nyuuzyou/wikis huggingface dataset:
Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution

== Theorem ==
The special case of Bessel's equation:
:$(1): \quad x^2 y' ' + x y' + \paren {x^2 - \dfrac 1 4} y = 0$
has a particular solution:
:$y = \dfrac {\sin x} {\sqrt x}$
== Proof ==
Note that:
Inserting into $(1)$, and gradually simplifying:
hence demonstrating that:
:$y_1 = \dfrac {\sin x} {\sqrt x}$
is a particular solution of $(1)$.
== Sources ==
* : $\S 3.16$: Problem $6$
Category:Examples of Bessel's Equation
Bessel's Inequality

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.
Let $E = \set {e_n: n \in \N}$ be a countably infinite orthonormal subset of $V$.
Then, for all $h \in V$:
:$\ds \sum_{n \mathop = 1}^\infty \size {\innerprod h {e_n} }^2 \le \norm h^2$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Note that for any natural number $n$ we have, applying sesquilinearity of the inner product: 
We have: 
Therefore: 
Since:
:$\ds \norm {h - \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k}^2 \ge 0$
we have: 
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 \le {\norm h}^2$
Since:
:$\size {\innerprod h {e_k} }^2 \ge 0$ for each $k$
we have that: 
:the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ is increasing.
So:
:the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ is bounded and increasing.
So from Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we have that: 
:the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ converges. 
Since: 
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 \le {\norm h}^2$ for each $n$
we then have from Limits Preserve Inequalities: 
:$\ds {\norm h}^2 \ge \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 = \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: $4.8$ Bessel's Inequality
Category:Inner Product Spaces
Category:Bessel's Inequality
Bessel's Inequality/Corollary 1

== Corollary to Bessel's Inequality ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space. 
Let $E$ be a orthonormal subset of $V$. 
Then, for each $h \in V$, the set: 
:$\set {e \in E : \innerprod h e \ne 0}$
is countable.
== Proof ==
Let:
:$X = \set {e \in E : \innerprod h e \ne 0}$
For each natural number $n$, define: 
:$\ds X_n = \set {e \in E : \size {\innerprod h e} > \frac 1 n}$
We have that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$
We can show that for each $n \in \N$, the set $X_n$ is finite. 
, suppose that for some $m \in \N$, the set $X_m$ is infinite. 
Then, there exists an countable orthonormal subset of $X_m$, say $\set {e_k : k \in \N}$, such that: 
:$\ds \size {\innerprod h {e_k} } > \frac 1 m$
for each $k$. 
By Bessel's Inequality, we have that: 
:$\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$ converges
and: 
:$\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 \le {\norm h}^2$ 
However, for each $n \in \N$, we have: 
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 > \frac n {m^2}$ 
So, for any natural number $n$ with $n > m^2 {\norm h}^2$, we have: 
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 > {\norm h}^2$
This contradicts Bessel's Inequality, so we must have that each $X_m$ is finite. 
We therefore see that:
:$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$
is the countable union of finite sets, so by Countable Union of Finite Sets is Countable:
:$X$ is countable
as required.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Corollary $4.9$
Category:Bessel's Inequality
Bessel's Inequality/Corollary 2

