, and:
:$t_1 < t_2 < \dotsb < t_v$
Let $\iota: E \to T$ be the inclusion mapping:
:$\forall q_i \in E: \map \iota {q_i} = q_i$
Let $\iota: F \to T$ be the inclusion mapping:
:$\forall s_i \in F: \map \iota {s_i} = s_i$
Then we have that:
and:
Thus $a$ and $b$ can be expressed as the product of powers of the same primes, on the understanding that one or more of the powers in either product may be zero.
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $4$
Category:Prime Decompositions
Expression for Integers as Powers of Same Primes/General Result

== Theorem ==
Let $a_1, a_2, \dotsc, a_n \in \Z$ be integers.
Let their prime decompositions be given by:
:$\ds a_i = \prod_{\substack {p_{i j} \mathop \divides a_i \\ \text {$p_{i j}$ is prime} } } {p_{i j} }^{e_{i j} }$
Then there exists a set $T$ of prime numbers:
:$T = \set {t_1, t_2, \dotsc, t_v}$
such that:
:$t_1 < t_2 < \dotsb < t_v$
:$\ds a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
::$\ds a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
=== Basis for the Induction ===
$\map P 2$ is the case:
there exist prime numbers $t_1 < t_2 < \dotsb < t_v$ such that:
This has been proved in Expression for Integers as Powers of Same Primes.
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:for all $a_i \in \set {a_1, a_2, \ldots, a_k}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
::$\ds a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
from which it is to be shown that:
:for all $a_i \in \set {a_1, a_2, \ldots, a_{k + 1} }$: there exists a set $T' = \set {t_1, t_2, \dotsc, t_w}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_w$ such that:
::$\ds a_i = \prod_{j \mathop = 1}^w {t_j}^{g_{i j} }$
=== Induction Step ===
This is the induction step:
Let $E = \set {q_i: q_i \divides a_{k + 1} , \text {$q_i$ is prime} }$.
Then let:
:$T' = E \cup T$
and let the elements of $T$ be renamed as:
:$T' = \set {t_1, t_2, \ldots, t_w}$
where all the $t_1, t_2, \dotsc, t_w$ are distinct, and:
:$t_1 < t_2 < \dotsb < t_w$
Then we have that:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore, for all $n \in \Z_{\ge 2}$:
:for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
::$\ds a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $10$
Category:Prime Decompositions