nyuuzyou/wikis huggingface dataset:
Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares

== Theorem ==
Let $X$ and $Y$ be random variables. 
Let the expectation of $X Y$, $\expect {X Y}$, exist and be finite. 
Then: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
== Proof ==
Note that: 
:$\map \Pr {Y^2 \ge 0} = 1$
so we have by Expectation of Non-Negative Random Variable is Non-Negative: 
:$\expect {Y^2} \ge 0$
First, take $\expect {Y^2} > 0$. 
Let $Z$ be a random variable with: 
:$Z = X - Y \dfrac {\expect {X Y} } {\expect {Y^2} }$
Note that we have: 
:$\map \Pr {Z^2 \ge 0} = 1$
so again applying Expectation of Non-Negative Random Variable is Non-Negative, we have: 
:$\expect {Z^2} \ge 0$
That is: 
We therefore have: 
:$\dfrac {\paren {\expect {X Y} }^2} {\expect {Y^2} } \le \expect {X^2}$
giving: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
as required. 
It remains to address the case $\expect {Y^2} = 0$. 
Note that since $\map \Pr {Y^2 \ge 0} = 1$, from Condition for Expectation of Non-Negative Random Variable to be Zero we necessarily have: 
:$\map \Pr {Y^2 = 0} = 1$
That is: 
:$\map \Pr {Y = 0} = 1$
This implies that the random variable $X Y$ has: 
:$\map \Pr {X Y = 0} = 1$
From which, we have that: 
:$\map \Pr {X Y \ge 0} = 1$
So, applying Expectation of Non-Negative Random Variable is Non-Negative again we have: 
:$\expect {X Y} = 0$
With that, we have: 
:$\paren {\expect {X Y} }^2 = 0$
and: 
:$\expect {X^2} \expect {Y^2} = \expect {X^2} \times 0 = 0$
So the inequality: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
also holds in the case $\expect {Y^2} = 0$, completing the proof.
== Also see ==
* Hölder's Inequality, of which this is an instantiation 
== Sources == 
* : $4.6$: Covariance and Correlation: Theorem $4.6.2$
Category:Expectation
Square of Golden Mean equals One plus Golden Mean

== Theorem ==
:$\phi^2 = \phi + 1$
where $\phi$ denotes the golden mean.
=== Decimal Expansion ===
== Proof ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $\S 1.2.1$: Mathematical Induction
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Golden Mean
Square of Golden Mean equals One plus Golden Mean/Decimal Expansion

== Decimal Expansion of Square of Golden Mean ==
The decimal expansion of $\phi^2$ is given as:
:$\phi^2 \approx 2 \cdotp 61803 \, 39887 \, 49894$
Thus the square of the golden mean is the unique number $n$ such that:
:$\sqrt n = n - 1$
== Sources ==
* : $2 \cdotp 618 \, 033 \ldots$
* : $2 \cdotp 61803 \, 3 \ldots$
Category:Golden Mean
Square of Hypotenuse of Pythagorean Triangle is Difference of two Cubes/Historical Note

== Historical Note on Square of Hypotenuse of Pythagorean Triangle is Difference of two Cubes ==
The statement:
appears in the $1997$ edition of by .
There is no record of where he may have found this statement, but it is demonstrably false.
== Sources ==
* : $13$
Category:Historical Notes
Square of Hypotenuse of Pythagorean Triangle is Difference of two Cubes/Mistake

== Source Work ==
:
:The Dictionary
::$13$
This statement does not appear in the first edition of of $1986$.
== Mistake ==
:The square of the hypotenuse of a right-angled triangle is also the difference of $2$ cubes; thus, $13^2 = 8^3 - 7^3$.
== Correction ==
This is not true for all right-angled triangles.
The most immediate counterexample is the $3-4-5$ triangle, whose hypotenuse is $5$.
The sequence of positive integers whose squares are the difference between $2$ cubes begins:
:$13, 28, 49, 104, 147, 181, 189, 224, 351, 361, \dotsc$
== Refutation ==
== Sources ==
* : $13$
Category:Mistakes/Curious and Interesting Numbers
Square of Hypotenuse of Pythagorean Triangle is Difference of two Cubes/Refutation