== Corollary to Bessel's Inequality ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space. 
Let $E$ be a orthonormal subset of $V$. 
Then, for all $h \in V$:
:$\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 \le \norm h^2$
== Proof ==
From , we have that: 
:$\innerprod h e \ne 0$ for only countably many $e \in E$.
So the set
:$X = \set {e \in E : \innerprod h e \ne 0}$
is countable.
Hence, there exists a sequence $\sequence {e_k}_{k \in \N}$ such that: 
:$X = \set {e_k : k \in \N}$
For brevity, write:
:$X_n = \set {e_k : k \le n}$
for each $n \in \N$. 
We have, by Bessel's Inequality, that: 
:$\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$ converges
with:
:$\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 \le {\norm h}^2$
We now argue that:
:$\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 = \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$
Let $\epsilon > 0$.
From the definition of net convergence, we aim to show that for each $\epsilon > 0$, there exists a finite set $S \subseteq X$ such that:
:$\ds \cmod {\sum_{e \mathop \in F} \size {\innerprod h e}^2 - \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2} < \epsilon$
for all finite sets $F \supseteq S$. 
Let $F$ be a finite set containing $X_n$. 
Since:
:$\innerprod h e = 0$ for $h \not \in X$ 
we have: 
:$\ds \sum_{e \mathop \in F} \size {\innerprod h e}^2 = \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2$
where $F_\ast$ is some subset of $\set {k \in \N : k \ge n + 1}$. 
Since $\size {\innerprod h {e_k} }^2 \ge 0$ for each $k \in \N$, we have:
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2 \le \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$
so that:
:$\ds \sum_{e \mathop \in F} \size {\innerprod h e}^2 \le \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$
For the lower bound, note that:
:$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2 \ge \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2$
Pick $N_\epsilon \in \N$ such that:
:$\ds \sum_{k \mathop = 1}^{N_\epsilon} \size {\innerprod h {e_k} }^2 \ge \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 - \epsilon$ 
So, for all finite sets $F \supseteq X_{N_\epsilon}$, we have:
:$\ds \cmod {\sum_{e \mathop \in F} \size {\innerprod h e}^2 - \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2} < \epsilon$
So:
:$\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2$ converges as a generalized sum to $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$
Since we have that:
:$\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 \le \norm h^2$
we are done. 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Corollary $4.10$
Category:Bessel's Inequality
Bessel Function of the First Kind/Instances/Order 0

== Specific Instance of Bessel Functions of the First Kind ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
== Proof ==
From Series Expansion of Bessel Function of the First Kind:
where $n$ is not a (strictly) negative integer.
$0$ fits that category, and so:
Or working directly upon the terms themselves:
== Sources ==
* : $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.5$
Category:Examples of Bessel Functions
Bessel Function of the First Kind for Imaginary Argument

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
:$\map {J_n } {i x} = i^{-n} \, \map {I_n} x$
where:
:$i$ denotes the imaginary unit
:$\map {I_n} x$ denotes the modified Bessel function of the first kind of order $n$.
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $5$
Category:Bessel Functions
Category:Modified Bessel Functions
Bessel Function of the First Kind of Negative Integer Order

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$, where $n$ is a positive integer.
Then:
:$\map {J_{-n} } x = \paren {-1}^n \map {J_n} x$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $1$
* : $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.4$
* : $\S 27$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $27.4.$
Category:Bessel Functions
Best Approximation from Below to 1 as Sum of Minimal Number of Unit Fractions

== Theorem ==
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {43} = 1 - \dfrac 1 {1806}$
This is the best approximation from below to $1$ as the minimal sum of unit fractions.
== Proof ==
== Sources ==
* : $43$
Category:Unit Fractions
Best Rational Approximations to Root 2 generate Pythagorean Triples

== Theorem ==
Consider the Sequence of Best Rational Approximations to Square Root of 2:
:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
Every odd term of $\sequence S$ can be expressed as:
:$\dfrac {2 a + 1} b$
such that:
:$a^2 + \paren {a + 1}^2 = b^2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Category:Pythagorean Triples
Category:Pell Numbers
Best Rational Approximations to Root 2 generate Pythagorean Triples/Proof 1

