nyuuzyou/wikis huggingface dataset:
Codomain of Composite Relation

== Theorem ==
Let $\RR_2 \circ \RR_1$ be a composite relation.
Then the codomain of $\RR_2 \circ \RR_1$ is the codomain of $\RR_2$:
:$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$
== Proof ==
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The codomain of $\RR_2$ is $S_3$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
From this definition:
:$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$
Thus the codomain of $\RR_2 \circ \RR_1$ is $S_3$.
Thus:
:$\Cdm {\RR_2 \circ \RR_1} = S_3 = \Cdm {\RR_2}$
Category:Composite Relations
Codomain of Internal Direct Isomorphism is Subset Product of Factors

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }$ and $\struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let $\struct {S, \circ}$ be the internal direct product of $A$ and $B$.
Then $S$ is the subset product of $A$ and $B$:
:$S = A \circ B$
== Proof ==
First we establish that from Set of Finite Subsets under Induced Operation is Closed:
:$A \times B \subseteq S$
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
:$\phi$ is a bijection
:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
By definition of internal direct product:
:the mapping $\phi: A \times B \to S$ defined as:
::$\forall a \in A, b \in B: \map \phi {a, b} = a \circ b$
:is an isomorphism from the (external) direct product $\struct {A, \circ {\restriction_A} } \times \struct {B, \circ {\restriction_B} }$ onto $\struct {S, \circ}$.
Thus $\phi$ is a bijection.
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection it follows that:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
That is:
:$S = \set {a \circ b: \tuple {a, b} \in A \times B}$
Hence $S = A \circ B$ by definition of subset product.
Category:Subset Products
Category:Internal Direct Products
Coefficient of Kinetic Friction is usually Less than Coefficient of Static Friction

== Physical Law ==
Let $B$ be a body resting on a rough surface $S$.
Let $\mu_k$ and $\mu_s$ denote the coefficient of kinetic fiction and coefficient of static fiction between $B$ and $S$ respectively.
Then it is typically the case that $\mu_k < \mu_s$.
== Proof ==
== Sources ==
* 
* 
Category:Coefficients of Friction
Coefficients in Linear Combination forming GCD are Coprime

== Theorem ==
Let $a$ and $b$ be integers.
Let there exist integers $x$ and $y$ such that:
:$a x + b y = \gcd \set {a, b}$
where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
Then:
:$x \perp y$
where $\perp$ denotes coprimality.
== Proof ==
Let $d = \gcd \set {a, b}$.
As $d$ is a divisor of both $a$ and $b$, both $\dfrac a d$ and $\dfrac b d$ are integers.
Hence, dividing through by $d$, we have:
:$\dfrac a d x + \dfrac b d y = \dfrac {\gcd \set {a, b} } d = 1$
Thus there exist integers $m = \dfrac a d$ and $n = \dfrac b d$ such that:
:$m x + n y = 1$
Hence from Integer Combination of Coprime Integers:
:$x \perp y$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $13 \ \text {(b)}$
Category:Integer Combinations
Category:Greatest Common Divisor
Category:Coprime Integers
Coefficients of Cosine Terms in Convergent Trigonometric Series

== Theorem ==
Let $\map S x$ be a trigonometric series which converges to $\map f x$ on the interval $\openint \alpha {\alpha + 2 \pi}$:
:$\map f x = \dfrac {a_0} 2 + \ds \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x}$
Then:
:$\forall n \in \Z_{\ge 0}: a_n = \dfrac 1 \pi \ds \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$
== Proof ==
Thus when $n = 0$ we have:
:$\ds \int_\alpha^{\alpha + 2 \pi} \map f x \rd x = \int_\alpha^{\alpha + 2 \pi} \map f x \cos 0 x \rd x = a_0 \pi$
and when $n \ne 0$ we have:
:$\ds \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x = a_n \pi$
Hence the result.
== Also see ==
* Coefficients of Sine Terms in Convergent Trigonometric Series
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(5 a)$, $(5 b)$
Category:Trigonometric Series
Category:Fourier Series
Coefficients of Polynomial Product