== Theorem ==
Let $h$ be the hypotenuse of a Pythagorean triangle.
Then it is not necessarily the case that:
:$h^2 = a^3 - b^3$
for some $a, b \in \Z_{>0}$.
== Refutation ==
Consider the equation:
:$(1): \quad h^2 = a^3 - b^3: a, b \in \Z_{>0}$
$h$ itself cannot be a cube, as this would be a counterexample to Fermat's Last Theorem.
Suppose we relax $\Z_{>0}$ to $\Z_{\ge 0}$.
It can be shown that $(1)$ does not hold even among the primitive Pythagorean triples.
We demonstrate that there is no solution for $h = 5$:
 there is.
We must have $\paren {a - b} \divides h^2$.
For $x \ge h$:
::$\paren {x + 1}^3 - x^3 = 3 x^2 + 3 x + 1 > h^2$
Hence $a \le 5$.
Combined with our divisibility condition, we must have:
:$a - b = 1$ or $5$
But:
:$2^3 < 25$
:$3^3 - 2^3 = 19$
:$4^3 - 3^3 > 25$
:$5^3 - 0^3 > 25$
So there is no solution for $a^3 - b^3 = 25$.
== Historical Note ==
Category:Pythagorean Triangles
Category:Cube Numbers
Square of Inner Product Norm of Sum

== Theorem ==
Let $\struct {X, \innerprod \cdot \cdot}$ be an inner product space. 
Let $\norm {\, \cdot \,}$ be the inner product norm. 
Let $x, y \in X$.
Then, we have:
:$\norm {x + y}^2 = \norm x^2 + 2 \map \Re {\innerprod x y} + \norm y^2$
== Proof ==
Category:Inner Product Spaces
Square of Metric does not necessarily form Metric

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $d_4: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \paren {\map d {x, y} }^2$
Then $d_4$ may or may not be a metric for $A$.
== Proof ==
Let $d$ be the standard discrete metric on $M$.
Then:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \map d {x, y}$
and indeed in this case $d_4$ is a metric for $A$.
Let $M = \struct {\R, d}$ be the real number line with the usual (Euclidean) metric.
Let $x = 1$, $y = \dfrac 1 2$ and $z = 0$.
We have:
So we have:
:$\map {d_4} {x, y} + \map {d_4} {y, z} < \map {d_4} {x, z}$
and it is seen that $d_4$ does not satisfy .
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 6$
Category:Examples of Metric Spaces
Square of Modulo less One equals One

== Theorem ==
Let $m \in \Z$ be an integer.
Let $\Z_m$ be the set of integers modulo $m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
Then:
:$\eqclass {m - 1} m \times_m \eqclass {m - 1} m = \eqclass 1 m$
where $\times_m$ denotes multiplication modulo $m$.
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.5$
Category:Modulo Arithmetic
Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is $0_D$.
Then:
:$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$
where $\map P {x \times x}$ denotes that $x \times x$ has the (strict) positivity property.
That is, the square of any element of an ordered integral domain is (strictly) positive that element is non-zero.
== Proof ==
Let $x = 0_D$.
Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero.
Thus by definition of strict positivity property:
:$\neg \map P {0_D \times 0_D}$
Now suppose $x \ne 0_D$.
One of two cases applies:
:$\map P x$
:$\neg \map P x$
Let $\map P x$.
Then by definition of (strict) positivity:
:$\map P {x \times x}$
Now suppose $\neg \map P x$.
Then by the trichotomy law of ordered integral domains:
:$\map P {-x}$
Then again by definition:
:$\map P {\paren {-x} \times \paren {-x} }$.
But by Product of Ring Negatives:
:$\paren {-x} \times \paren {-x} = x \times x$
So again:
:$\map P {x \times x}$
Hence the result.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $7 \ \text{(i)}$
Category:Ordered Integral Domains
Square of Non-Zero Real Number is Strictly Positive

== Theorem ==
:$\forall x \in \R: x \ne 0 \implies x^2 > 0$
== Proof ==
There are two cases to consider:
:$(1): \quad x > 0$
:$(2): \quad x < 0$
Let $x > 0$.
Then:
Let $x < 0$.
Then:
== Also see ==
* Square of Real Number is Non-Negative
== Sources ==
* 
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(f)}$
* 
Category:Square Function
Category:Real Numbers
Square of Norm of Vector Cross Product

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\times$ denote the vector cross product.
Then:
:$\norm {\mathbf a \times \mathbf b}^2 = \norm {\mathbf a}^2 \norm {\mathbf b}^2 - \paren {\mathbf a \cdot \mathbf b}^2$
== Proof ==
Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
Then:
== Sources ==
* : $\S 7.5$
Category:Vector Cross Product
Square of Number Always Exists

== Theorem ==
Let $x$ be a number.
Then its square $x^2$ is guaranteed to exist.
== Proof ==
Whatever flavour of number under discussion, the algebraic structure $\struct {\mathbb K, +, \times}$ in which this number sits is at least a semiring.
The binary operation that is multiplication is therefore closed on that algebraic structure.
Therefore:
: $\forall x \in \mathbb K: x \times x \in \mathbb K$
Category:Numbers
Square of Odd Multiple of 3 is Difference between Triangular Numbers