== Theorem ==
== Proof ==
From Parity of Best Rational Approximations to Root 2‎:
:The numerators of the terms of $\sequence S$ are all odd.
:For all $n$, the parity of the denominator of term $S_n$ is the same as the parity of $n$.
Thus it follows that every other term of $\sequence S$ has a numerator and a denominator which are both odd.
This proof proceeds by induction.
=== Basis for the Induction ===
$\dfrac 1 1$ can be expressed as $\dfrac {2(0) + 1} 1$, and:
:$0^2 + \paren {0 + 1}^2 = 1^2$
=== Induction Hypothesis ===
This is our induction hypothesis:
The best rational approximation $\dfrac {p_k} {q_k}$ of $\sqrt 2$, when expressed as:
:$\dfrac {2 a + 1} b$
gives the relation:
:$a^2 + \paren {a + 1}^2 = b^2$
We need to show that the best rational approximation $\dfrac {p_{k + 2} } {q_{k + 2} }$ of $\sqrt 2$, when expressed as:
:$\dfrac {2 a' + 1} {b'}$
also give the relation:
:$a'^2 + \paren {a' + 1}^2 = b'^2$
=== Induction Step ===
This is our induction step:
From the induction hypothesis we have:
:$a = \dfrac {p_k - 1} 2, b = q_k$
and thus:
:$\paren {\dfrac {p_k - 1} 2}^2 + \paren {\dfrac {p_k - 1} 2 + 1}^2 = q_k^2$
Expanding, we have:
:$\dfrac {p_k^2} 2 + \dfrac 1 2 = q_k^2$
Now by Relation between Adjacent Best Rational Approximations to Root 2, we have:
:$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$
We check that, via GCD with Remainder:
:$\gcd \set {p_k + 2 q_k, p_k + q_k} = \gcd \set {q_k, p_k + q_k} = \gcd \set {q_k, p_k} = 1$
Since both fractions are in canonical form and Canonical Form of Rational Number is Unique, we can write:
:$p_{k + 1} = p_k + 2 q_k$
:$q_{k + 1} = p_k + q_k$
Therefore:
:$p_{k + 2} = 3 p_k + 4 q_k$
:$q_{k + 2} = 2 p_k + 3 q_k$
We need to show that:
:$\paren {\dfrac {p_{k + 2} - 1} 2}^2 + \paren {\dfrac {p_{k + 2} - 1} 2 + 1}^2 = q_k^2$
or:
:$\dfrac {p_{k + 2}^2} 2 + \dfrac 1 2 = q_{k + 2}^2$
We have:
The result follows by induction.
Category:Square Root of 2
Category:Pythagorean Triples
Best Rational Approximations to Root 2 generate Pythagorean Triples/Proof 2

== Theorem ==
== Proof ==
From Pell Number as Sum of Squares, we have:
:$P_{2 n + 1} = P_{n + 1}^2 + P_n^2$
Therefore:
Category:Pell Numbers
Category:Pythagorean Triples
Beta Function as Integral of Power of t by Power of 1 minus t over Power of r plus t

== Theorem ==
:$\ds \map \Beta {x, y} := r^y \paren {r + 1}^x \int_{\mathop \to 0}^{\mathop \to 1} \frac {t^{x - 1} \paren {1 - t}^{y - 1} } {\paren {r + t}^{x + y} } \rd t$
where $\Beta$ denotes the Beta function.
== Proof ==
Let:
Then:
== Sources ==
* : $\S 17$: The Beta Function:Some Important Results: $17.6$
* : $\S 26$: The Beta Function: Some Important Results: $26.6.$
Category:Beta Function
Beta Function as Integral of Power of t over Power of t plus 1

== Theorem ==
:$\ds \map \Beta {x, y} = \int_{\mathop \to 0}^{\mathop \to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t$
where $\Beta$ denotes the Beta function.
== Proof ==
Consider the substitution $s = \dfrac t {1 + t}$.
We have the following:
:$\dfrac 1 {1 + t} = 1 - s$
:$t \to 0, s \to 0$
:$t \to \infty, s \to 1$
:$\rd s = \dfrac 1 {\paren {1 + t}^2} \rd t$
Then:
== Sources ==
* : $\S 17$: The Beta Function: Some Important Results: $17.5$
* : $\S 26$: The Beta Function: Some Important Results: $26.5.$
Category:Beta Function
Beta Function expressed using Gamma Functions

