nyuuzyou/wikis huggingface dataset:
Square Modulo 3/Corollary 1

== Corollary to Square Modulo 3 ==
Let $x, y \in \Z$ be integers.
Then:
:$3 \divides \paren {x^2 + y^2} \iff 3 \divides x \land 3 \divides y$
where $3 \divides x$ denotes that $3$ divides $x$.
== Proof ==
=== Sufficient Condition ===
Let $3 \divides x \land 3 \divides y$.
Then by definition of divisibility:
:$x \equiv 0 \pmod 3$
and
:$y \equiv 0 \pmod 3$
From Square Modulo 3:
:$x^2 \equiv 0 \pmod 3$
and
:$y^2 \equiv 0 \pmod 3$
Then from Modulo Addition is Well-Defined:
:$\paren {x^2 + y^2} \equiv 0 \pmod 3$
Thus:
:$3 \divides \paren {x^2 + y^2}$
=== Necessary Condition ===
Now suppose $3 \divides \paren {x^2 + y^2}$.
Then by definition of divisibility:
:$\paren {x^2 + y^2} \pmod 3$
From Square Modulo 3:
: $x^2 \equiv 0 \pmod 3$ or $x^2 \equiv 1 \pmod 3$
and
: $y^2 \equiv 0 \pmod 3$ or $y^2 \equiv 1 \pmod 3$
Thus the only way $\paren {x^2 + y^2} \equiv 0 \pmod 3$ is for:
: $x^2 \equiv 0 \pmod 3$
and:
: $y^2 \equiv 0 \pmod 3$
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.9$
Category:Square Modulo 3
Square Modulo 3/Corollary 2

== Corollary to Square Modulo 3 ==
Let $x, y, z \in \Z$ be integers.
Then:
:$x^2 + y^2 = 3 z^2 \iff x = y = z = 0$
== Proof ==
Proof by the Method of Infinite Descent:
Suppose $u, v, w$ are the smallest non-zero positive integers such that $u^2 + v^2 = 3 w^2$.
Then from Square Modulo 3: Corollary 1 each of $u, v, w$ are multiples of $3$.
So we have $u', v', w'$ such that:
:$3 u' = u, 3 v' = v, 3 w' = w$
Then:
:$\paren {3 u'}^2 + \paren {3 v'}^2 = 3 \paren {3 w'}^2$
which leads to:
:$\paren {u'}^2 + \paren {v'}^2 = 3 \paren {w'}^2$
contradicting the supposition that $u, v, w$ are the smallest non-zero positive integers such that $u^2 + v^2 = 3 w^2$.
Thus:
: $x^2 + y^2 = 3 z^2 \implies x = y = z = 0$
If $x = y = z = 0$ then it trivially follows that $x^2 + y^2 = 3 z^2$.
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.9$
Category:Square Modulo 3
Square Modulo 3/Corollary 3

== Corollary to Square Modulo 3 ==
Let $n \in \Z$ be an integer such that:
:$3 \nmid n$
where $\nmid$ denotes non-divisibility.
Then:
:$3 \divides n^2 - 1$
where $\divides$ denotes divisibility.
== Proof ==
From Square Modulo 3:
:$n \equiv 0 \pmod 3 \iff n^2 \equiv 0 \pmod 3$
Hence also from Square Modulo 3:
:$n \not \equiv 0 \pmod 3 \iff n^2 \equiv 1 \pmod 3$
That is:
$3 \nmid n \iff 3 \divides n^2 - 1$
== Sources ==
* : $\text {3-2}$ Fermat's Little Theorem: Exercise $5$
Category:Square Modulo 3
Square Modulo 4

== Theorem ==
Let $x \in \Z$ be an integer.
Then $x$ is:
:even $x^2 \equiv 0 \pmod 4$
:odd $x^2 \equiv 1 \pmod 4$
== Proof ==
== Also see ==
* Square Modulo 3
* Square Modulo 8
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm
Category:Modulo Arithmetic
Category:Square Numbers
Square Modulo 5