== Theorem ==
Let $J$ be a set.
Let $p_1, \ldots p_n$ be polynomial forms in the indeterminates $\set {X_j : j \in J}$ over a commutative ring $R$.
Suppose that for each $i$ with $1 \le i \le n$, we have, for appropriate $a_{i, k} \in R$:
:$p_i = \ds \sum_{k \mathop \in Z} a_{i, k} X^k$
where $Z$ comprises the multiindices of natural numbers over $J$.
Then:
:$\ds \prod_{i \mathop = 1}^n p_i = \sum_{k \mathop \in Z} b_k X^k$
where:
:$\ds b_k := \sum_{k_1 + \cdots + k_n \mathop = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$
== Proof ==
We proceed by induction over $n \ge 1$.
=== Basis for the Induction ===
If $n = 1$ the result is trivially true.
This establishes the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\ds \prod_{i \mathop = 1}^{n - 1} p_i = \sum_{k \mathop \in Z} c_d X^k$
where:
:$\ds c_d := \sum_{k_1 + \cdots + k_{n - 1} = d} \paren {\prod_{i \mathop = 1}^{n - 1} a_{i, k_i} }$
Now we need to show that the result is true for the product $\ds \prod_{i \mathop = 1}^n p_i$.
=== Induction Step ===
This is our induction step:
Let $b_k$ be the coefficient of $X^k$ in $\ds \prod_{i \mathop = 1}^n p_i$.
Then:
The result follows by induction.
Category:Polynomial Theory
Category:Proofs by Induction
Coefficients of Polynomial add to 0 iff 1 is a Root

== Theorem ==
Let $\map E x$ be the equation in $x$ represented as:
:$\ds \sum_{j \mathop = 0}^n a_j x^j = 0$
where the $a_j$s are constants.
Then $x$ is a root of $\map E x$ :
:$\ds \sum_{j \mathop = 0}^n a_j = 0$
That is, $x$ is a root of $\map E x$ all the coefficients of the polynomial in $x$ sum to zero.
== Proof ==
Letting $x = 1$ in $E$;
Category:Algebra
Coefficients of Product of Two Polynomials

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $f, g \in R \sqbrk x$ be polynomials over $R$.
For a natural number $n \ge 0$, let:
:$a_n$ be the coefficient of the monomial $x^n$ in $f$.
:$b_n$ be the coefficient of the monomial $x^n$ in $g$.
=== As an indexed summation ===
The coefficient $c_n$ of $x^n$ in $f g$ is the sum:
:$c_n = \ds \sum_{k \mathop = 0}^n a_k b_{n - k}$
=== As an indexed summation bounded by degrees ===
Let $\deg f$ and $\deg g$ be their degrees.
The coefficient $c_n$ of $x^n$ in $fg$ is the sum:
:$c_n = \ds \sum_{k \mathop = n - \deg g}^{\deg f} a_k b_{n - k}$
== Proof ==
== Also see ==
* Degree of Product of Polynomials
Category:Polynomial Theory
Coefficients of Sine Terms in Convergent Trigonometric Series

== Theorem ==
Let $\map S x$ be a trigonometric series which converges to $\map f x$ on the interval $\openint \alpha {\alpha + 2 \pi}$:
:$\map f x = \dfrac {a_0} 2 + \ds \sum_{m \mathop = 1}^\infty \left({a_m \cos m x + b_m \sin m x}\right)$
Then:
:$\forall n \in \Z_{\ge 0}: b_n = \dfrac 1 \pi \ds \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$
== Proof ==
== Also see ==
* Coefficients of Cosine Terms in Convergent Trigonometric Series
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: $(5 c)$
Category:Trigonometric Series
Category:Fourier Series
Coequalizer is Epimorphism

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $q: D \to Q$ be the coequalizer of two morphisms $f, g: C \to D$.
Then $q$ is an epimorphism.
== Proof ==
Follows directly from Equalizer is Monomorphism and the Duality Principle.
== Sources ==
* : $\S 3.4$: Proposition $3.19$
Category:Category Theory
Cofactor/Examples/Arbitrary Example 1

== Example of Cofactor ==
Let $D$ be the determinant defined as:
:$D = \begin {vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end {vmatrix}$
Then the cofactor of $2$ is defined as:
== Sources ==
* 
Category:Examples of Cofactors
Cofactor/Examples/Arbitrary Example 2

== Example of Cofactor ==
Let $D$ be the determinant defined as:
$\quad D = \begin {vmatrix} a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}\end {vmatrix}$
Then the cofactor of $a_{2 1}$ is defined as:
== Sources ==
* 
* 
Category:Examples of Cofactors
Cofactor/Examples/Arbitrary Example 3