== Theorem ==
Let $\map \Beta {x, y}$ denote the Beta function.
Then:
:$\map \Beta {x, y} = \dfrac {\map \Gamma x \, \map \Gamma y} {\map \Gamma {x + y} }$
where $\Gamma$ is the Gamma function:
== Proof ==
From Beta Function of x with y+m+1:
:$\map \Beta {x, y} = \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}$
where $\Gamma_m$ is the partial Gamma function:
:$\map {\Gamma_m} y := \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \dotsm \paren {y + m} }$
From Partial Gamma Function expressed as Integral:
Thus:
:$\ds \lim_{m \mathop \to \infty} m^x \, \map \Beta {x, m + 1} = \map \Gamma x$
As $m^x$ is monotone, it does not matter if $m$ is integer or real.
Thus:
:$\ds \lim_{m \mathop \to \infty} \paren {m + y}^x \, \map \Beta {x, m + y + 1} = \map \Gamma x$
Hence:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $41 \ \text{(b)}$
Category:Beta Function
Category:Gamma Function
Beta Function is Continuous and Positive on Positive Reals

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $\map \Beta {x, y}$ be the Beta function:
:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Let $y \in \R_{>0}$ be given.
Then $\map \Beta {x, y}$ is a positive and continuous function of $x$ on $\R_{>0}$.
== Proof ==
For each $x > 0$, we have for all $t$ with $0 < t < 1$ that:
:$t^{x - 1} \paren {1 - t}^{y - 1} > 0$
from which it is immediate that $\map \Beta {x, y} > 0$.
== Sources ==
* : $\S 17.7 \ (4)$
Category:Beta Function
Beta Function is Defined for Positive Reals

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $\map \Beta {x, y}$ be the Beta function:
:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Then $\map \Beta {x, y}$ exists provided that $x, y > 0$.
== Proof ==
Consider the following inequalities, valid for $0 < t < 1$:
Then:
and similarly:
The result follows from the Comparison Test for Improper Integral.
== Sources ==
* : $\S 17.7 \ (4)$
Category:Beta Function
Beta Function of Half with Half/Proof 1

== Theorem ==
== Proof ==
By definition of the Beta function:
:$\ds \map \Beta {x, y} := 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta$
Thus:
Category:Beta Function of Half with Half
Beta Function of Half with Half/Proof 2

== Theorem ==
== Proof ==
By definition of the Beta function:
:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Thus:
Let $t = \sin^2 \theta$.
Then:
:$\rd t = 2 \sin \theta \cos \theta \rd \theta$
and:
and so:
== Sources ==
* : $\S 17.7 \ (6)$
Category:Beta Function of Half with Half
Beta Function of Half with Half/Proof 3

== Theorem ==
== Proof ==
By definition of the Beta function:
:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Thus:
Let $t = u^2$.
Then:
:$\rd t = 2 u \rd u$
and:
and so:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $43$
Category:Beta Function of Half with Half
Beta Function of Real Number with 1

== Theorem ==
Let $\map \Beta {x, y}$ denote the Beta function.
Then:
:$\map \Beta {x, 1} = \map \Beta {1, x} = \dfrac 1 x$
== Proof ==
By definition:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $40 \ \text{(a)}$
Category:Beta Function
Beta Function of x+1 with y plus Beta Function of x with y+1

== Theorem ==
Let $\map \Beta {x, y}$ denote the Beta function.
Then:
:$\map \Beta {x + 1, y} + \map \Beta {x, y + 1} = \map \Beta {x, y}$
== Proof ==
By definition:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $40 \ \text{(b)}$
Category:Beta Function
Beta Function of x with y+1 by x+y over y

== Theorem ==
Let $\map \Beta {x, y}$ denote the Beta function.
Then:
:$\map \Beta {x, y} = \dfrac {x + y} y \map \Beta {x, y + 1}$
== Proof ==
By definition of Beta function:
:$\ds \map \Beta {x + 1, y} = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t$
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Hence:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $40 \ \text{(c)}$
Category:Beta Function
Beta Function of x with y+m+1