== Theorem ==
Let $x \in \Z$ be an integer.
Then one of the following holds:
=== Corollary ===
== Proof ==
Let $x$ be an integer.
Using Congruence of Powers throughout, we make use of $x \equiv y \pmod 5 \implies x^2 \equiv y^2 \pmod 5$.
There are five cases to consider:
: $x \equiv 0 \pmod 5$: we have $x^2 \equiv 0^2 \pmod 5 \equiv 0 \pmod 5$.
: $x \equiv 1 \pmod 5$: we have $x^2 \equiv 1^2 \pmod 5 \equiv 1 \pmod 5$.
: $x \equiv 2 \pmod 5$: we have $x^2 \equiv 2^2 \pmod 5 \equiv 4 \pmod 5$.
: $x \equiv 3 \pmod 5$: we have $x^2 \equiv 3^2 \pmod 5 \equiv 4 \pmod 5$.
: $x \equiv 4 \pmod 5$: we have $x^2 \equiv 4^2 \pmod 5 \equiv 1 \pmod 5$.
== Sources ==
* : $\S 14$: Worked Example $1$: Congruence modulo $m$ ($m \in \N$)
Category:Modulo Arithmetic
Category:Square Numbers
Square Modulo 5/Corollary

== Theorem ==
When written in conventional base $10$ notation, no square number ends in one of $2, 3, 7, 8$.
== Proof ==
The absence of $2$ and $3$ from the digit that can end a square follows directly from Square Modulo 5.
As $7 \equiv 2 \pmod 5$ and $8 \equiv 3 \pmod 5$, the result for $7$ and $8$ follows directly.
== Sources ==
* : $\S 14$: Worked Example $1$: Congruence modulo $m$ ($m \in \N$)
Category:Modulo Arithmetic
Category:Square Numbers
Square Modulo 8

== Theorem ==
Let $x \in \Z$ be an integer.
: If $x$ is even then:
::: $x^2 \equiv 0 \pmod 8$ or $x^2 \equiv 4 \pmod 8$
: If $x$ is odd then:
::: $x^2 \equiv 1 \pmod 8$
== Proof ==
=== Proof for Even Integer ===
Let $x \in \Z$ be even.
Then from Square Modulo 4:
: $x^2 \equiv 0 \pmod 4$
Hence there are two possibilities for $x^2$:
: $x^2 \equiv 0 \pmod 8$
: $x^2 \equiv 4 \pmod 8$
The fact that there do exist such squares can be demonstrated by example:
: $2^2 = 4 \equiv 4 \pmod 8$
: $4^2 = 16 \equiv 0 \pmod 8$
=== Proof for Odd Integer ===
Let $x \in \Z$ be odd.
Then from Odd Square Modulo 8:
: $x^2 \equiv 1 \pmod 8$
Category:Modulo Arithmetic
Category:Square Numbers
Square Number/Examples

== Examples of Square Numbers ==
The first few square numbers are as follows:
:
=== Sequence of Square Numbers ===
== Sources ==
* : Fig. $160 \ \text{(b)}$
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $\text {A}.13$: Fermat ()
* 
* 
Category:Square Numbers
Square Number/Sequence

== Definition ==
The sequence of square numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, \ldots$
== Sources ==
* : Tables: $1$ The First $100$ Triangular Numbers, Squares and Cubes
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $\text {A}.13$: Fermat ()
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Tables: $1$ The First $100$ Triangular Numbers, Squares and Cubes
* 
* 
Category:Square Numbers
Square Number Less than One

== Theorem ==
Let $x$ be a real number such that $x^2 < 1$.
Then:
:$x \in \openint {-1} 1$
where $\openint {-1} 1$ is the open interval $\left\{{x \in \R: -1 < x < 1}\right\}$.
== Proof ==
First note that from Square of Real Number is Non-Negative:
:$x^2 \ge 0$
From Ordering of Squares in Reals:
:$(1): \quad x > 1 \implies x^2 > 1$
:$(2): \quad x < 1 \implies x^2 < 1$
From Identity Element of Multiplication on Numbers:
:$1^2 = 1$
so it is clear that the strict inequalities apply above.
For clarity, therefore, the case where $x = \pm 1$ can be ignored.
Suppose $x \notin \openint {-1} 1$.
Then either:
:$x < -1$
or:
:$x > 1$
If $x > 1$ then:
:$x^2 > 1$
If $x < -1$ then:
:$-x > 1$
From Square of Real Number is Non-Negative:
:$\paren {-x}^2 = x^2$
So again $x^2 > 1$.
Thus:
:$x \notin \openint {-1} 1 \implies x^2 > 1$
So by the Rule of Transposition:
:$x^2 < 1 \implies x \in \openint {-1} 1$
Category:Real Analysis
Square Number ending in 9 Digits in Reverse Order

