nyuuzyou/wikis huggingface dataset:
Parity Addition is Associative/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & : x \text { is even} \\
1 & : x \text { is odd} \end{cases}$
Thus an element of $R$ can be expressed as an arbitrary integer of the form:
:$x = 2 k + \map f x$
where:
:$k \in \Z$ is an integer
:$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd.
Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities.
Then:
The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$:
:$\paren {a + b} + c = a + \paren {b + c}$
Category:Parity Addition is Associative
Parity Addition is Commutative

== Theorem ==
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is commutative:
:$\forall a, b \in R: a + b = b + a$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$
Category:Parity Ring
Category:Parity Addition is Commutative
Parity Addition is Commutative/Proof 1

== Theorem ==
== Proof ==
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Addition is Commutative
and:
:Isomorphism Preserves Commutativity.
Category:Parity Addition is Commutative
Parity Addition is Commutative/Proof 2

== Theorem ==
== Proof ==
Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & : x \text { is even} \\
1 & : x \text { is odd} \end{cases}$
Thus an element of $R$ can be expressed as an arbitrary integer of the form:
:$x = 2 k + \map f x$
where:
:$k \in \Z$ is an integer
:$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd.
Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities.
Then:
The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$:
:$a + b = b + a$
Category:Parity Addition is Commutative
Parity Function is Homomorphism

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\map \sgn \pi$ be the sign of $\pi$.
Let the parity function of $\pi$ be defined as:
:Parity of $\pi = \begin {cases} \mathrm {Even} & : \map \sgn \pi = 1 \\ \mathrm {Odd} & : \map \sgn \pi = -1 \end {cases}$
The mapping $\sgn: S_n \to C_2$, where $C_2$ is the cyclic group of order 2, is a homomorphism.
== Proof ==
We need to show that:
:$\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$
Let $\Delta_n$ be an arbitrary product of differences.
As $\struct {\set {1, -1}, \times}$ is the parity group, the result follows immediately.
== Sources ==
* : $\S 7.4$. Kernel and image: Example $142$
* : $\S 1.10$: Example $34$
* : Chapter $9$: Permutations: Proposition $9.16$
Category:Symmetric Groups
Parity Group/Cayley Table

== Cayley Table for Parity Group ==
We can completely describe the parity group by showing its Cayley table:
:$\begin{array} {r|rr}
\struct {\set {1, -1} , \times} & 1 & -1\\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array}
\qquad
\begin{array} {r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}
\qquad
\begin{array}{r|rr}
+ & \text{even} & \text{odd} \\
\hline
\text{even} & \text{even} & \text{odd} \\
\text{odd} & \text{odd} & \text{even} \\
\end{array}$ 
Category:Examples of Cayley Tables
Category:Parity Group
Parity Group is Group

== Theorem ==
The parity group is in fact a group.
== Proof ==
We can completely describe the parity group by showing its Cayley table:
:$\begin{array}{r|rr}
\struct {\set {1, -1}, \times} & 1 & -1\\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array} \qquad
\begin{array}{r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}$
From Prime Group is Cyclic, there is only one group of order $2$, up to isomorphism.
Thus both instantiations of the parity group are isomorphic to $C_2$, the cyclic group of order $2$.
== Sources ==
* : $\S 1.3$: Example $11$
* : Chapter $\text{II}$: Groups: A Little Number Theory
Category:Parity Group
Category:Groups of Order 2
Parity Group is Only Group with 2 Elements

== Theorem ==
Let $\struct {G, \circ}$ be a group with exactly $2$ elements.
Then $\struct {G, \circ}$ is isomorphic to the parity group, which can be exemplified $\struct {\Z_2, +_2}$.
That is, the additive group of integers modulo $2$.
== Proof ==
We have that $2$ is a prime number.
Hence $\struct {\Z_2, +_2}$ is a prime group.
The result follows from Prime Groups of Same Order are Isomorphic.
== Cayley Table ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.4$
Category:Parity Group
Category:Groups of Order 2
Parity Multiplication is Associative