== Example of Cofactor ==
Let $D$ be the determinant defined as:
$\quad D = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{vmatrix}$
Let $\map D {2, 3 \mid 2, 4}$ be an order-$2$ minor of $D$.
Then the cofactor of $\map D {2, 3 \mid 2, 4}$ is given by:
Category:Examples of Cofactors
Cofactor/Examples/Arbitrary Example 4

== Example of Cofactor ==
Let $D$ be the determinant defined as:
:$D = \begin {vmatrix} a & b & c \\ d & e & f \\ g & h & i \end {vmatrix}$
The cofactor of $e$ is defined as:
The cofactor of $d$ is defined as:
== Sources ==
* 
* 
Category:Examples of Cofactors
Cofactor Sum Identity

== Theorem ==
Let $J_n$ be the $n \times n$ matrix of all ones.
Let $A$ be an $n \times n$ matrix.
Let $A_{ij}$ denote the cofactor of element $\tuple {i, j}$ in $\map \det A$, $1 \le i, j \le n$.
Then:
:$\ds \map \det {A -J_n} = \map \det A - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n A_{ij} $
== Proof ==
Let $P_j$ equal matrix $A$ with column $j$ replaced by ones, $1\le j \le n$.
Then by the Expansion Theorem for Determinants applied to column $j$ of $P_j$:
:$\ds \sum_{j \mathop = 1}^n \map \det {P_j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = 1}^n A_{ij}$
To complete the proof, it suffices to prove the equivalent identity:
:$\ds \map \det {A - J_n} = \map \det A - \sum_{j \mathop = 1}^n \map \det {P_j}$
Expansion of $\map \det {A - J_n}$ for the $2 \times 2$ case illustrates how determinant theorems will be used:
Let $A$ be an $n \times n$ matrix.
Let matrix $Q_m$ equal ones matrix $J_n$ with zeros replacing all entries in columns $1$ to $m$. 
For example, for $n = 5$ and $m = 2$:
:$Q_2 = \begin {pmatrix}
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
\end {pmatrix}$
Induction on $m$ will be applied to prove the induction identity:
:$\ds \map \det {A - J_n} = \map \det {A - Q_m} - \sum_{j \mathop = 1}^m \map \det {P_j}$
for $1 \le m \le n$.
;Induction step $m = 1$:
;Induction step $m = k$ and $k < n$ implies $m = k + 1$:
;Conclusion:
Matrix $A - Q_n$ equals $A$ because $Q_n$ is the zero matrix.
Let $m = n$ in the induction identity, then:
:$\ds \map \det {A - J_n} = \map \det A - \sum_{j \mathop = 1}^n \map \det {P_j}$
Coffee Mug and Doughnut are Homeomorphic

== Theorem ==
A and a are homeomorphic.
== Proof ==
They are both solid objects with one hole.
A classical is in the shape of a torus.
That is, it is a solid figure with one hole.
The classic coffee-cup shape is a receptacle with a closed loop of china forming a handle.
The hole through which you put your fingers is homeomorphic to the hole in the doughnut.
== Also see ==
* ProofWiki:Jokes/Coffee and Doughnuts
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(e)}$
Category:Examples of Homeomorphisms
Cofinal Limit Ordinals