== Theorem ==
The next integer after $1 \, 111 \, 111 \, 111$ whose square ends in $987 \, 654 \, 321$ is:
:$2 \, 380 \, 642 \, 361^2 = 5 \, 667 \, 458 \, 050 \, 987 \, 654 \, 321$
== Proof ==
== Historical Note ==
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Square Numbers
Square Number ending in 9 Digits in Reverse Order/Historical Note

== Historical Note on Square Number ending in 9 Digits in Reverse Order ==
According to , in his of $1986$, this factoid was noted by , but this needs to be corroborated.
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Historical Notes
Square Numbers which are Divisor Sum values

== Theorem ==
The sequence of square numbers which are the divisor sum value of a (strictly) positive integer begins:
:$1, 4, 36, 121, 144, 256, 324, 400, 576, 784, 900, 961, \ldots$
== Proof ==
== Also see ==
* Numbers whose Divisor Sum is Square
== Sources ==
* : $36$
Category:Divisor Sum Function
Category:Square Numbers
Square Numbers which are Divisor Sum values/Mistake

== Source Work ==
:
:The Dictionary
::$36$
== Mistake ==
:The sequence of squares which are values of $\map \sigma n$ goes: $1 \ \ 4 \ \ 36 \ \ 121 \ \ 144 \ \ 256 \ \ 576 \ \ldots$
== Correction ==
$400$ is missing from this list.
== Sources ==
* : $36$
Category:Mistakes/Curious and Interesting Numbers
Square Numbers which are Sum of Consecutive Powers

== Theorem ==
The only two square numbers which are the sum of consecutive powers of a positive integer are $121$ and $400$:
:$121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 = 11^2$
:$400 = 7^0 + 7^1 + 7^2 + 7^3 = 20^2$
== Proof ==
:$121 = 1 + 3 + 9 + 27 + 81$
:$400 = 1 + 7 + 49 + 343$
== Sources ==
* : $121$
* : $121$
Category:Square Numbers
Category:121
Category:400
Square Numbers which are Sum of Consecutive Powers/Mistake

== Source Work ==
:
:The Dictionary
::$121$
:
:The Dictionary
::$121$
== Mistake ==
:$121$ is the only square number that is the sum of consecutive powers from $1$: $121 = 1 + 3 + 9 + 27 + 81$.
== Correction ==
:$400$ also has this property: $400 = 1 + 7 + 49 + 343$.
== Sources ==
* : $121$
* : $121$
Category:Mistakes/Curious and Interesting Numbers
Square Numbers which are Sum of Sequence of Odd Cubes

== Theorem ==
The sequence of square numbers which can be expressed as the sum of a sequence of odd cubes from $1$ begins:
:$1, 1225, 1 \, 413 \, 721, 1 \, 631 \, 432 \, 881, \dotsc$
The sequence of square roots of this sequence is:
:$1, 35, 1189, 40 \, 391, \dotsc$
== Proof ==
We have that:
From Sum of Sequence of Odd Cubes we have:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$
Thus we need to find all $n$ such that $2 n^2 − 1$ is square.
This corresponds to the Pellian equation $x^2 - 2 y^2 = -1$, which has the positive integral solutions:
:$\begin {array} {r|r} x & y \\ \hline
1 & 1 \\
7 & 5 \\
41 & 29 \\
239 & 169 \\
1393 & 985 \\
\end {array}$
and so on.
By substituting $y = n$ and $x = \sqrt {2 n^2 - 1}$, we see that the products of $x$ and $y$ values give the required sequence of square roots.
== Sources ==
* : $1225$
Category:Square Numbers
Category:Sums of Sequences
Category:Cube Numbers
Square Numbers which are Sum of Sequence of Odd Cubes/Mistake

== Source Work ==
:
:The Dictionary
::$1225$
== Mistake ==
:$35^2 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3$. The next such sum is $1^3 + \ldots 29^3$.
== Correction ==
This is wrong, as can be seen by the fact that:
:$1^3 + \cdots + 29^3 = 101 \, 025$
which is not a square number.
In fact, the next such sum is $1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3$.
That is:
:$1^3 + \paren {2 \times 2 - 1}^3 + \paren {2 \times 3 - 1}^3 + \dotsb + \paren {2 \times 28 - 1}^3 + \paren {2 \times 29 - 1}^3$
What seems to have done (and is seems to be a common confusion) is confuse:
:$\ds \sum_{j \mathop = 1}^{29} \paren {2 j - 1}^3$
with:
:$1^3 + 3^3 + \dotsb + 27^3 + 29^3$
which is not the same thing at all.
== Sources ==
* : $1225$
Category:Mistakes/Curious and Interesting Numbers
Square Numbers whose Divisor Sum is Square