== Theorem ==
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is associative:
:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$
Category:Parity Ring
Category:Parity Multiplication is Associative
Parity Multiplication is Associative/Proof 1

== Theorem ==
== Proof ==
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Multiplication is Associative
and:
:Isomorphism Preserves Associativity.
Category:Parity Multiplication is Associative
Parity Multiplication is Associative/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c \in R$.
That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
:$\forall x \in R: \map f x := \begin{cases}
0 & : x \text { is even} \\
1 & : x \text { is odd} \end{cases}$
Thus an element of $R$ can be expressed as an arbitrary integer of the form:
:$x = 2 k + \map f x$
where:
:$k \in \Z$ is an integer
:$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd.
Then:
Category:Parity Multiplication is Associative
Parity Multiplication is Commutative

== Theorem ==
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $\times$ is commutative:
:$\forall a, b \in R: a \times b = b \times a$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$
Category:Parity Ring
Category:Parity Multiplication is Commutative
Parity Multiplication is Commutative/Proof 1

== Theorem ==
== Proof ==
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Multiplication is Commutative
and:
:Isomorphism Preserves Associativity.
Category:Parity Multiplication is Commutative
Parity Multiplication is Commutative/Proof 2

== Theorem ==
== Proof ==
Let $a, b \in R$.
That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
By definition of odd:
:$\text{odd} = 2 m + 1$
for some $m \in \Z$.
By definition of even:
:$\text{even} = 2 n + 0$
for some $n \in \Z$.
Thus we can define the mapping $f: R \to \Z$ as:
$\quad \forall x \in R: \map f x := \begin{cases}
0 & : x \text { is even} \\
1 & : x \text { is odd} \end{cases}$
Thus an element of $R$ can be expressed as an arbitrary integer of the form:
:$x = 2 k + \map f x$
where:
:$k \in \Z$ is an integer
:$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd.
Then:
Category:Parity Multiplication is Commutative
Parity Ring/Cayley Tables

== Cayley Tables for Parity Ring ==
The parity ring can be described completely by showing its Cayley tables:
:$\begin{array}{r|rr}
+ & \text{even} & \text{odd} \\
\hline
\text{even} & \text{even} & \text{odd} \\
\text{odd} & \text{odd} & \text{even} \\
\end{array}
\qquad
\begin{array}{r|rr}
\times & \text{even} & \text{odd} \\
\hline
\text{even} & \text{even} & \text{even} \\
\text{odd} & \text{even} & \text{odd} \\
\end{array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$
Category:Examples of Cayley Tables
Category:Parity Ring
Parity of Best Rational Approximations to Root 2

== Theorem ==
Consider the Sequence of Best Rational Approximations to Square Root of 2:
:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
where $S_1 := \dfrac 1 1$.
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 $n$.
== Proof ==
First the parity of the numerators of the terms of $\sequence S$ is established.
Let $\dfrac {p_n} {q_n}$ be a general term of $\sequence S$.
By Relation between Adjacent Best Rational Approximations to Root 2:
:$p_{n + 1} = p_n + 2 q_n$
Thus if $p_n$ is odd then so is $p_{n + 1}$.
But $p_1 = 1$ is odd.
So $p_n$ is odd for all $n$, by Principle of Mathematical Induction.
The denominators are the Pell numbers.
The result follows from Parity of Pell Numbers.
Category:Square Root of 2
Parity of Conjugate of Permutation

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Then:
:$\forall \pi, \rho \in S_n: \map \sgn {\pi \rho \pi^{-1} } = \map \sgn \rho$
where $\map \sgn \pi$ is the sign of $\pi$.
== Proof ==
As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of commutativity and associativity:
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.17$
Category:Symmetric Groups
Parity of Integer equals Parity of Positive Power

