nyuuzyou/wikis huggingface dataset:
Rational Numbers form Metric Subspace of Real Numbers under Euclidean Metric

== Theorem ==
Let $\struct {\Q, d_\Q}$ be the set of rational numbers under the function $d_\Q: \Q \times \Q \to \R$ defined as:
:$\forall x, y \in \Q: \map {d_\Q} {x, y} = \size {x - y}$
Let $\struct {\R, d}$ denote the real number line with the usual (Euclidean) metric.
Then $\struct {\Q, d_\Q}$ is a metric subspace of $\struct {\R, d}$.
== Proof ==
From Rational Numbers form Subfield of Real Numbers:
:$\Q \subseteq \R$
By definition of the Euclidean metric on $\R$:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
and so $d_\Q$ is a restriction of $d$:
:$d_\Q = d {\restriction}_{\Q \times \Q}$
From Euclidean Metric on Real Number Line is Metric, $d$ is a metric on $\R$.
The result follows by definition of metric subspace.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Example $1$
Category:Rational Number Space
Category:Real Number Line with Euclidean Metric
Rational Numbers form Ordered Integral Domain

== Theorem ==
The rational numbers $\Q$ form an ordered integral domain under addition and multiplication.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $9$
Category:Rational Numbers
Category:Examples of Ordered Integral Domains
Category:Rational Numbers form Ordered Integral Domain
Rational Numbers form Ordered Integral Domain/Proof 2

== Theorem ==
== Proof ==
We have that the rational numbers $\struct {\Q, +, \times}$ form an integral domain.
It remains to specify a property $P$ on $\Q$ such that:
:$(1): \quad \forall a, b \in \Q: \map P a \land \map P b \implies \map P {a + b}$
:$(2): \quad \forall a, b \in \Q: \map P a \land \map P b \implies \map P {a \times b}$
:$(3): \quad \forall a \in \Q: \map P a \lor \map P {-a} \lor a = 0$
We have that the integers $\struct {\Z, +, \times}$ form an ordered integral domain.
Let $P'$ be the (strict) positivity property on $\struct {\Z, +, \times}$.
Let us define the property $P$ on $\Q$ as:
:$\forall a \in \Q: \map P a \iff a = \dfrac p q: \map {P'} p, \map {P'} q$
That is, an element $a = \dfrac p q$ has $P$ both $p$ and $q$ have the (strict) positivity property in $\Z$.
Now let $a = \dfrac p q$ and $b = \dfrac r s$ such that $\map P a$ and $\map P b$.
Then by definition of rational addition:
:$\dfrac p q + \dfrac r s = \dfrac {p s + r q} {q s}$
and rational multiplication:
:$\dfrac p q \times \dfrac r s = \dfrac {p r} {q s}$
It can be seen from the definition of (strict) positivity $P'$ on $\Z$ that $\map P {a + b}$ and $\map P {a \times b}$.
It can be seen that if $\map P a$ then $\neg \map P {-a}$ and vice versa.
Also we note that $\neg \map P 0$ and of course $\neg \map P {-0}$.
So the property $P$ we defined fulfils the criteria for the (strict) positivity property.
Hence the result.
Category:Rational Numbers form Ordered Integral Domain
Rational Numbers form Prime Field

== Theorem ==
The field of rational numbers $\struct {\Q, +, \times}$ is a prime field.
That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\set 0$, and vacuously $\O$.
== Proof ==
Let $F$ be a subfield of $\struct {\Q, +, \times}$.
Then $F$ is by definition a field.
Thus by definition:
:$\struct {F, +}$ is an abelian group
and:
:$\struct {F^*, \times}$ is an abelian group, where $F^* = F \setminus \set 0$.
and $\times$ is distributive over $+$:
:$\forall a, b, c \in F: a \times \paren {b + c} = a \times b + a \times c$
Let $a \in F^*$.
Then:
We have $1 \in F$.
Suppose $n \in \Z$ such that $n \in F$.
Then:
:$n + 1 \in F$
So by the Principle of Mathematical Induction, all positive integers are in $F$.
By in $\struct {F, +}$:
:$m \in F \implies -m \in F$
So all negative integers are also in $F$.
That is:
:$\forall m \in \Z: m \in \F$
Recall in $\struct {F^*, \times}$:
:$m \in F^* \implies m^{-1} \in F^*$
So:
:$\forall p \in \Z, q \in \Z_{\ne 0}: p \times q^{-1} \in F$
and it follows by the Subfield Test that $F = \Q$.
== Also see ==
* Number Field has Rational Numbers as Subfield, a direct consequence
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties: Example $1$
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $3$
Category:Rational Numbers
Category:Prime Fields
Rational Numbers form Ring