== Theorem ==
The sequence of square numbers whose divisor sum is square starts as follows:
== Examples ==
== Also see ==
* Numbers whose Divisor Sum is Square
* Square Numbers which are Divisor Sum values
== Historical Note ==
== Sources ==
Category:Divisor Sum Function
Category:Square Numbers
Category:Numbers whose Divisor Sum is Square
Category:Square Numbers whose Divisor Sum is Square
Square Numbers whose Divisor Sum is Square/Examples

== Examples of Square Numbers whose Divisor Sum is Square ==
=== $\sigma_1$ of $9^2$ is Square ===
=== $\sigma_1$ of $20^2$ is Square ===
=== $\sigma_1$ of $180^2$ is Square ===
=== $\sigma_1$ of $1306^2$ is Square ===
Category:Square Numbers whose Divisor Sum is Square
Square Numbers whose Divisor Sum is Square/Examples/1306

== Example of Square Number whose Divisor Sum is Square ==
::$\map {\sigma_1} {1306^2} = 1729^2$
== Proof ==
We have that $1306^2 = 1 \, 705 \, 636$.
Then from :
which equals $1729^2$.
Category:Square Numbers whose Divisor Sum is Square
Category:1306
Square Numbers whose Divisor Sum is Square/Examples/20

== Example of Square Number whose Divisor Sum is Square ==
::$\map {\sigma_1} {20^2} = 31^2$
== Proof ==
We have that $20^2 = 400$.
Then from :
which equals $31^2$.
== Sources ==
* : $81$
Category:Square Numbers whose Divisor Sum is Square
Category:20
Square Order 2 Matrices over Real Numbers form Ring with Unity

== Theorem ==
Let $S$ denote the set of square matrices of order $2$ whose entries are the set of real numbers.
Then $S$ forms a non-commutative ring with unity whose unity is the matrix $\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$.
== Proof ==
This is an instance of Ring of Square Matrices over Field is Ring with Unity.
== Sources ==
* : $\S 55 \ (2)$. Special types of ring and ring elements
* 
* 
Category:Examples of Rings with Unity
Square Product 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 $T_n \times T_{n + 1} \times T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 3$
== Proof ==
Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.
We see that:
Thus $n = 3$ appears to satisfy the conditions.
It remains for us to check:
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $5 \ \text {(b)}$
Category:Triangular Numbers
Category:Square Numbers
Square Pyramidal Number also Square

== Theorem ==
$4900$ is the only square pyramidal number which is also square:
:$4900 = 70^2 = \ds \sum_{k \mathop = 1}^{24} k^2 = \dfrac {24 \paren {24 + 1} \paren {2 \times 24 + 1} } 6$
== Proof ==
== Sources ==
* : $4900$
* : $4900$
Category:Pyramidal Numbers
Category:Square Numbers
Category:4900
Square Pyramidal and Triangular Numbers

== Theorem ==
The only positive integers which are simultaneously square pyramidal and triangular are:
:$1, 55, 91, 208 \, 335$
== Proof ==
== Sources ==
* 
* : $55$
* : $196,560$
* 
* : $55$
* : $208,335$
Category:Triangular Numbers
Category:Pyramidal Numbers
Square Root/Examples

== Examples of Square Roots ==
=== Square Root of $9$ ===
The decimal expansions of the square roots of various numbers are as follows:
=== Square Root of $2$ ===
=== Square Root of $3$ ===
=== Square Root of $5$ ===
=== Square Root of $10$ ===
=== Square Root of $e$ ===
Category:Examples of Square Roots
Square Root/Examples/10