== Theorem ==
Let $p \in \Z$ be an integer.
Let $n \in \Z_{>0}$ be a strictly positive integer.
Then $p$ is even $p^n$ is even.
That is, the parity of an integer equals the parity of all its (strictly) positive powers.
== Proof ==
Proof by induction:
For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
:For all $p \in \Z$, $p$ is even $p^n$ is even.
First it is worth confirming that $P \left({0}\right)$ does not hold:
:$\forall p \in \Z: p^0 = 1$
which is not even whatever the parity of $p$.
$P \left({1}\right)$ is true, as this just says:
: $p$ is even $p$ is even
which is a tautology.
=== Basis for the Induction ===
$P \left({2}\right)$ is the case:
: $p$ is even $p^2$ is even
which is demonstrated in Parity of Integer equals Parity of its Square.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
: $p$ is even $p^k$ is even
Then we need to show:
: $p$ is even $p^{k+1}$ is even
=== Induction Step ===
This is our induction step:
Let $p$ be even.
By the induction hypothesis, $p^k$ is also even.
Then:
and so:
So $p^{k+1}$ is even.
Now suppose $p$ is not even (that is, odd).
By the induction hypothesis, $p^k$ is also not even (that is, odd).
Then:
and so:
So $p^{k+1}$ is not even (that is, odd).
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:For all $n \in \Z_{>0}$, for all $p \in \Z$, $p$ is even $p^n$ is even.
Category:Integers
Parity of Integer equals Parity of its Square

== Theorem ==
Let $p \in \Z$ be an integer.
Then $p$ is even $p^2$ is even.
== Proof ==
Let $p$ be an integer.
By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd.
=== Even case ===
=== Odd case ===
Therefore, if it is not the case that an integer is even, then it is not the case that its square is even.
Conversely, if it is the case that an integer is even (and also a perfect square), then it is the case that its square root is even.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $4$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* 
* 
Category:Parity of Integer equals Parity of its Square
Category:Square Numbers
Parity of Integer equals Parity of its Square/Even

== Theorem ==
Let $p \in \Z$ be an integer.
Let $p$ be even.
Then $p^2$ is also even.
== Proof ==
Let $p$ be an integer.
By the Division Theorem, there exist unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd.
Let $r = 0$, so:
:$p = 2 k$
Then:
:$p^2 = \paren {2 k}^2 = 4 k^2 = 2 \paren {2 k^2}$
and so $p^2$ is even.
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
Category:Parity of Integer equals Parity of its Square
Parity of Integer equals Parity of its Square/Odd

== Theorem ==
Let $p \in \Z$ be an integer.
Let $p$ be odd.
Then $p^2$ is also odd.
== Proof ==
Let $p$ be an integer.
By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where:
:$r = 0$ corresponds to the case of $p$ being even
:$r = 1$ corresponds to the case of $p$ being odd.
Let $r = 1$, so:
:$p = 2 k + 1$
Then:
:$p^2 = \paren {2 k + 1}^2 = 4 k^2 + 4 k + 1 = 2 \paren {2 k^2 + 2 k} + 1$
and so $p^2$ is odd.
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $\S 1.14$: Exercise $18 \ (3)$
Category:Parity of Integer equals Parity of its Square
Parity of Inverse of Permutation

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Then:
:$\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$
== Proof ==
From Parity Function is Homomorphism:
:$\map \sgn {I_{S_n} } = 1$
Thus:
:$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$
The result follows immediately.
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.17$
Category:Symmetric Groups
Parity of K-Cycle

== Theorem ==
Let $\pi$ be a $k$-cycle.
Then:
:$\map \sgn \pi = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$
Thus:
:$\map \sgn \pi = \paren {-1}^{k - 1}$
or equivalently:
:$\map \sgn \pi = \paren {-1}^{k + 1}$
== Proof ==
From Transposition is of Odd Parity, any transposition is of odd parity.
From K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $k - 1$ transpositions.
Thus $\pi$ is even $k$ is odd.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 81 \alpha$
* : Chapter $9$: Permutations: Corollary $9.18$
Category:Cyclic Permutations‎
Parity of Numbers in Amicable Pair

== Open Question ==
It is not known whether it is possible for the numbers in an amicable pair to have opposite parity.
All known amicable pairs consist of numbers which are either both odd or both even.
== Progress ==
If there do exist an amicable pair whose numbers are of opposite parity, then:
:the even number is either square or $2$ times a square
:the odd number is square.
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Category:Open Questions
Parity of Pell Numbers

