nyuuzyou/wikis huggingface dataset:
Square Root of 2 is Irrational/Classic Proof

== Theorem ==
== Proof ==
First we note that, from Parity of Integer equals Parity of its Square, if an integer is even, its square root, if an integer, is also even.
Thus it follows that:
:$(1): \quad 2 \divides p^2 \implies 2 \divides p$
where $2 \divides p$ indicates that $2$ is a divisor of $p$.
 that $\sqrt 2$ is rational.
So:
:$\sqrt 2 = \dfrac p q$
for some $p, q \in \Z$, and:
:$\gcd \set {p, q} = 1$
where $\gcd$ denotes the greatest common divisor.
Squaring both sides yields:
:$2 = \dfrac {p^2} {q^2} \iff p^2 = 2 q^2$
Therefore from $(1)$:
:$2 \divides p^2 \implies 2 \divides p$
That is, $p$ is an even integer.
So $p = 2 k$ for some $k \in \Z$.
Thus:
:$2 q^2 = p^2 = \paren {2 k}^2 = 4 k^2 \implies q^2 = 2 k^2$
so by the same reasoning:
:$2 \divides q^2 \implies 2 \divides q$
This contradicts our assumption that $\gcd \set {p, q} = 1$, since $2 \divides p, q$.
Therefore, from Proof by Contradiction, $\sqrt 2$ cannot be rational.
== Historical Note ==
== Sources ==
* : Book $\text{X}$. Introductory Note
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies: Footnote $1$
* : $\S 1.1$. Number Systems
* : Chapter $1$: The Real and Complex Number Systems: Introduction
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory: Exercise $1.10$
* : Chapter $3$: The Integers: $\S 13$. Unique Factorisation: Theorem $24$
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* : $\S 1$: Real Numbers: $\S 1.13$: Irrational Numbers
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $2$. Graphs: Paradox
* : $\S 1.14$: Exercise $18 \ (1)$
* : $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 5$
* : $\S 2.3$
* 
* 
Category:Square Root of 2 is Irrational
Square Root of 2 is Irrational/Historical Note

== Historical Note on Square Root of 2 is Irrational ==
This result Square Root of 2 is Irrational is attributed to , or to a student of his.
Some legends have it that it is due to who was thrown off a boat by his angry fellow Pythagoreans and drowned.
The ancient Greeks prior to believed that irrational numbers did not exist in the real world.
However, from the Pythagorean Theorem, a square with sides of length $1$ has a diagonal of length $\sqrt 2$.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {I}$: Number: $1.1$ Introduction
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $10$
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $10$
* : Chapter $2$: The Logic of Shape: Pythagoras
Category:Historical Notes
Square Root of 2 is Irrational/Proof 3

== Theorem ==
== Proof ==
 that $\sqrt 2$ is rational.
Then $\sqrt 2 = \dfrac p q$ for some $p, q \in \Z_{>0}$
Consider the quantity $\paren {\sqrt 2 - 1}$:
Now, observe that for any $n \in \Z_{>0}$:
By Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$
where $\lim$ denotes limit.
Recall the definition of $a_n$ and $b_n$.
By Lower and Upper Bounds for Sequences:
:$0 = \ds \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$
which is a contradiction.
Category:Square Root of 2 is Irrational
Square Root of 2 is Irrational/Proof 4

== Theorem ==
== Proof ==
 that $\sqrt 2$ is rational.
Let $n$ be the smallest positive integer such that:
:$\sqrt 2 = \dfrac m n$
for some $m \in \Z_{>0}$
Then:
:$m = n \sqrt2 > n$
so:
:$(1): \quad m - n > 0$
We also have:
:$m = n \sqrt2 < 2 n$
so:
:$m < 2 n$
and therefore:
:$(2): \quad m - n < n$
Finally, we have
It follows that:
:$\dfrac {2 n - m} {m - n} = \sqrt2$
By $(1)$, the denominator of $\dfrac {2 n - m} {m - n}$ is positive.
By $(2)$, the denominator of $\dfrac {2 n - m} {m - n}$ is less than $n$.
We have thus written $\sqrt2$ as a fraction with a smaller denominator than $n$, which is a contradiction.
Category:Square Root of 2 is Irrational
Square Root of Complex Number in Cartesian Form