== Decimal Expansion ==
The decimal expansion of $\sqrt {10}$ starts:
:$\sqrt 10 \approx 3 \cdotp 16227 \, 76601 \, 68379 \, 33199 \, 88935 \, 44432 \, 71853 \, 3719 \ldots$
== Sources ==
* : $3 \cdotp 162 \, 277 \, 660 \, 168 \, 379 \, 331 \, 198 \, 893 \, 544 \, 432 \, 718 \, 533 \, 719 \ldots$
* : $3 \cdotp 16227 \, 76601 \, 68379 \, 33119 \, 88935 \, 44432 \, 71853 \, 3719 \ldots$
Category:Examples of Square Roots
Category:10
Square Root/Examples/17

== Decimal Expansion ==
The decimal expansion of $\sqrt {17}$ starts:
:$\sqrt 17 \approx 4 \cdotp 12310 \, 56256 \, 17660 \, 5498 \ldots$
== Historical Note ==
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $4 \cdotp 123 \, 105 \ldots$
* : $4 \cdotp 12310 \, 5 \ldots$
Category:Examples of Square Roots
Category:17
Square Root/Examples/2

== Decimal Expansion ==
The decimal expansion of the square root of $2$ starts:
:$\sqrt 2 \approx 1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
== Also known as ==
:$\sqrt 2$ is sometimes known as 's constant, for .
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.3$
* : Chapter $1$: Variables and Functions: The Set of Real Numbers
* : $1$: Complex Numbers: The Real Number System: $4$
* : $1,41421 35623 73 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Category:Examples of Square Roots
Category:Square Root of 2 is Irrational
Square Root/Examples/3

== Decimal Expansion ==
The decimal expansion of $\sqrt 3$ starts:
:$\sqrt 3 \approx 1 \cdotp 73205 \, 08075 \, 68877 \, 2935 \ldots$
== Historical Note ==
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.4$
* : $1 \cdotp 732 \, 050 \, 807 \, 568 \, 877 \, 293 \, 527 \, 446 \, 341 \, 505 \, 872 \, 366 \, 942 \ldots$
* : $1 \cdotp 73205 \, 08075 \, 68877 \, 29352 \, 74463 \, 41505 \, 87236 \, 69428 \ldots$
Category:Examples of Square Roots
Square Root/Examples/3/Historical Note

== Historical Note on Square Root of $3$ ==
The square root of $3$ was the second number, after the square root of $2$, to be identified as being irrational.
This was achieved by .
 provided the approximation:
:$\dfrac {1351} {780} < \sqrt 3 < \dfrac {265} {153}$
What he actually demonstrated was:
:$26 - \dfrac 1 {52} < 15 \sqrt 3 < 26 - \dfrac 1 {51}$
These can be achieved by interpreting Pell's equation, to obtain:
:$1351^2 - 3 \times 780^2 = 1$
:$265^2 - 3 \times 153^2 = -2$
Thus it appears that was familiar with Pell's equation.
== Sources ==
* : $1 \cdotp 732 \, 050 \, 807 \, 568 \, 877 \, 293 \, 527 \, 446 \, 341 \, 505 \, 872 \, 366 \, 942 \ldots$
* : $1 \cdotp 73205 \, 08075 \, 68877 \, 29352 \, 74463 \, 41505 \, 87236 \, 69428 \ldots$
Category:Historical Notes
Square Root/Examples/5

== Decimal Expansion ==
The decimal expansion of $\sqrt 5$ starts:
:$\sqrt 5 \approx 2 \cdotp 23606 \, 79774 \, 99789 \, 6964 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.5$
* : $2 \cdotp 236 \, 067 \ldots$
* 
* 
Category:Examples of Square Roots
Square Root/Examples/Euler's Number

== Decimal Expansion ==
The decimal expansion of Euler's Number $e$ starts:
:$\sqrt e \approx 1 \cdotp 64872 \, 12707 \, 00128 \, 1468 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.22$
* : $\S 1$: Special Constants: $1.5.$
Category:Examples of Square Roots
Category:Euler's Number
Square Root is Strictly Increasing

== Theorem ==
The positive square root function is strictly increasing, that is:
:$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$
== Proof ==
Let $x$ and $y$ be positive real numbers such that $x < y$.
 $\sqrt x \ge \sqrt y$.
Thus a contradiction is created.
Therefore:
:$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$
Category:Real Numbers
Category:Square Roots
Category:Examples of Strictly Increasing Real Functions
Square Root is not Lipschitz Continuous