== Theorem ==
Consider the Pell numbers $P_0, P_1, P_2, \ldots$
:$0, 1, 2, 5, 12, 29, \ldots$
$P_n$ has the same parity as $n$.
That is:
:if $n$ is odd then $P_n$ is odd
:if $n$ is even then $P_n$ is even.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$P_n$ has the same parity as $n$.
=== Basis for the Induction ===
$\map P 0$ is the case:
:$P_0 = 0$
which is even.
$\map P 1$ is the case:
:$P_1 = 1$
which is odd.
Thus $\map P n$ holds for $n = 0$ and $n = 1$.
These comprise the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true for all $j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\forall j \le k: P_j$ has the same parity as $j$
from which it is to be shown that:
:$P_{k + 1}$ has the same parity as $k + 1$.
=== Induction Step ===
This is the induction step:
By definition of Pell numbers:
:$P_{k + 1} = 2 P_k + P_{k - 1}$
$2 P_k$ is even by definition.
By Sum of Even Integers is Even, if $P_{k - 1}$ is even then so is $P_{k + 1}$.
Similarly, if $P_{k - 1}$ is odd then so is $P_{k + 1}$.
Thus the parity of $P_{k + 1}$ matches the parity of $P_{k - 1}$.
But by the induction hypothesis, the parity of $P_{k - 1}$ matches $k - 1$.
We have that:
:$k + 1 = 2 + k - 1$
and so the parity of $k + 1$ matches the parity of $k - 1$.
Thus the parity of $P_{k + 1}$ matches $k + 1$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Strong Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: P_n$ has the same parity as $n$.
Category:Pell Numbers
Parity of Smaller Elements of Primitive Pythagorean Triple

== Theorem ==
Let $\left({x, y, z}\right)$ be a Pythagorean triple, that is, integers such that $x^2 + y^2 = z^2$.
Then $x$ and $y$ are of opposite parity.
== Proof ==
From Smaller Elements of Pythagorean Triple not both Odd, $x$ and $y$ are not both odd.
 $x$ and $y$ are both even.
Then by definition they have $2$ as a common divisor.
But from Elements of Primitive Pythagorean Triple are Pairwise Coprime this cannot be the case.
So by Proof by Contradiction, $x$ and $y$ cannot both be even.
It follows that $x$ and $y$ are of opposite parity.
Category:Pythagorean Triples
Parseval's Theorem/Formulation 1

== Theorem == 
Let $f$ be a real function which is square-integrable over the interval $\openint {-\pi} \pi$.
Let $f$ be expressed by the Fourier series: 
:$\map f x \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
Then:
:$\ds \frac 1 \pi \int_{-\pi}^\pi \size {\map {f^2} x} \rd x = \frac { {a_0}^2} 2 + \sum_{n \mathop = 1}^\infty \paren { {a_n}^2 + {b_n}^2}$
== Proof ==
Category:Parseval's Theorem
Parseval's Theorem/Formulation 2

== Theorem ==
Let $f$ be a real function which is square-integrable over the interval $\openint {-\pi} \pi$.
Let $f$ be expressed by the Fourier series:
:$\map f x = \ds \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}$
where:
:$c_n = \ds \frac 1 {2 \pi} \int_{-\pi}^\pi \map f t e^{-i n t} \rd t$
Then:
:$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \size {\map f x}^2 \rd x = \sum_{n \mathop = -\infty}^\infty \size {c_n}^2$
== Proof == 
Category:Parseval's Theorem
Part-Time Carpenter

== Classic Problem ==
A carpenter agrees to work, on the condition that:
:he is paid $\pounds 2$ for every day he works
:he forfeits $\pounds 3$ for every day he does not work.
At the end of $30$ days he finds he has paid out exactly as much as he has received.
== Solution ==
$18$ days.
== Proof ==
Let $x$ be the number of days he worked.
Then:
Checking our work:
For $18$ days and at $\pounds 2$ per day, he earns $\pounds 2 \times 18 = \pounds 36$.
For $12$ days, forfeiting $\pounds 3$ per day, he forfeits $\pounds 3 \times 12 = \pounds 36$.
== Sources ==
* 
* : p. $235$
* : The Nuns in their Cells: $96$
Category:Classic Problems
Category:Part-Time Carpenter
Partial Correlation Coefficient/Examples/Arbitrary Example 1