== Theorem ==
The set of rational numbers $\Q$ forms a ring under addition and multiplication: $\struct {\Q, +, \times}$.
== Proof ==
Recall that $\struct {\Q, +, \times}$ is a field.
As a field is also by definition a division ring, which is an example of a ring, the result follows.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $4$
Category:Examples of Rings
Category:Rational Numbers
Rational Numbers form Subfield of Complex Numbers

== Theorem ==
Let $\struct {\Q, +, \times}$ denote the field of rational numbers.
Let $\struct {\C, +, \times}$ denote the field of complex numbers.
$\struct {\Q, +, \times}$ is a subfield of $\struct {\C, +, \times}$.
== Proof ==
From Rational Numbers form Subfield of Real Numbers, $\struct {\Q, +, \times}$ is a subfield of $\struct {\R, + \times}$.
From Real Numbers form Subfield of Complex Numbers, $\struct {\R, +, \times}$ is a subfield of $\struct {\C, + \times}$.
Thus from Subfield of Subfield is Subfield $\struct {\Q, +, \times}$ is a subfield of $\struct {\C, + \times}$.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Example $21$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
Category:Rational Numbers
Category:Complex Numbers
Category:Examples of Subfields
Rational Numbers form Subfield of Real Numbers

== Theorem ==
The (ordered) field $\struct {\Q, +, \times, \le}$ of rational numbers forms a subfield of the field of real numbers $\struct {\R, +, \times, \le}$.
That is, the field of real numbers $\struct {\R, +, \times, \le}$ is an extension of the rational numbers $\struct {\Q, +, \times, \le}$.
== Proof ==
Recall that Rational Numbers form Totally Ordered Field.
Then from Rational Numbers form Subset of Real Numbers:
:$\Q \subseteq \R$
Hence the result by definition of subfield.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Example $21$
* : $\S 2.1$: Subrings: Examples $1$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
* 
* 
Category:Rational Numbers
Category:Real Numbers
Category:Examples of Subfields
Rational Numbers form Subset of Real Numbers

== Theorem ==
The set $\Q$ of rational numbers forms a subset of the real numbers $\R$.
== Proof ==
Let $x \in \Q$, where $\Q$ denotes the set of rational numbers.
Consider the rational sequence:
:$x, x, x, \ldots$
This sequence is trivially Cauchy.
Thus there exists a Cauchy sequence $\eqclass {\sequence {x_n} } {}$ which is identified with a rational number $x \in \Q$ such that:
So by the definition of a real number:
:$x \in \R$
where $\R$ denotes the set of real numbers.
Thus, by definition of subset:
:$\Q \subseteq \R$
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.31$. Definition
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
Category:Rational Numbers
Category:Real Numbers
Rational Numbers form Totally Ordered Field

== Theorem ==
The set of rational numbers $\Q$ forms a totally ordered field under addition and multiplication: $\struct {\Q, +, \times, \le}$.
== Proof ==
Recall that by Integers form Ordered Integral Domain, $\struct {\Z, +, \times, \le}$ is an ordered integral domain
By Rational Numbers form Field, $\struct {\Q, +, \times}$ is a field.
In the formal definition of rational numbers, $\struct {\Q, +, \times}$ is the field of quotients of $\struct {\Z, +, \times, \le}$
By Total Ordering on Field of Quotients is Unique, it follows that $\struct {\Q, +, \times}$ has a unique total ordering on it that is compatible with its ring structure.
Thus $\struct {\Q, +, \times, \le}$ is a totally ordered field.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Examples of Fields
Category:Rational Numbers
Rational Numbers form Vector Space

== Theorem ==
Let $\Q$ be the set of rational numbers.
Then the $\Q$-module $\Q^n$ is a vector space.
It follows directly, by setting $n = 1$, that the $\Q$-module $\Q$ itself can also be regarded as a vector space.
== Proof ==
From the definition, a vector space is a unitary module whose scalar ring is a field.
From Rational Numbers form Field, we have that $\Q$ is a field.
So the $\Q$-module $\Q^n$ fits the description.
Category:Linear Algebra
Category:Rational Numbers
Rational Numbers in Real Euclidean Plus Space are Open Set