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $T_n$ denote the $n$th triangular number.
Let $m = 2 n + 1$ be an odd integer
Then:
:$\paren {3 m}^2 = T_{9 n + 4} - T_{3 n + 1}$
== Proof ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $4$
Category:Triangular Numbers
Category:Square Numbers
Square of Odd Number as Difference between Triangular Numbers

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\exists a, b \in \Z_{\ge 0}: \paren {2 n + 1}^2 = T_a - T_b$
where:
:$T_a$ and $T_b$ are triangular numbers
:$T_a$ and $T_b$ are coprime.
That is, the square of every odd number is the difference between two coprime triangular numbers.
== Proof ==
Let $a = 3b + 1$
The square of every odd number can be made through this method.
Both triangular numbers are coprime.
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Square Numbers
Category:Coprime Integers
Square of Ones Matrix

== Theorem ==
Let $\mathbf J = \sqbrk 1_n$ be a square ones matrix of order $n$.
Then $\mathbf J^2 = n \mathbf J$.
That is:
:$\begin{bmatrix}
1 & 1 & \cdots & 1 \\
1 & 1 & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \cdots & 1
\end{bmatrix}^2 = \begin{bmatrix}
n & n & \cdots & n \\
n & n & \cdots & n \\
\vdots & \vdots & \ddots & \vdots \\
n & n & \cdots & n
\end{bmatrix}$
== Proof ==
Follows directly from the definition of matrix multiplication:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
In this case, $m = n$ and $a_{i k} = b_{k j} = 1$.
Hence:
:$\ds c_{i j} = \sum_{k \mathop = 1}^n 1 \times 1 = n$
Category:Matrix Algebra
Square of Pythagorean Prime is Hypotenuse of Pythagorean Triangle

== Theorem ==
Let $p$ be a Pythagorean prime.
Then $p^2$ is the hypotenuse of a Pythagorean triangle.
== Proof ==
By Fermat's Two Squares Theorem, a Pythagorean prime, $p$ can be expressed in the form:
:$p = m^2 + n^2$
where $m$ and $n$ are (strictly) positive integers.
:$(1): \quad m \ne n$, otherwise $p$ would be of the form $2 m^2$ and so even and therefore not a prime.
:$(2): \quad m \perp n$, otherwise $\exists c \in \Z: p = c^2 \paren {p^2 + q^2}$ for some $p, q \in \Z$ and therefore not a prime.
:$(3): \quad m$ and $n$ are of opposite parity, otherwise $p = m^2 + n^2$ is even and therefore not a prime.
, let $m > n$.
From Solutions of Pythagorean Equation, the triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ forms a primitive Pythagorean triple such that $m^2 + n^2$ forms the hypotenuse of a primitive Pythagorean triangle.
Thus we have that $p^2 = r^2 + s^2$ for some (strictly) positive integers $r$ and $s$.
Similarly, we have that $r \ne s$, otherwise $p^2$ would be of the form $2 r^2$ and so not a square number.
, let $r > s$.
From Solutions of Pythagorean Equation, the triple $\tuple {2 r s, r^2 - s^2, r^2 + s^2}$ forms a Pythagorean triple such that $p^2 = r^2 + s^2$ forms the hypotenuse of a Pythagorean triangle.
 
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Square of Quadratic Gauss Sum

== Theorem ==
Let $p$ be an odd prime.
Let $a$ be an integer coprime to $p$.
Let $\map g {a, p}$ denote the quadratic Gauss sum of $a$ and $p$.
Then:
:$\map g {a, p}^2 = \paren {\dfrac {-1} p} \cdot p$
== Proof ==
Category:Number Theory
Square of Random Variable with t-Distribution has Snedecor's F-Distribution

== Theorem ==
Let $k$ be a strictly positive integer. 
Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.
Then: 
:$X^2 \sim F_{1, k}$
where $F_{1, k}$ is Snedecor's $F$-distribution with $\tuple {1, k}$ degrees of freedom.
== Proof ==
Let $Y \sim F_{1, k}$. 
We aim to show that: 
:$\map \Pr {Y < x^2} = \map \Pr {\size X < x}$
for all $x \ge 0$. 
That is: 
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all $x \ge 0$. 
We have: 
== Sources ==
* 
* 
Category:Student's t-Distribution
Category:Snedecor's F-Distribution
Square of Real Number is Non-Negative