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $\mathrm {cof}$ denote the cofinal relation.
Let $K_{II}$ denote the class of all limit ordinals.
Then:
:$\map {\mathrm {cof} } {x, y} \implies \paren {x \in K_{II} \iff y \in K_{II} }$
== Proof ==
=== Necessary Condition ===
Suppose $y \in K_{II}$.
$x \ne 0$ by Cofinal to Zero iff Ordinal is Zero.
If $x = z^+$ for some $z$, then $z = \bigcup x$ by Union of Successor Ordinal.
Therefore, $z$ would be the least upper bound of $x$.
Since $\map {\mathrm {cof} } {x, y}$, it follows by the definition of cofinal that:
:$\forall a \in x: \exists b \in y: \map f b \ge a$
So $\map f b \ge z$ for some $b \in y$.
But since $y$ is a limit ordinal, $b^+ \in y$ by Successor of Ordinal Smaller than Limit Ordinal is also Smaller.
Therefore, $\map f {b^+} \in x$ and $\map f {b^+} > \map f b \ge z$, which contradicts the fact that $z$ is an upper bound of $x$.
Therefore, $x \ne z^+$ for any $z$ and $x \in K_{II}$.
=== Sufficient Condition ===
Suppose $x \in K_{II}$.
$y \ne 0$ by Cofinal to Zero iff Ordinal is Zero.
Moreover, if $y = z^+$ for some ordinal $z$, then $z = \bigcup y$ by Union of Successor Ordinal.
Therefore, $z$ would be the least upper bound of $y$.
Since $\map {\mathrm {cof} } {x, y}$, it follows by the definition of cofinal that:
:$\forall a \in x: \exists b \in y: \map f b \ge a$
But $z \in y$, so by the definition of $f$:
:$\map f z \in x$
Therefore, from Successor of Ordinal Smaller than Limit Ordinal is also Smaller:
:$\map f z^+ \in x$
This means that:
:$\exists b \in y: \map f b \ge \map f z^+$
This would mean that by Ordinal is Less than Successor:
:$\map f b > \map f z$
But this means that $b \ge z$ since otherwise, by the definition of strictly increasing mapping:
:$\map f b < \map f z$
Furthermore, $b \ne z$, since otherwise, by Substitutivity of Equality:
:$\map f b = \map f z$.
It follows that $b > z$.
This contradicts the fact that from Union of Ordinals is Least Upper Bound, $z$ is an upper bound of $y$.
Therefore, $y \ne z^+$ for any $z$ and $y \in K_{II}$.
== Sources ==
* : $\S 10.55$
Category:Ordinals
Cofinal Ordinal Relation is Reflexive

== Theorem ==
Let $x$ be an ordinal.
Then $x$ is cofinal to itself.
That is:
:$\operatorname{cof} \left({x, x}\right)$
== Proof ==
Each of the conditions for cofinal ordinals shall be verified:
:$x \le x$ follows by Set is Subset of Itself.
The mapping $f: x \to x$ can simply be the identity mapping $I_x$:
:$I_x: x \to x$
Moreover, $a < b \implies I_x \left({a}\right) < I_x \left({b}\right)$ by the definition of the identity mapping.
Therefore:
:$I_x$ is strictly increasing.
Finally, $I_x \left({a}\right) \ge a$ by definition of identity mapping.
It follows by Existential Generalisation that:
:$\forall a \in x: \exists b \in x: I_x \left({b}\right) \ge a$
== Sources ==
* : $\S 10.52 \ (1)$
Category:Ordinals
Cofinal Ordinal Relation is Transitive

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Let $\operatorname {cof}$ denote the cofinal relation.
Then:
:$\map {\operatorname {cof} } {x, y} \land \map {\operatorname {cof} } {y, z} \implies \map {\operatorname {cof} } {x, z}$
== Proof ==
The conditions for $z$ being cofinal with $x$ shall be verified individually.
Let $\le$ denote the subset relation.
Since $x \le y$ and $y \le z$, it follows that $x \le z$ by Subset Relation is Transitive
Let $f : y \to x$ be the strictly increasing mapping that satisfies:
:$\forall a \in x: \exists b \in y: \map f b \ge a$
Let $g : z \to y$ be the strictly increasing mapping that satisfies:
:$\forall a \in y: \exists b \in z: \map g b \ge a$
By Composite of Strictly Increasing Mappings is Strictly Increasing:
:$f \circ g$ is a strictly increasing mapping.
Moreover, take any $a \in x$.
There is some $b \in y$ such that:
:$\map f b \ge a$
But $b \in y$, so there is some $c \in z$ such that:
:$\map g c \ge b$
Therefore:
:$\map {f \circ g} c \ge a$ 
for some $c \in z$.
== Sources ==
* : $\S 10.52 \ (2)$
Category:Ordinals
Cofinal to Zero iff Ordinal is Zero

== Theorem ==
Let $x$ be an ordinal.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $0$ denote the zero ordinal.
Then the following are equivalent:
:$\map {\operatorname{cof} } {x, 0}$
:$\map {\operatorname{cof} } {0, x}$
:$x = 0$
== Proof ==
=== $\map {\operatorname{cof} } {x, 0} \implies x = 0$ ===
If $\map {\operatorname{cof} } {x, 0}$, then there is a function $f : x \to 0$.
If $x \ne 0$, then $x$ has an element $a$.
But then, $\map f a \in 0$, which contradicts the definition of the empty set.
Therefore, $x = 0$.
=== $x = 0 \implies \map {\operatorname{cof} } {0, x}$ ===
Follows directly from Cofinal Ordinal Relation is Reflexive.
=== $\map {\operatorname{cof} } {0, x} \implies \map {\operatorname{cof} } {x, 0}$ ===
If $\map {\operatorname{cof} } {0, x}$, then $x \le 0$ by the definition of cofinal.
By Subset of Empty Set, $x = 0$.
$\map {\operatorname{cof} } {x, 0}$ follows directly from Cofinal Ordinal Relation is Reflexive.
== Sources ==
* : $\S 10.53 \ (1)$
Category:Ordinals
Cohen's Kappa Statistic/Examples/Medical Diagnosis