== Theorem ==
Let $\R$ be the set of real numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Let $\Q$ be the set of rational numbers.
Then $\Q$ is an open set of $\struct {\R, d}$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $5$
Category:Euclidean Plus Metric
Category:Open Balls
Rational Numbers under Addition form Infinite Abelian Group

== Theorem ==
Let $\Q$ be the set of rational numbers.
The structure $\struct {\Q, +}$ is a countably infinite abelian group.
== Proof ==
The rational numbers are, by definition, the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence by definition, $\struct {\Q, +, \times}$ is a field.
The fact that $\struct {\Q, +}$ forms an abelian group follows directly from the definition of a field.
From Rational Numbers are Countably Infinite, we have that $\struct {\Q, +}$ is countably infinite.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : $\S 1$: Some examples of groups: Example $1.02$
* : Chapter $2$: The Definition of Group Structure: $\S 29 \alpha \ (1)$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(i)}$
* : Chapter $1$: Definitions and Examples: Example $1.4$
Category:Rational Addition
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Rational Numbers under Addition form Monoid

== Theorem ==
The set of rational numbers under addition $\struct {\Q, +}$ forms a monoid.
== Proof ==
Follows directly from Rational Numbers under Addition form Infinite Abelian Group.
By definition, an abelian group is a group.
Also by definition, a group is a monoid.
Hence the result.
== Sources ==
* : $\S 3.1$: Monoids
Category:Rational Addition
Category:Examples of Monoids
Rational Numbers under Multiplication do not form Group

== Theorem ==
The algebraic structure $\struct {\Q, \times}$ consisting of the set of rational numbers $\Q$ under multiplication $\times$ is not a group.
== Proof ==
 that $\struct {\Q, \times}$ is a group.
By the definition of the number $0 \in \Q$:
:$\forall x \in \Q: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\Q, \times}$ is the trivial group.
But $\Q$ contains other elements besides $0$.
From this contradiction it follows that $\struct {\Q, \times}$ is not a group.
== Also see ==
* Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
Category:Rational Multiplication
Category:Examples of Groups
Rational Numbers under Multiplication form Commutative Monoid

== Theorem ==
The set of rational numbers under multiplication $\struct {\Q, \times}$ forms a countably infinite commutative monoid.
== Proof ==
From Rational Numbers under Multiplication form Monoid, $\struct {\Q, \times}$ is a monoid.
Then:
:from Rational Multiplication is Commutative we have that $\times$ is commutative on $\Q$
:from Rational Numbers are Countably Infinite we have that $\Q$ is a countably infinite set.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Rational Multiplication
Category:Examples of Commutative Monoids
Rational Numbers under Multiplication form Monoid

== Theorem ==
The set of rational numbers under multiplication $\struct {\Q, \times}$ forms a monoid.
== Proof ==
Taking the monoid axioms in turn:
=== ===
Rational Multiplication is Closed.
=== ===
Rational Multiplication is Associative.
=== ===
Rational Multiplication Identity is $1$.
Hence the result.
== Sources ==
* : $\S 3.1$: Monoids
Category:Rational Multiplication
Category:Examples of Monoids
Rational Numbers whose Denominators are not Divisible by 4 do not form Ring

== Theorem ==
Let $S$ be the set defined as:
:$S = \set {\dfrac m n : m, n \in \Z, m \perp n, 4 \nmid n}$
That is, $S$ is defined as the set of rational numbers such that, when expressed in canonical form, their denominators are not divisible by $4$.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
== Proof ==
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $x = \dfrac 1 2$ and $y = \dfrac 3 2$.
Both $x$ and $y$ are in $S$, as both are rational numbers expressed in canonical form whose denominators are not divisible by $4$.
But then:
:$x \times y = \dfrac 1 2 \times \dfrac 3 2 = \dfrac 3 4$
which is a rational numbers expressed in canonical form whose denominator is divisible by $4$.
Hence $x \times y \notin S$ and so $\struct {S, \times}$ is not closed.
The result follows.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(b)}$
Category:Rational Numbers
Category:Examples of Rings
Rational Numbers with Denominator Power of Two form Integral Domain

