avewright/proofwiki-math huggingface dataset:
Theorem: Closed Ball contains Smaller Closed Ball

Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
:$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$

Proof:
{{begin-eqn}}
{{eqn | l = x \in \map {B^-_\epsilon} a 
      | o = \leadsto
      | r = \map d {x, a} \le \epsilon
      | c = {{Defof|Closed Ball}}
}}
{{eqn | o = \leadsto
      | r = \map d {x, a} \le \delta
      | c = As $\epsilon \le \delta$
}}
{{eqn | o = \leadsto
      | r = x \in \map {B^-_\delta} a 
      | c = {{Defof|Closed Ball}}
}}
{{end-eqn}}
By definition of subset:
:$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
{{qed}}
Category:Metric Spaces
i3895kegu0xmjo9zyhdjqdtsqt5p3y1
Theorem: Open Ball contains Strictly Smaller Closed Ball

Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{>0}$ such that $\epsilon < \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B_\delta} a$ be the open $\delta$-ball on $a$.
Then:
:$\map {B^-_\epsilon} a \subseteq \map {B_\delta} a$

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \map {B^-_\epsilon} a 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \map \cl {\map {B_\epsilon} a} 
      | c = {{Defof|Closed Ball of Metric Space}}
}}
{{eqn | ll= \leadsto
      | l = \map d {x, a}
      | o = \le
      | r = \epsilon
      | c = Closure of Open Ball in Metric Space
}}
{{eqn | ll= \leadsto
      | l = \map d {x, a}
      | o = <
      | r = \delta
      | c = As $\epsilon < \delta$
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \map {B_\delta} a
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{end-eqn}}
By definition of subset:
:$\map {B^-_\epsilon} a \subseteq \map {B_\delta} a$
{{qed}}
Category:Open Balls
Category:Closed Balls
m9uwhkov7cxa5ikwkengfqr3i0t9ebw
Theorem: Closed Ball contains Smaller Open Ball

Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
:$\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$

Proof:
{{begin-eqn}}
{{eqn | l = x \in \map {B_\epsilon} a 
      | o = \leadsto
      | r = \map d {x, a} < \epsilon
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{eqn | o = \leadsto
      | r = \map d {x, a} < \delta
      | c = As $\epsilon \le \delta$
}}
{{eqn | o = \leadsto
      | r = \map d {x, a} \le \delta
      | c = {{Defof|Total Ordering}}
}}
{{eqn | o = \leadsto
      | r = x \in \map {B^-_\delta} a 
      | c = {{Defof|Closed Ball of Metric Space}}
}}
{{end-eqn}}
By definition of subset:
:$\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$
{{qed}}
Category:Metric Spaces
9d2zaxlmsyqndb3jval5krm6nmwpn3t
Theorem: Sample Matrix Independence Test

Let $V$ be a vector space of real or complex-valued functions on a set $J$.
Let $f_1, \ldots, f_n$ be functions in $V$.
Let $x_1, \ldots, x_n$ from $J$ be given.
Define the '''sample matrix''':
:<nowiki>$S = \begin{bmatrix}
\map {f_1} {x_1} & \cdots & \map {f_n} {x_1} \\
\vdots           & \ddots & \vdots \\
\map {f_1} {x_n} & \cdots & \map {f_n} {x_n} \\
\end{bmatrix}$</nowiki>
Let $S$ be nonsingular.
Then $f_1, \ldots, f_n$ are linearly independent in $V$.

Proof:
The definition of linear independence is applied.
Assume a linear combination of the functions $f_1, \ldots, f_n$ is the zero function:
{{begin-eqn}}
{{eqn | n = 1
      | l = \sum_{i \mathop = 1}^n c_i \map {f_i} x 
      | r = 0
      | c = for all $x$
}}
{{end-eqn}}
Let $\vec c$ have components $c_1, \ldots, c_n$.
For $i = 1, \ldots, n$ replace $x = x_i$ in $(1)$.
There are $n$ linear homogeneous algebraic equations, written as:
:$S \vec c = \vec 0$
Because $S$ is nonsingular:
:$\vec c = \vec 0$
The functions are linearly independent.
{{qed}}
Theorem: Open and Closed Balls in P-adic Numbers are Compact Subspaces/P-adic Integers

Then the set of $p$-adic integers $\Z_p$ is compact.

Proof:
By definition the $p$-adic integers $\Z_p$ is the closed ball $\map {B^-_1} 0$.
From Open and Closed Balls in P-adic Numbers are Compact Subspaces, $\map {B^-_1} 0$ is compact.
{{qed}}
Theorem: Open Balls form Local Basis for Point of Metric Space

Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
Let $\BB_x$ be the set of all open balls of $M$ centered on $x$.
That is:
:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
Then $\BB$ is a local basis of $x$.