== Theorem ==
Let $z \in \C$ be a complex number.
Let $z = x + i y$ where $x, y \in \R$ are real numbers.
Let $z$ not be wholly real, that is, such that $y \ne 0$.
Then the square root of $z$ is given by:
:$z^{1/2} = \pm \paren {a + i b}$
where:
== Proof ==
Let $a + i b \in z^{1/2}$.
Then:
Equating imaginary parts in $(1)$:
Equating real parts in $(1)$:
Note that in $(3)$, only the positive square root of the discriminant $x^2 + y^2$ is used.
This is because the negative square root of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$.
As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$.
Hence $\sqrt {x^2 + y^2} > 0$.
Then:
But from $(2)$ we have:
:$b = \dfrac y {2 a}$
and so having picked either the positive square root or negative square root of either $a^2$ or $b^2$, the root of the other is forced.
So:
:if $y > 0$, then $a$ and $b$ are both of the same sign.
Thus:
:$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
:if $y < 0$, then $a$ and $b$ are of opposite sign.
Thus:
:$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
Hence:
:$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
== Examples ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $80 \ \text{(b)}$
Category:Square Root of Complex Number in Cartesian Form
Category:Complex Analysis
Category:Square Roots
Category:Complex Numbers
Square Root of Complex Number in Cartesian Form/Examples

== Examples of Square Root of Complex Number in Cartesian Form ==
=== Example: $i$ ===
=== Example: $2 + 2 i$ ===
=== Example: $3 + 4 i$ ===
=== Example: $-8 + 6 i$ ===
=== Example: $5 - 12 i$ ===
=== Example: $-15 - 8 i$ ===
=== Example: $8 + 4 \sqrt 5 i$ ===
Category:Square Root of Complex Number in Cartesian Form
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 1

== Example of Square Root of Complex Number in Cartesian Form ==
== Proof ==
We have that:
:$-15 - 8 i = 17 \map \cis {\theta + 2 k \pi}$
where:
:$\cos \theta = -\dfrac {15} {17}$
:$\sin \theta = -\dfrac 8 {17}$
Then the square roots of $-15 - 8 i$ are:
We have that:
and:
As $\theta$ is in the third quadrant, $\dfrac \theta 2$ is in the second quadrant.
Hence:
Hence from $(1)$ and $(2)$:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $30$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 2

== Example of Square Root of Complex Number in Cartesian Form ==
== Proof ==
Let the required square roots be $p = i q$ for real $p$ and $q$.
Then:
The result follows by substituting $+1$ and $-1$ into $2$, in turn.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $30$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/-8+6i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt {-8 + 6 i} = \pm \paren {1 + 3 i}$
== Proof ==
As $2 x y = 6$ it follows that the two solutions are:
:$1 + 3 i$
:$-1 - 3 i$
== Sources ==
* : $\S 4.6$. The Logarithm: Examples: $\text {(iii)}$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/2+2i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt {2 + 2 i} = \pm \left({\sqrt {1 + \sqrt 2} + i \sqrt {\sqrt 2 - 1}}\right)$
== Proof ==
As $2 x y = 2$ it follows that the two solutions are:
:$\sqrt {1 + \sqrt 2} + i \sqrt {\sqrt 2 - 1}$
:$-\sqrt {1 + \sqrt 2} - i \sqrt {\sqrt 2 - 1}$
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $6 \ \text {(ii)}$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/3+4i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt {3 + 4 i} = \pm \left({2 + i}\right)$
== Proof ==
As $2 x y = 4$ it follows that the two solutions are:
:$2 + i$
:$-2 - i$
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $6 \ \text {(iii)}$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/5-12i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt {5 - 12 i} = \pm \paren {3 - 2 i}$
== Proof ==
As $2 x y = -12$ it follows that the two solutions are:
:$3 - 2 i$
:$-3 + 2 i$
== Sources ==
* : $\S 1.2$. The Algebraic Theory: Example
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $98 \ \text{(a)}$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/8 + 4 root 5 i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt {8 + 4 \sqrt 5 i} = \pm \paren {\sqrt {10} + \sqrt 2 i}$
== Proof ==
As $2 x y = 4 \sqrt 5$ it follows that the two solutions are:
:$\sqrt {10} + \sqrt 2 i$
:$-\sqrt {10} - \sqrt 2 i$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $98 \ \text{(b)}$
Category:Examples of Square Roots
Square Root of Complex Number in Cartesian Form/Examples/i

== Example of Square Root of Complex Number in Cartesian Form ==
:$\sqrt i = \pm \left({\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i}\right)$
== Proof ==
As $2 x y = 1$ it follows that the two solutions are:
:$\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i$
:$-\dfrac {\sqrt 2} 2 - \dfrac {\sqrt 2} 2 i$
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $6 \ \text {(i)}$
Category:Examples of Square Roots
Square Root of Function under Derivative