== Theorem ==
Let $\Q$ denote the set of rational numbers.
Let $S \subseteq \Q$ denote the set of set of rational numbers of the form $\dfrac p q$ where $q$ is a power of $2$:
:$S = \set {\dfrac p q: p \in \Z, q \in \set {2^m: m \in \Z_{\ge 0} } }$
Then $\struct {S, +, \times}$ is an integral domain.
== Proof ==
From Rational Numbers form Integral Domain we have that $\struct {\Q, +, \times}$ is an integral domain.
Hence to demonstrate that $\struct {S, +, \times}$ is an integral domain, we can use the Subdomain Test.
We have that the unity of $\struct {\Q, +, \times}$ is $1$.
Then we note:
:$1 = \dfrac 1 1$
and:
:$1 = 2^0$
and so $1 \in S$.
Thus property $(2)$ of the Subdomain Test is fulfilled.
It remains to demonstrate that $\struct {S, +, \times}$ is a subring of $\struct {\Q, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test.
Hence we use the Subring Test.
We note that $S \ne \O$ as $1 \in S$.
This fulfils property $(1)$ of the Subring Test.
Let $x, y \in S$.
Then:
This fulfils property $(2)$ of the Subring Test.
Then: 
This fulfils property $(3)$ of the Subring Test.
Hence the result.
== Sources ==
* : Exercises: Chapter $1$: Exercise $1 \ \text{(iii)}$
Category:Rational Numbers
Category:Integral Domains
Rational Numbers with Denominators Coprime to Prime under Addition form Group

== Theorem ==
Let $p$ be a prime number.
Let $\Q_p$ denote the set:
:$\set {\dfrac r s : s \perp p}$
where $s \perp p$ denotes that $s$ is coprime to $p$.
Then $\struct {\Q_p, +}$ is a group.
== Proof ==
Taking each of the group axioms in turn:
=== ===
=== ===
As $\Q_p \subseteq \Q$ the result follows directly from Rational Addition is Associative and Restriction of Associative Operation is Associative.
=== ===
By Integer is Coprime to 1:
: $\dfrac 0 1 \in \Q_p$
regardless of our choice of $p$.
By the definition of addition on $\Q$:
:$\dfrac a b + \dfrac 0 1 = \dfrac a b$
and
:$\dfrac 0 1 + \dfrac a b = \dfrac a b$
for all $\dfrac a b \in \Q$.
Hence $\dfrac 0 1$ is the identity.
=== ===
For $\dfrac a b$ we have that $\dfrac {-a} b$ is the inverse of $\dfrac a b$.
As it has the same denominator as $\dfrac a b$ we have that $\dfrac {-a} b \in \Q_p$ as well.
== Sources ==
* 
Category:Examples of Groups
Category:Rational Addition
Rational Points on Graph of Cosine Function

== Theorem ==
Consider the graph of the cosine function in the real Cartesian plane $\R^2$:
:$f := \left\{ {\left({x, y}\right) \in \R^2: y = \cos x}\right\}$
The only rational point of $f$ is $\left({0, 1}\right)$.
== Proof ==
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Sine Function
Rational Points on Graph of Exponential Function

== Theorem ==
Consider the graph $f$ of the exponential function in the real Cartesian plane $\R^2$:
:$f := \set {\tuple {x, y} \in \R^2: y = e^x}$
The only rational point of $f$ is $\tuple {0, 1}$.
== Proof ==
From Exponential of Rational Number is Irrational:
:$r \in \Q_{\ne 0} \implies e^r \in \R - \Q$
Thus, apart from the point $\tuple {0, 1}$, when $x$ is rational, $e^x$ is not.
Hence the result.
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Exponential Function
Rational Points on Graph of Logarithm Function

== Theorem ==
Consider the graph of the logarithm function in the real Cartesian plane $\R^2$:
:$f := \set {\tuple {x, y} \in \R^2: y = \ln x}$
The only rational point of $f$ is $\tuple {1, 0}$.
== Proof ==
Consider the graph of the exponential function in the real Cartesian plane $\R^2$:
:$g := \set {\tuple {x, y} \in \R^2: y = e^x}$
From Rational Points on Graph of Exponential Function, the only rational point of $g$ is $\tuple {0, 1}$.
By definition of the exponential function, $f$ and $g$ are inverses.
Thus:
:$\tuple {x, y} \in g \iff \tuple {y, x} \in f$
Thus for $\tuple {x, y} \in g$ to be rational, $\tuple {y, x} = \tuple {0, 1}$.
Hence the result.
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Logarithms
Rational Points on Graph of Sine Function

== Theorem ==
Consider the graph of the sine function in the real Cartesian plane $\R^2$:
:$f := \set {\tuple {x, y} \in \R^2: y = \sin x}$
The only rational point of $f$ is $\tuple {0, 0}$.
== Proof ==
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Sine Function
Rational Polynomial is Content Times Primitive Polynomial