== Theorem ==
Let $x \in \R$.
Then:
: $0 \le x^2$
== Proof ==
There are two cases to consider:
: $(1): \quad x = 0$
: $(2): \quad x \ne 0$
Let $x = 0$.
Then:
:$x^2 = 0$
and thus
:$0 \le x^2$
Let $x \ne 0$.
From Square of Non-Zero Real Number is Strictly Positive it follows that:
:$0 < x^2$
and so by definition:
:$0 \le x^2$
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (1)$
Category:Square Function
Category:Real Analysis
Square of Repunit times Sum of Digits

== Theorem ==
The following pattern emerges:
and so on, up until $999 \, 999 \, 999^2$ after which the pattern breaks down.
== Proof ==
From Square of Repunit:
and so on.
Then from 1+2+...+n+(n-1)+...+1 = n^2:
and so on.
Then:
The pattern breaks down after $9$:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
== Sources ==
* : $1,234,321$
* : $1,234,321$
Category:Repunits
Category:Square Numbers
Square of Reversal of Small-Digit Number

== Theorem ==
Let $n$ be an integer whose decimal representation consists of sufficiently small digits.
Then the reversal of the square of $n$ is the square of the reversal of $n$.
== Proof ==
== Examples ==
== Sources ==
* : $12$
* : $12$
Category:Recreational Mathematics
Category:Square of Reversal of Small-Digit Number
Category:Square Numbers
Category:Reversals
Square of Reversal of Small-Digit Number/Examples/12

== Example of Square of Reversal of Small-Digit Number ==
:$12^2 = 144$
:$21^2 = 441$
== Proof ==
== Sources ==
* : $12$
* : $144$
* : $12$
* : $144$
Category:Square of Reversal of Small-Digit Number
Category:12
Square of Reversal of Small-Digit Number/Examples/13

== Example of Square of Reversal of Small-Digit Number ==
:$13^2 = 169$
:$31^2 = 961$
== Proof ==
== Sources ==
* : $12$
* : $169$
* : $12$
* : $169$
Category:Square of Reversal of Small-Digit Number
Category:13
Square of Riemann Zeta Function

== Theorem ==
:$\ds \map {\zeta^2} z = \sum_{k \mathop = 1}^\infty \frac {\map {\sigma_0} k} {k^z}$
where:
:$\zeta$ is the Riemann zeta function
:$\sigma_0$ is the divisor count function.
== Proof ==
Expanding this product, we get:
We see that each $\dfrac 1 {n^z}$ term in this sum will occur as many times as there are ways represent $n$ as $a b$, counting order.
But this is precisely the number of divisors of $n$, since each way of representing $n = a b$ corresponds to the first term $a$ of the product.
Hence this sum is:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\sigma_0} n} {z^n}$
as desired.
Category:Riemann Zeta Function
Square of Sample Correlation Coefficient is no greater than 1

== Definition ==
Let $S$ be a set of paired observations of a statistic from a population.
The sample correlation coefficient $r$ of $S$ satisfies the inequality:
:$r^2 \le 1$
== Proof ==
Let $\ds \bar x = \frac 1 n \sum_{i \mathop = 1}^n x_i$ and $\ds \bar y = \frac 1 n \sum_{i \mathop = 1}^n y_i$.
== Sources ==
* : $(2)$
* : $(2)$
Category:Sample Correlation Coefficients
Category:Statistics
Square of Small-Digit Palindromic Number is Palindromic

== Theorem ==
Let $n$ be an integer such that the sum of the squares of the digits of $n$ in decimal representation is less than $10$.
Let $n$ be palindromic.
Then $n^2$ is also palindromic.
The sequence of such numbers begins:
:$0, 1, 2, 3, 11, 22, 101, 111, 121, 202, 212, 1001, 1111, \dots$
== Proof ==
Let $\ds n = \sum_{k \mathop = 0}^m a_k 10^k$ be a number satisfying the conditions above.
Then:
Consider $\ds n^2 = \paren {\sum_{k \mathop = 0}^m a_k 10^k}^2$.
From definition of Multiplication of Polynomials, the coefficient of $10^l$ in the product is:
so no carries occur in the multiplication, and this form satisfies Basis Representation Theorem.
Moreover:
so the coefficient of $10^{2 m - l}$ is equal to the coefficient of $10^l$ in the expansion of $n^2$.
This shows that $n^2$ is palindromic.
== Examples ==
== Sources ==
* : $22$
* : $22$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:Palindromic Numbers
Category:Square Numbers
Category:Recreational Mathematics
Square of Small-Digit Palindromic Number is Palindromic/Examples