== Theorem ==
Let $\map \Beta {x, y}$ denote the Beta function.
Then:
:$\map \Beta {x, y} = \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}$
where $\Gamma_m$ is the partial Gamma function:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $41 \ \text{(a)}$
Category:Beta Function of x with y+m+1
Category:Beta Function
Between Two Sets Exists Injection or Surjection

== Theorem ==
Let $S$ and $T$ be sets.
Then either or both of the following cases hold:
:$(1):$ There exists a mapping $f: S \to T$ such that $f$ is an injection
:$(2):$ There exists a mapping $f: S \to T$ such that $f$ is a surjection.
== Proof ==
From Zermelo's Theorem, exactly one of the following cases holds:
:$(1): \quad S \le T$, that is, $T$ dominates $S$
:$(2): \quad S \equiv T$, that is, $T$ is equivalent to $S$
:$(3): \quad S \ge T$, that is, $S$ dominates $T$.
Suppose $(1)$ holds, that is, $T$ dominates $S$.
By definition, there exists an injection from $S$ to $T$.
Hence in this case the result holds.
Suppose $(2)$ holds, that is, $T$ is equivalent to $S$.
By definition, there then exists a bijection from $S$ to $T$.
A bijection is, by definition, a mapping which is both an injection and a surjection.
Hence in this case the result holds.
Suppose $(3)$ holds, that is, $S$ dominates $T$.
By definition, there exists an injection from $S$ to $T$.
By Injection has Surjective Left Inverse Mapping, $f$ has a left inverse $g: T \to S$ which is surjective.
Hence in this case the result holds.
All cases have been shown to lead to the same conclusion, that is, that there exists either an injection or a surjection or both between $S$ and $T$.
Hence the result, by Proof by Cases.
Category:Injections
Category:Surjections
Between two Cubes exist two Mean Proportionals

== Theorem ==
== Proof ==
Let $a^3$ and $b^3$ be cube numbers.
From the corollary to Form of Geometric Sequence of Integers:
:$\left({a^3, a^2 b, a b^2, b^3}\right)$
is a geometric sequence.
It follows by definition that $a^2 b$ and $a b^2$ are mean proportionals of $a^3$ and $b^3$.
Then:
:$\left({\dfrac a b}\right)^3 = \dfrac {a^3} {b^3}$
By definition, it follows that $a^3$ has to $b^3$ the triplicate ratio that $a$ has to $b$.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Between two Rational Numbers exists Irrational Number

== Theorem ==
Let $a, b \in \Q$ where $a < b$. 
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$
=== Lemma 1 ===
=== Lemma 2 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.14$: Exercise $18 \ (2)$
Category:Real Analysis
Category:Between two Rational Numbers exists Irrational Number
Between two Rational Numbers exists Irrational Number/Lemma 1

== Lemma for Between two Rational Numbers exists Irrational Number ==
Let $\alpha \in \Q \setminus \set 0$ and $\beta \in \R \setminus \Q$. 
Then: 
:$\alpha \cdot \beta \in \R \setminus \Q$
== Proof ==
 $\alpha \cdot \beta \in \Q$.
By the definition of rational numbers:
:$\exists n, m, p, q \in \Z: \alpha = \dfrac n m$
:$\alpha \cdot \beta = \dfrac p q$
Thus:
:$\beta = \dfrac p q \cdot \dfrac 1 \alpha = \dfrac p q \cdot \dfrac m n$
By Rational Multiplication is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$.
Therefore $\alpha \cdot \beta \in \R \setminus \Q$.
Category:Between two Rational Numbers exists Irrational Number
Between two Rational Numbers exists Irrational Number/Lemma 2