== Theorem ==
:$\ds \int \dfrac {\map {f'} x} {\sqrt {\map f x} } \rd x = 2 \sqrt {\map f x} + C$
== Proof ==
Let $u = \map f x$.
Then:
:$\dfrac {\d u} {\d x} = \map {f'} x$
Hence:
== Sources ==
* : $\text {II}$. Calculus: Integration
Category:Primitives
Square Root of Number Minus Square Root

== Theorem ==
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
:$\ds \sqrt {a - \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$
== Proof 1 ==
== Proof 2 ==
Category:Square Root of Number Plus or Minus Square Root
Square Root of Number Minus Square Root/Proof 1

== Theorem ==
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
== Proof ==
We are given that $a^2 - b > 0$.
Then:
:$a > \sqrt b$
and so $\ds \sqrt {a - \sqrt b}$ is defined on the real numbers.
Let $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Observe that:
:$\ds 0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$
Squaring both sides gives:
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
From $\sqrt b = 2 \sqrt {x y}$ we get:
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
:$z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
Because we have that $x > y$:
Subsituting into $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$:
Category:Square Root of Number Plus or Minus Square Root
Square Root of Number Plus Square Root

== Theorem ==
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
:$\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$
== Proof 1 ==
== Proof 2 ==
Category:Square Root of Number Plus or Minus Square Root
Square Root of Number Plus Square Root/Proof 1

== Theorem ==
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
== Proof ==
We are given that $a^2 - b > 0$.
Then:
:$a > \sqrt b$
and so $\ds \sqrt {a + \sqrt b}$ is defined on the real numbers.
Let $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Squaring both sides gives:
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
From $\sqrt b = 2 \sqrt {x y}$ we get:
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
:$z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
:
Subsituting into $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:
Category:Square Root of Number Plus or Minus Square Root
Square Root of Prime is Irrational

== Theorem ==
The square root of a prime number is irrational.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* The special case of $p = 2$ is a well-known mathematical proof.
Category:Square Root of Prime is Irrational
Category:Prime Numbers
Category:Irrational Numbers
Category:Square Roots
Category:Irrationality Proofs
Square Root of Prime is Irrational/Proof 1

== Theorem ==
== Proof ==
Let $p$ be prime.
 that $\sqrt p$ is rational.
Then there exist natural numbers $m$ and $n$ such that:
Any prime in the prime factorizations of $n^2$ and $m^2$ must occur an even number of times because they are squares.
Thus, $p$ must occur in the prime factorization of $n^2 p$ an odd number of times.
Therefore, $p$ occurs as a factor of $m^2$ an odd number of times, which is a contradiction.
So $\sqrt p$ must be irrational.
Category:Square Root of Prime is Irrational
Square Root of Prime is Irrational/Proof 2

== Theorem ==
== Proof ==
Let $p \in \Z$ be a prime number.
Consider the polynomial:
:$\map P x = x^2 - p$
over the ring of polynomials $\Q \sqbrk X$ over the rational numbers.
From Difference of Two Squares:
:$x^2 - p = \paren {x + \sqrt p} \paren {x - \sqrt p}$
Because $p$ is prime, $\sqrt p$ is not an integer.
From Polynomial which is Irreducible over Integers is Irreducible over Rationals it follows that $\paren {x + \sqrt p}$ and $\paren {x - \sqrt p}$ do not have rational coefficients.
That is:
:$\sqrt p$ is not rational.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Example $60$
Category:Square Root of Prime is Irrational
Square Root of Sum as Sum of Square Roots

== Theorem ==
Let $a, b \in \R, a \ge b$.
Then:
:$\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$
== Proof 1 ==
== Proof 2 ==
Category:Algebra
Category:Square Root of Sum as Sum of Square Roots
Square Root of Sum as Sum of Square Roots/Proof 1

== Theorem ==
== Proof ==
Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$.
From Square of Sum:
: $a + b = c + d + 2 \sqrt {c d}$
We now need to solve the simultaneous equations:
: $a = c + d$
: $b = 2 \sqrt {c d}$
First:
Solving for $c$:
Solving for $d$:
From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$.
Category:Square Root of Sum as Sum of Square Roots
Square Root of Sum as Sum of Square Roots/Proof 2