== Examples of Partial Correlation Coefficients ==
Consider a population $P$ of children.
Let $X_1$, $X_2$ and $X_3$ be random variables such that:
:$X_1$ represents the height
:$X_2$ represents the weight
:$X_3$ represents the age
of the members of $P$.
A high positive correlation between $X_1$ and $X_2$ may partly reflect the high positive correlation of each with $X_3$.
The partial correlation coefficient $r_{12 \cdotp 3}$ eliminates the effect of age.
== Sources ==
* 
* 
Category:Examples of Partial Correlation Coefficients
Partial Denominators of Continued Fraction Expansion of Irrational Square Root

== Theorem ==
Let $n \in \Z$ such that $n$ is not a square.
Let the continued fraction expansion of $\sqrt n$ be expressed as:
:$\sqbrk {a_0, a_1, a_2, \ldots}$
Then the partial denominators of this continued fraction expansion can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt n} + P_r} {Q_r} }$
where:
:$P_r = \begin{cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end{cases}$
:$Q_r = \begin{cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end{cases}$
== Proof ==
The proof proceeds by strong induction.
For all $r \in \Z_{\ge 0}$, let $\map P r$ be the proposition:
:$a_r = \floor {\dfrac {\floor {\sqrt n} + P_r} {Q_r} }$
where:
:$P_r = \begin{cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end{cases}$
:$Q_r = \begin{cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end{cases}$
=== Edge Cases ===
By definition of the structure of a continued fraction:
:$\sqrt n = a_0 + \dfrac 1 {x_1}$
for some $a_0 \in \Z$ and $x_1 \in \R_{>0}$.
Thus:
Thus $\map P 0$ is seen to hold.
We have that:
:$\sqrt n = a_0 + \dfrac 1 {x_1}$
Then:
By definition of the structure of a continued fraction:
:$x_1 = a_1 + \dfrac 1 {x_2}$
for some $a_1 \in \Z$ and $x_2 \in \R_{>0}$.
Hence:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
We have that:
:$x_1 = a_1 + \dfrac 1 {x_2}$
Then:
By definition of the structure of a continued fraction:
:$x_2 = a_2 + \dfrac 1 {x_3}$
for some $a_2 \in \Z$ and $x_3 \in \R_{>0}$.
Thus:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P {k - 1}$ and $\map P k$ are true, for all $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$a_k = \floor {\dfrac {a_0 + P_k} {Q_k} }$
where:
:$P_k = a_{k - 1} Q_{k - 1} - P_{k - 1}$
:$Q_k = \dfrac {n - {P_k}^2} {Q_{k - 1} }$
and:
:$x_k = \dfrac{\sqrt n + P_k} {Q_k}$
from which it is to be shown that:
:$a_{k + 1} = \floor {\dfrac{a_0 + P_{k + 1} } {Q_{k + 1} } }$
where:
:$P_{k + 1} = a_k Q_k - P_k$
:$Q_{k + 1} = \dfrac {n - {P_{k + 1} }^2} {Q_k}$
and:
:$x_{k + 1} = \dfrac{\sqrt n + P_{k + 1} } {Q_{k + 1} }$
=== Induction Step ===
This is the induction step:
By definition of the structure of a continued fraction:
:$x_k = a_k + \dfrac 1 {x_{k+1} }$
where:
:$x_{k+1} \in \R_{>0}$
:$a_k = \floor {\dfrac{a_0 + P_k} {Q_k} }$
by the induction hypothesis.
Then:
By definition of the structure of a continued fraction:
:$x_{k + 1} = a_{k + 1} + \dfrac 1 {x_{k + 2} }$
for some $a_{k + 1} \in \Z$ and $x_{k + 2} \in \R_{>0}$.
Thus:
So $\map P k \implies P \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Hence the result.
== Sources ==
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Category:Continued Fractions