== Example of Use of Cohen's Kappa Statistic ==
Let there be $80$ patients claiming to suffer from depression
Let there be $2$ doctors who are to assess whether or not it is appropriate to treat each patient with a particular antidepressant drug.
In $32$ cases, both agree that treatment is appropriate.
In $35$ cases, both agree that treatment is not appropriate.
In the remaining $13$ cases, they disagree: one doctor believes treatment is appropriate, while the other does not.
Then Cohen's kappa statistic $\kappa$ is evaluated to be:
:$\kappa = 0 \cdotp 675$
== Proof ==
Here we have:
== Sources ==
* 
Category:Examples of Cohen's Kappa Statistic
Coherent Sequence Converges to P-adic Integer

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence {\alpha_n}$ be a coherent sequence. 
Let $\Z_p$ be the $p$-adic integers.
Then the sequence $\sequence {\alpha_n}$ converges to some $x \in \Z_p$.
That is, there exists $x \in \Z_p$ such that:
:$\ds \lim_{n \mathop \to \infty} \alpha_n = x$
== Proof ==
From Coherent Sequence is Partial Sum of P-adic Expansion there exists a unique $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
such that:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
From P-adic Expansion Converges to P-adic Number:
:$\exists x \in \Q_p : \ds \lim_{n \mathop \to \infty} \alpha_n = x$
By definition of a coherent sequence:
:$\forall n \in \N: \alpha_n \in \Z \subseteq \Z_p$
By definition, the $p$-adic integers $\Z_p$ is the closed ball $\map {B^-_1} 0$.
From Closed Ball of Non-Archimedean Division Ring is Clopen, $\Z_p$ is closed in $\norm{\,\cdot\,}_p$.
From Subset of Metric Space contains Limits of Sequences iff Closed:
:$x \in \Z_p$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$
Category:P-adic Number Theory
Coherent Sequence is Partial Sum of P-adic Expansion

== Theorem ==
Let $p$ be a prime number.
Let $\sequence {\alpha_n}$ be a coherent sequence.
Then there exists a unique $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
such that:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
== Informal Proof ==
== Proof ==
By definition of a coherent sequence:
:$\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
From Zero Padded Basis Representation, for all $n \in \N$ there exists a sequence $\sequence {b_{j, n} }_{0 \le j \le n} :$
:$(1) \quad \ds \alpha_n = \sum_{j \mathop = 0}^n b_{j, n} p^j$
:$(2) \quad \forall j \in \closedint 0 n : 0 \le b_{j, n} < p$
=== Lemma ===
For all $n \in \N$, let:
:$d_n = b_{n, n}$
Then:
:$\forall n \in \N : 0 \le d_n < p$
By definition:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$ 
is a $p$-adic expansion.
From the lemma:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$
Category:P-adic Number Theory
Category:Coherent Sequence is Partial Sum of P-adic Expansion
Coherent Sequence is Partial Sum of P-adic Expansion/Informal Proof

== Theorem ==
== Informal Proof ==
Consider the $\ds \alpha_n$ written in base $p$. 
To reduce an integer modulo $p^n$, it is simply a matter of stripping off all but the last $n$ digits.
So the coherence condition:
:$\alpha_{n + 1} \equiv \alpha_n \mod{p^{n+1}}$
means that the last $n + 1$ digits of both integers are the same.
So the sequence $\sequence {\alpha_n}$ can be written:
Putting this together, we get the $p$-adic expansion:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
such that:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.8$
Category:Coherent Sequence is Partial Sum of P-adic Expansion
Coherent Sequence is Partial Sum of P-adic Expansion/Lemma