== Examples of Square of Small-Digit Palindromic Number is Palindromic ==
=== Square of $11$ ===
=== Square of $22$ ===
=== Square of $111$ ===
=== Square of $121$ ===
=== Square of $212$ ===
=== Square of $1111$ ===
=== Square of $11 \, 111$ ===
=== Square of $111 \, 111$ ===
Category:Square of Small-Digit Palindromic Number is Palindromic
Square of Small-Digit Palindromic Number is Palindromic/Examples/11

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$11^2 = 121$
== Proof ==
== Sources ==
* : $22$
* : $121$
* : $22$
* : $121$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:11
Category:121
Square of Small-Digit Palindromic Number is Palindromic/Examples/11,111

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$11 \, 111^2 = 123 \, 454 \, 321$
== Proof ==
 11111
x 11111
---------
 11111
 111110
 1111100
 11111000
111110000
---------
123454321
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:11,111
Square of Small-Digit Palindromic Number is Palindromic/Examples/111

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$111^2 = 12321$
== Proof ==
 111
x 111
-----
 111
 1110
11100
-----
12321
== Sources ==
* : $22$
* : $22$
* : $12,321$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:111
Square of Small-Digit Palindromic Number is Palindromic/Examples/111,111

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$111 \, 111^2 = 12 \, 345 \, 654 \, 321$
== Proof ==
 111111
x 111111
----------
 111111
 1111110
 11111100
 111111000
 1111110000
11111100000
-----------
12345654321
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:111,111
Square of Small-Digit Palindromic Number is Palindromic/Examples/1111

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$1111^2 = 1234321$
== Proof ==
 1111
x 1111
-------
 1111
 11110
 111100
1111000
-------
1234321
== Sources ==
* : $22$
* : $22$
* : $12,321$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:1111
Square of Small-Digit Palindromic Number is Palindromic/Examples/121

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$121^2 = 14641$
== Proof ==
 121
x 121
-----
 121
 2420
12100
-------
14641
== Sources ==
* : $22$
* : $22$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:121
Square of Small-Digit Palindromic Number is Palindromic/Examples/212

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$212^2 = 44944$
== Proof ==
 212
x 212
-----
 424
 2120
42400
-------
44944
== Sources ==
* : $22$
* : $22$
* : $44,944$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:212
Square of Small-Digit Palindromic Number is Palindromic/Examples/22

== Example of Square of Small-Digit Palindromic Number is Palindromic ==
::$22^2 = 484$
== Proof ==
== Sources ==
* : $22$
* : $484$
* : $22$
* : $484$
Category:Square of Small-Digit Palindromic Number is Palindromic
Category:22
Category:484
Square of Small Repunit is Palindromic

== Theorem ==
The squares of repunits with up to $9$ digits are palindromic.
== Proof ==
but:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunits
Category:Square Numbers
Square of Small Repunit is Palindromic/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$1,111,111,111,111,111,111$
== Mistake ==
:The squares of repunits make a pretty pattern:
:For example: ... $1111111111111^2 = 12345678900987654321$
== Correction ==
That should read:
:$1111111111^2 = 1234567900987654321$
The repunit in question is of index $13$, not $10$, and there is an extra $8$ in its square.
The first of these mistakes has been corrected in of $1997$, but not the second.
== Sources ==
* : $1,111,111,111,111,111,111$
Category:Mistakes/Curious and Interesting Numbers
Square of Small Repunit is Palindromic/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$1,111,111,111,111,111,111$
== Mistake ==
:The squares of repunits make a pretty pattern:
:For example: ... $1,111,111,111^2 = 12,345,678,900,987,654,321$
== Correction ==
That should read:
:$1,111,111,111^2 = 1,234,567,900,987,654,321$
There is an extra $8$ in the square where it should not be.
== Sources ==
* : $1,111,111,111,111,111,111$
Category:Mistakes/Curious and Interesting Numbers
Square of Standard Normal Random Variable has Chi-Squared Distribution

== Theorem ==
Let $X \sim \Gaussian 0 1$ where $\Gaussian 0 1$ is the standard normal distribution. 
Then $X^2 \sim \chi^2_1$ where $\chi^2_1$ is the chi-square distribution with $1$ degree of freedom.
== Proof ==
Let $Y \sim \chi^2_1$. 
We aim to show that: 
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all real $x \ge 0$.
We have: 
Category:Normal Distribution
Category:Chi-Squared Distribution
Square of Sum

== 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: $2.1$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* 
* 
* : $\S 2$: Special Products and Factors: $2.1.$
Category:Square of Sum
Category:Square Function
Square of Sum/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 $n = 2$.
== Sources ==
* : $\S 20$: Binomial Series: $20.5$
* : $\S 22$: Taylor Series: Binomial Series: $22.5.$
Category:Square of Sum