== Theorem ==
Let $\sqrt {\size x} : \R \to \R_{\ge 0}$ be a real function.
$\sqrt {\size x}$ is not Lipschitz continuous.
== Proof ==
 $\sqrt {\size x}$ is Lipschitz continuous.
Then:
:$\exists L \in \R_{> 0} : \forall x, y \in \R : \size {\sqrt {\size x} - \sqrt {\size y} } \le L \size {x - y}$
Suppose $x = \dfrac 1 {n^2}$ with $n \in \N$ and $y = 0$.
Then:
:$\dfrac 1 n \le \dfrac L {n^2}$
In other words:
:$\forall n \in \N : n \le L$
However, $L$ was assumed to be finite.
Hence the contradiction.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Lipschitz Continuous Functions
Square Root is of Exponential Order Epsilon

== Theorem == 
The positive square root function:
:$t \mapsto \sqrt t$
is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
== Proof ==
Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a' t}$ for any $a' > 0$, arbitrarily small in magnitude.
Therefore the inequality $\sqrt t < K e^{a t}$ has solutions of the same nature.
Category:Exponential Order
Square Root of 2/Binary Expansion

== Binary Expansion ==
The binary expansion of the square root of $2$ starts:
:$\sqrt 2 \approx 1 \cdotp 01101 \, 01000 \, 00 \ldots$
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Category:Square Root of 2 is Irrational
Square Root of 2 as Sum of Egyptian Fractions

== Theorem ==
The square root of $2$ can be approximated by the following sequence of Egyptian fractions:
:$\sqrt 2 = 1 + \dfrac 1 3 + \dfrac 1 {13} + \dfrac 1 {253} + \dfrac 1 {218 \, 201} + \dfrac 1 {61 \, 323 \, 543 \, 802} + \cdots$
== Proof ==
We have by definition of the square root of $2$ that:
:$\sqrt 2 - 1 \approx 0 \cdotp 41421 \, 35623 \, 73095 \, 04880 \ldots$
By inspection:
:$\dfrac 1 3 < \sqrt 2 - 1 < \dfrac 1 2$
Thus:
:$\sqrt 2 - 1 - \dfrac 1 3 \approx 0 \cdotp 08088 \, 02290 \, 39761 \, 71546 \ldots$
Then:
:$\dfrac 1 {13} = 0 \cdotp 07692 \, 3 \ldots$ repeating
and:
:$\dfrac 1 {12} = 0 \cdotp 08333 \ldots$ repeating
and so:
:$\sqrt 2 - 1 - \dfrac 1 3 - \dfrac 1 {13} \approx 0 \cdotp 00395 \, 71521 \, 16684 \, 79239 \cdots$
 
Thus one can generate this sequence of denominators $\sequence {D_n}$ by:
:$\ds D_n = \ceiling {\paren {\sqrt 2 - \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Square Root of 2 is Algebraic of Degree 2

== Theorem ==
The square root of $2$ is an algebraic number of degree $2$.
== Proof ==
Suppose $\sqrt 2$ could be expressed as the root of the linear function:
:$a_1 x + a_0 = 0$
for some $a_0, a_1 \in \Q$.
Then:
:$\sqrt 2 = -\dfrac {a_0} {a_1}$
and would be rational.
But as Square Root of 2 is Irrational, this is not the case.
However, $\sqrt 2$ is a root of the polynomial of degree $2$:
:$x^2 - 2 = 0$
Hence the result by definition of degree of algebraic number.
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Algebraic Numbers
Category:Square Root of 2
Square Root of 2 is Irrational

== Theorem ==
:$\sqrt 2$ is irrational.
== Classic Proof ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
By the rational root theorem, every rational root of the polynomial $n^2-2$ would have a numerator that divides $2$ and a denominator that divides $1$, i.e., the only possible rational roots: $n=\pm1,\pm2$. But $(\pm1)^2-2\neq0\neq(\pm2)^2-2$. So, its root $n=\sqrt2$ is not a rational number.
== Decimal Expansion ==
== Binary Expansion ==
== Historical Note ==
== Sources ==
* : Chapter $\text {I}$: Number: $1.1$ Introduction
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $2$
* : Chapter $1$: Numbers: Real Numbers: $4$
* : $\S 1$: Real Numbers: $\S 1.2$: The set of real numbers
* : Chapter $1$: Complex Numbers: The Real Number System: $4$
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Chapter $2$: The Logic of Shape: Pythagoras
Category:Square Root of 2 is Irrational
Category:Square Root of 2
Category:Irrationality Proofs