== Theorem ==
Let $p$ be a prime number.
Let $\sequence {\alpha_n}$ be a coherent sequence.
For all $n \in \N$, let $\sequence {b_{j, n} }_{0 \le j \le n}$ be a sequence such that:
:$(1) \quad \ds \alpha_n = \sum_{j \mathop = 0}^n b_{j,n} p^j$
:$(2) \quad \forall j: 0 \le j \le n: 0 \le b_{j, n} < p$
Then:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n b_{i, i} p^i$
== Proof ==
The theorem is proved by induction:
=== Basis for the Induction ===
$n = 0$ 
By definition:
$\alpha_0 = b_{0, 0}$
This is the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\alpha_k = \ds \sum_{i \mathop = 0}^k b_{i, i} p^i$
Now we need to show true for $n = k + 1$:
:$\alpha_{k + 1} = \ds \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$
=== Induction Step ===
This is our induction step.
From Difference of Consecutive terms of Coherent Sequence there exists $c_{k + 1} \in \Z$:
:$(a) \quad 0 \le c_{k + 1} < p$
:$(b) \quad \alpha_{k + 1} = c_{k + 1} p^{k + 1} + \alpha_k$
From the induction hypothesis:
:$\alpha_{k + 1} = c_{k + 1} p^{k + 1} + \sum_{j \mathop = 0}^k b_{j, j} p^j$
From the Zero Padded Basis Representation it follows that the two sequences:
:$\sequence {b_{j, k + 1} }_{0 \le j \le k + 1}$
and
:$\sequence {b_{0, 0}, b_{1, 1}, \dots, b_{k - 1, k - 1}, b_{k, k}, c_{k + 1} }$
are equal.
Then:
:$\forall j \in \closedint 0 {k + 1} : b_{j, k + 1} = b_{j, j}$
It follows that:
:$\alpha_{k + 1} = \ds \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$
By induction:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n b_{i, i} p^i$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.8$
Category:Coherent Sequence is Partial Sum of P-adic Expansion
Coin-Tossing Modeled by Bernoulli Trial

== Theorem ==
The act of coin-tossing can be modeled as a Bernoulli trial.
This applies whether the coin is fair or biased.
== Proof ==
When a coin is tossed, it can land either with the "head" side up, or the "tail" side up.
Not taking into account tricks and teases in which the coin is given due consideration for landing on its edge, these are the only two possible outcomes.
Let the probability that it lands heads-up be $p$.
Let $\EE$ be the experiment of tossing a coin up into the air and seeing which side lands uppermost.
Let $\EE$ be modeled as the probability space $\struct {\Omega, \Sigma, \Pr}$.
We have that:
:$\Omega = \set {H, T}$
where $H$ and $T$ are the elementary events "Heads land uppermost" and "Tails land uppermost" respectively.
We define $\Pr: \Omega \to \R$ as follows:
:$\forall \omega \in \Omega: \map \Pr\omega = \begin{cases}
p & : \omega = H \\
1 - p & : \omega = T
\end{cases}$
in accordance with the definition of the experiment.
Finally, we let $\Sigma = \powerset \Omega$.
This is, by definition, clearly a Bernoulli trial.
Category:Bernoulli Distribution
Coincidence of Hour and Minute Hands

== Problem ==
:The hour and minute hands of a watch coincide at noon.
:When will they once again coincide, during the next $12$ hours?
== Solution ==
The hands coincide at the approximate times, in hours, minutes and seconds:
:$13:05:27$
:$14:10:55$
:$15:16:22$
:$16:22:49$
:$17:27:16$
:$18:32:44$
:$19:38:11$
:$20:43:38$
:$21:49:05$
:$22:54:33$
:Midnight
== Proof ==
We have that:
:the minute hand takes $1$ hour to go $360 \degrees$ around the dial.
:the hour hand takes $1$ hour to go $30 \degrees$ around the dial.
Let the minute hand and hour hand be coincident at some point in time.
Let $t$ hours be the time elapsed when they are next coincident.
Let $\theta$ degrees be the angle the hour hand has moved in time $t$.
During $t$, the minute hand has moved $\theta + 360 \degrees$.
Hence we have:
That is, the hour hand and minute hand coincide every $1 \tfrac 1 {11}$ hours.
Hence that will be at:
:$1 \tfrac 1 {11}, 2 \tfrac 2 {11}, 2 \tfrac 3 {11}, \ldots$
The result follows by converting the fractional hours into minutes and seconds.
== Sources ==
* 
* : From Ozanam to Hutton: $165$
Category:Clock Puzzles