== Theorem ==
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial.
For a given polynomial $\map f X$, both $\cont f$ and $\map {f^*} X$ are unique.
== Proof ==
=== Proof of Existence ===
=== Proof of Uniqueness ===
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $61$
Category:Polynomial Theory
Category:Rational Polynomial is Content Times Primitive Polynomial
Rational Polynomial is Content Times Primitive Polynomial/Existence

== Theorem ==
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial.
== Proof ==
Consider the coefficients of $f$ expressed as fractions.
Let $k$ be any positive integer that is divisible by the denominators of all the coefficients of $f$.
Such a number is bound to exist: just multiply all those denominators together, for example.
Then $\map f X$ is a polynomial equal to $\dfrac 1 k$ multiplied by a polynomial with integral coefficients.
Let $d$ be the GCD of all these integral coefficients.
Then $\map f X$ is equal to $\dfrac h k$ multiplied by a primitive polynomial.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $61$
Category:Rational Polynomial is Content Times Primitive Polynomial
Rational Polynomial is Content Times Primitive Polynomial/Uniqueness

== Theorem ==
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$ be given.
Then there exist unique content $\cont f$ of $\map f X$ and primitive polynomial $\map {f^*} X$ such that:
:$\map f X = \cont f \, \map {f^*} X$
== Proof ==
Existence is proved in Rational Polynomial is Content Times Primitive Polynomial: Existence.
Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are primitive.
Then:
:$\map g X = \dfrac a b \map f X$
where $\dfrac a b$ is some rational number which can be expressed as $\dfrac m n$ where $m$ and $n$ are coprime.
Then:
:$\map g X = \dfrac m n \map f X$
that is:
:$m \cdot \map f X = n \cdot \map g X$
Suppose $m > 1$.
Then from Euclid's Lemma $m$ has a divisor $p$ which does not divide $n$ (as $m \perp n$).
So $m$ must divide every coefficient of $g$.
But this can not be so, as $g$ is primitive, so $m = 1$.
In a similar way, $n = 1$.
So $f = g$ and $a = b$, so demonstrating uniqueness.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $61$
Category:Rational Polynomial is Content Times Primitive Polynomial
Rational Power is of Exponential Order Epsilon

== Theorem == 
Let $r = \dfrac p q$ be a rational number, with $p, q \in \Z: q \ne 0, r > 0$.
Then: 
:$t \mapsto t^r$
is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
== Proof ==
Write $t^r = t^{p/q}$, and set $t > 1$.
Recall from Polynomial is of Exponential Order Epsilon, $t^p < K'e^{a't}$ for any $a' > 0$, arbitrarily small in magnitude.
Therefore the inequality $t^{p/q} < K e^{a t}$ has solutions of the same nature.
Category:Exponential Order
Rational Power of Product of Real Numbers

== Theorem ==
Let $r, s \in \R_{> 0}$ be (strictly) positive real numbers.
<onlyinclude>
Let $x \in \Q$ be a rational number.
Let $r^x$ be defined as $r$ to the power of $x$.
Then:
:$\paren {r s}^x = r^x s^x$
== Proof ==
Let $x = \dfrac p q$ where $p, q \in \Z$ and $q > 0$.
We have:
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (3) \, \text{(iii)}$
Category:Powers
Rational Root Theorem

== Theorem ==
Let $\map P x$ be a polynomial whose coefficients are all integers:
:$\map P x = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$
where $a_n \ne 0$ and $a_0 \ne 0$.
Let the polynomial equation $\map P x = 0$ have a root which is a rational number expressed in canonical form as $\dfrac p q$.
Then:
:the leading coefficent $a_n$ of $\map P x$ is divisible by $q$
:the constant term $a_0$ of $\map P x$ is divisible by $p$.
== Proof ==
We have that $\dfrac p q$ is a root of $P$.
Hence:
By the closure of addition and multiplication over the integers:
:$a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} \in \Z$
By Euclid's Lemma, either:
:$p \divides q^n$
or:
:$p \divides a_0$
where $\divides$ denotes divisibility.
Because $p \nmid q$ we have that:
:$p \nmid q^n$
Hence:
:$p \divides a_0$
Then:
Using the same argument as for $p \divides a_0$, it follows that:
:$q \divides a_n$
== Sources ==
* 
* 
Category:Polynomial Equations
Category:Named Theorems