== Lemma for Between two Rational Numbers exists Irrational Number ==
Let $\alpha \in \Q$ and $\beta \in \R \setminus \Q$. 
Then: 
:$\alpha + \beta \in \R \setminus \Q$
== Proof ==
 $\alpha + \beta \in \Q$.
By the definition of rational numbers:
:$\exists p, q \in \Z: \alpha + \beta = \dfrac p q$
Thus:
:$\beta = \dfrac p q - \alpha$
By Rational Subtraction is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$.
Therefore:
:$\alpha + \beta \in \R \setminus \Q$
Category:Between two Rational Numbers exists Irrational Number
Between two Rational Numbers exists Irrational Number/Proof 1

== Theorem ==
=== Lemma 1 ===
=== Lemma 2 ===
== Proof ==
Let $d = b - a$.
As $a, b \in \Q: a < b$ it follows from Rational Numbers form Ordered Integral Domain that $d \in \Q: d > 0$.
From Square Root of 2 is Irrational, $\sqrt 2$ is not a rational number, so it is an element of $\R \setminus \Q$.
From Square Number Less than One, for any given real number $x$:
:$x^2 < 1 \implies x \in \openint {-1} 1$
Let $k = \dfrac {\sqrt 2} 2$.
Then from Lemma 1:
:$k \in \R \setminus \Q$
As $k^2 = \dfrac 1 2$, it follows that $0 < k < 1$.
Let $\xi = a + k d$.
Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows from Lemma 1 and Lemma 2 that:
:$\xi \in \R \setminus \Q$
$d > 0$ and $k > 0$, so:
:$\xi = a + k d > a + 0 \cdot 0 = a$
$k < 1$, so:
:$\xi = a + k d < a + 1 \cdot d < a + \paren {b - a} = b$
We thus have:
:$\xi \in \R \setminus \Q: \xi \in \openint a b$
Hence the result.
Category:Between two Rational Numbers exists Irrational Number
Between two Rational Numbers exists Irrational Number/Proof 2

== Theorem ==
=== Lemma ===
== Proof ==
From Between two Real Numbers exists Rational Number, there exists $x \in \Q$ such that:
:$a - \sqrt2 < x < b - \sqrt2$
Since Square Root of 2 is Irrational, by the Lemma:
:$x + \sqrt 2$ is irrational.
But:
:$a < x + \sqrt 2 < b$
which is what we wanted to show.
Category:Between two Rational Numbers exists Irrational Number
Between two Real Numbers exists Irrational Number

== Theorem ==
Let $a, b \in \R$ be real numbers where $a < b$. 
Then there exists an irrational number $\xi \in \R \setminus \Q$ such that:
:$a < \xi < b$
== Proof ==
From Number of Type Rational r plus s Root 2 is Irrational we have that a real number of the form $r \sqrt 2$, where $r \ne 0$ is rational, is irrational.
From Between two Real Numbers exists Rational Number there exists a rational number $r$ such that:
:$\dfrac a {\sqrt 2} < r < \dfrac b {\sqrt 2}$
and so:
:$a < r \sqrt 2 < b$
giving us $\xi = r \sqrt 2$ as the irrational number we seek.
If $a < 0$ and $b > 0$, it may be that the $r$ that we find happens to be zero
Then we locate a second rational number $s$ such that:
:$0 < s < \dfrac b {\sqrt 2}$
which leads us to:
:$a < 0 < s \sqrt 2 < b$
and so in this case $\xi = s \sqrt 2$ is the irrational number we seek.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Corollary $1.1.7$: Remark
* : $1$: Review of some real analysis: Exercise $1.5: 7$
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (6)$
Category:Real Analysis
Between two Real Numbers exists Rational Number

== Theorem ==
Let $a, b \in \R$ be real numbers such that $a < b$.
Then:
:$\exists r \in \Q: a < r < b$
=== Corollary: Real Numbers are Densely Ordered ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Real Analysis
Category:Between two Real Numbers exists Rational Number