== Theorem ==
== Proof ==
From Sum of Square Roots as Square Root of Sum:
:$\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$
Let
:$p = \dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2$,
:$q = \dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2$.
Then
Therefore:
:$\sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2} = \sqrt {a + b}$
Category:Square Root of Sum as Sum of Square Roots
Square Roots of Primes

== Square Roots of Primes ==
From Square Root of Prime is Irrational, no prime number can be expressed precisely by a rational fraction.
The decimal expansions of the square roots of some of the first few primes are as follows:
=== Square Root of $2$ ===
=== Square Root of $3$ ===
=== Square Root of $5$ ===
=== Square Root of $7$ ===
=== Square Root of $11$ ===
=== Square Root of $13$ ===
=== Square Root of $17$ ===
=== Square Root of $19$ ===
=== Square Root of $23$ ===
=== Square Root of $29$ ===
=== Square Root of $31$ ===
=== Square Root of $37$ ===
=== Square Root of $41$ ===
=== Square Root of $43$ ===
=== Square Root of $47$ ===
=== Square Root of $53$ ===
=== Square Root of $59$ ===
=== Square Root of $61$ ===
=== Square Root of $67$ ===
=== Square Root of $71$ ===
=== Square Root of $73$ ===
=== Square Root of $79$ ===
=== Square Root of $83$ ===
=== Square Root of $89$ ===
=== Square Root of $97$ ===
Category:Examples of Square Roots
Square Sum of Three Consecutive Triangular Numbers

== Theorem ==
Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n + T_{n + 1} + T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 5$
== Proof ==
Let $T_n + T_{n + 1} + T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
Thus we need to find $n$ such that $\dfrac {3 n^2 + 9 n + 8} 2$ is a square number.
We see that:
Thus $n$ appears to satisfy the conditions.
It remains for us to check:
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $5 \ \text {(c)}$
Category:Triangular Numbers
Category:Square Numbers
Square equals Sum of Squares implies Right Triangle

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle such that the square on $BC$ equals the sum of the squares on $AB$ and $AC$.
Construct $AD$ perpendicular to $AC$.
Make $AD$ equal to $AB$, and join $CD$.
Since $DA = AB$, the square on $AD$ equals the square on $AB$.
Add the square on $AC$ to each.
Then the squares on $AD$ and $AC$ equal the squares on $AB$ and $AC$.
But $\angle DAC$ is a right angle.
Therefore, by Pythagoras's Theorem, the square on $DC$ equals the sum of the squares on $AD$ and $AC$.
Also, the square on $BC$ equals the sum of the squares on $AB$ and $AC$, by hypothesis.
So the square on $DC$ equals the square on $BC$, and so $DC = BC$.
So from Triangle Side-Side-Side Congruence, $\triangle ABC = \triangle DAC$.
But as $\triangle DAC$ is a right triangle, then so must $\triangle ABC$ be.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Hippocrates' Quadrature of the Lune ( )
* 
Category:Triangles
Square is Sum of Two Rectangles

== Theorem ==
== Proof ==
:
Let $AB$ be the given straight line cut at random at the point $C$.
Construct the square $ABED$ on $AB$.
Construct $CF$ parallel to $AD$.
Then $\Box ABED = \Box ACFD + \Box CBEF$.
Now $\Box ABED$ is the square on $AB$.
Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
: $\Box ACDF$ is the rectangle contained by $AB$ and $AC$, as $AB = AD$
: $\Box CBFE$ is the rectangle contained by $AB$ and $BC$, as $AB = AD$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Rectangles
Category:Squares
Square of 1 Less than Number Base

== Theorem ==
Let $b \in \Z$ be an integer such that $b > 2$.
Let $n = b - 1$.
The square of $n$ is expressed in base $b$ as:
:$n^2 = \left[{c1}\right]_b$
where $c = b - 2$.
== Proof ==
The result follows by definition of number base.
== Examples ==
Category:Square Numbers
Category:Square of 1 Less than Number Base
Square of Chi Random Variable has Chi-Squared Distribution

== Theorem == 
Let $n$ be a strictly positive integer.
Let $X \sim \chi_n$ where $\chi_n$ is the chi distribution with $n$ degrees of freedom. 
Then $X^2 \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.
== Proof == 
Let $Y \sim \chi^2_n$. 
We aim to show that: 
:$\map \Pr {Y < x^2} = \map \Pr {X < x}$
for all $x \in \hointr 0 \infty$.
We have:
Category:Chi Distribution
Category:Chi-Squared Distribution
Square of Complex Conjugate is Complex Conjugate of Square

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then:
: $\overline {z^2} = \left({\overline z}\right)^2$
== Proof ==
A direct consequence of Product of Complex Conjugates:
: $\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$
for two complex numbers $z_1, z_2 \in \C$.
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Conjugates
Category:Complex Multiplication
Square of Complex Modulus equals Complex Modulus of Square

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\cmod z$ be the modulus of $z$.
Then:
: $\cmod {z^2} = \cmod z^2$
== Proof ==
From Complex Modulus of Product of Complex Numbers:
: $\cmod {z_1 z_2} = \cmod {z_1} \cmod {z_2}$
for $z_1, z_2 \in \C$.
Set $z = z_1 = z_2$ and the result follows.
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Modulus
Category:Complex Multiplication
Square of Complex Number

== Theorem ==
Let $z = a + i b$ be a complex number.
Then its square is given by:
:$z^2 = a^2 - b^2 + i \paren {2 a b}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.18$
Category:Complex Numbers
Category:Square Function
Square of Coprime Number is Coprime

== Theorem ==
Let $a$ and $b$ be coprime integers:
:$a, b \in \Z: a \perp b$
Then:
:$a^2 \perp b$
== Proof ==
Let $a \perp b$.
Let $a^2 = c$.
Let $d = a$.
As $a \perp b$ it follows that $d \perp b$.
From :
:$a d \perp b$
But $a d = c = a^2$.
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Square of Covariance is Less Than or Equal to Product of Variances

== Theorem ==
Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ exist and be finite. 
Then: 
:$\paren {\cov {X, Y} }^2 \le \var X \, \var Y$
where $\cov {X, Y}$ denotes the covariance of $X$ and $Y$. 
 
== Proof == 
We have, by the definition of variance, that both: 
:$\expect {\paren {X - \expect X}^2}$
and: 
:$\expect {\paren {Y - \expect Y}^2}$
exist and are finite. 
Therefore:
== Sources == 
* : $4.6$: Covariance and Correlation: Theorem $4.6.3$
Category:Covariance
Category:Variance
Square of Cube Number is Cube

== Theorem ==
Let $a \in \N$ be a natural number.
Let $a$ be a cube number.
Then $a^2$ is also a cube number.
 
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Category:Square of Cube Number is Cube
Square of Difference

== Theorem ==
:$\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$
== Algebraic Proof 1 ==
== Algebraic Proof 2 ==
== Geometric Proof ==
== Sources ==
* : $\S 2$: Special Products and Factors: $2.2$
* : $\S 2$: Special Products and Factors: $2.2.$
Category:Square Function
Category:Square of Difference
Square of Difference/Algebraic Proof 2

== Theorem ==
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2$ and $y = -y$.
Category:Square of Difference
Square of Difference/Geometric Proof

== Theorem ==
== Proof ==
:
That is: 
:$x^2 + y^2 = \paren {x - y}^2 + 2 x y$
Let the straight line $AB$ be cut at random at $C$.
Construct the square $ADEB$ on $AB$ and join $DB$.
Construct $CN$ parallel to $AD$ through $C$ and let it cross $DB$ at $G$.
Construct $HF$ parallel to $AB$ through $G$.
From Complements of Parallelograms are Equal, $\Box ACGH = \Box FGNE$, so add $\Box CBFG$ to each.
So the whole of $\Box ABFH$ equals the whole of $\Box CBEN$.
So $2 \Box ABFH = \Box ABFH + \Box CBEN$.
But $\Box ABFH + \Box CBEN$ equals the area of the gnomon $KLM$ together with $\Box CBFG$.
So the gnomon $KLM$ together with $\Box CBFG$ equals $2 \Box ABFH$.
But twice the rectangle contained by $AB$ and $BC$ is also equal to $2 \Box ABFH$, as $BG = BC$.
So the gnomon $KLM$ together with $\Box CBFG$ equals twice the rectangle contained by $AB$ and $BC$.
Add $\Box DHGN$ to each.
Note that $\Box DHGN$ equals the square on $AC$.
Then the gnomon $KLM$ together with $\Box CBFG$ and $\Box DHGN$ equals the whole of $\Box ABED$ and $\Box CBFG$, which are the squares on $AB$ and $BC$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Square of Difference
Square of Element Less than Unity in Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
Let $x \in D$ such that $0 < x < 1$.
Then:
: $0 < x \times x < x$
== Proof ==
We have that $0 < x < 1$.
From Relation Induced by Strict Positivity Property is Compatible with Multiplication:
:$0 \times x < x \times x < 1 \times x$
Hence the result.
Category:Ordered Integral Domains