Proof:
Let $U$ be an open set of $M$ which has $x$ as an element.
Then by definition of an open set:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
From Open Ball of Metric Space is Open Set, $\BB_x$ is a set of open set which have $x$ as an element.
By definition of a local basis, $\BB_x$ is a local basis of $x$.
{{qed}}
Category:Open Balls
dig179y4pboam1jrlfv6t34iualnlwj
Theorem: Local Basis of P-adic Number/Cosets

Let $\Z_p$ be the $p$-adic integers.
Then the set $\set {a + p^n \Z_p: n \in Z}$ is a local basis of $a$ consisting of clopen sets.

Proof:
From Local Basis of P-adic Number the set $\set { \map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of clopen sets.
From Open Balls of P-adic Number:
:$\set {\map {B_{p^{-n} } } a : n \in \Z} = \set {a + p^{n + 1} \Z_p : n \in \Z} = \set {a + p^n \Z_p : n \in \Z}$
The result follows.
{{qed}}
Theorem: Local Basis of P-adic Number/Closed Balls

Then the set of closed balls $\set {\map {B^-_{p^{-n} } } a: n \in Z}$ is a local basis of $a$ consisting of clopen sets.

Proof:
From Local Basis of P-adic Number the set $\set {\map {B_{p^{-n } } } a: n \in \Z}$ is a local basis of clopen sets.
From Open Ball in P-adic Numbers is Closed Ball:
:$\set {\map {B_{p^{-n} } } a: n \in \Z} = \set {\map {B^-_{p^{-\paren{n + 1} } } } a: n \in \Z} = \set {\map {B^-_{p^{-n} } } a : n \in \Z}$
The result follows.
{{qed}}
Category:Topology of P-adic Numbers
5beebr9uelonbubljw0qclgzrf2ocmv
Theorem: Measurement of Terrestrial Latitude/Pole Star

An observer's (terrestrial) latitude can be identified by measuring the (celestial) altitude of the Pole Star.

Proof:
From Altitude of North Celestial Pole equals Latitude of Observer, all the observer needs to do is to measure the (celestial) altitude of the north celestial pole.
The star Polaris is situated at practically exactly that point.
Hence the result.
{{qed}}
Category:Measurement of Terrestrial Latitude
9houihdiff95zu5sxgfziz5k0u9rlw1
Theorem: Measurement of Terrestrial Latitude/Sun at Noon

An observer's (terrestrial) latitude can be identified by measuring the (celestial) altitude of the Sun at noon.

Proof:
{{ProofWanted|Needs to reference the fact that the time of year needs to be known, and then tables can be used.}}
{{qed}}
Theorem: Coherent Sequence is Partial Sum of P-adic Expansion

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:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$

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 $\sequence {b_{j, n} }_{0 \le j \le n} :$
:$(1) \quad \ds \alpha_n = \sum_{j \mathop = 0}^n b_{j, n} p^j$
:$(2) \quad \forall j \in \closedint 0 n : 0 \le b_{j, n} < p$
Theorem: Difference of Consecutive terms of Coherent Sequence

Let $p$ be a prime number.
Let $\sequence {\alpha_n}$ be a coherent sequence.
Then:
:for all $n \in \N_{>0}$ there exists $c_n \in \N$ such that:
::$0 \le c_n < p$
::$\alpha_n - \alpha_{n - 1} = c_n p^n$

Proof:
By definition of a coherent sequence:
:$\forall n \in \N_{>0}: \alpha_n \equiv \alpha_{n - 1} \pmod {p^n}$
That is:
:$\forall n \in \N_{>0}: \exists c_n \in \Z : \alpha_n - \alpha_{n - 1} = c_n p^n$
So it remains to show that:
:$\forall n \in \N_{>0} : 0 \le c_n < p$
{{AimForCont}} for some $N \in \N_{>0}$:
:$c_N \ge p$
Then:
:$c_N p^N \ge p^{N + 1}$
By definition of a coherent sequence:
:$\alpha_{N - 1} \ge 0$
Then:
:$\alpha_N = c_N p^N + \alpha_{N - 1} \ge p^{N + 1}$
This contradicts the coherent sequence condition:
:$\alpha_N \le p^{N + 1} - 1$
So:
:$\forall n \in \N_{>0} : c_n < p$
{{AimForCont}} for some $M \in \N_{>0}$:
:$c_M < 0$
Then:
:$-c_M \ge 1$
Now:
:$c_M p^M + \alpha_{M - 1} = \alpha_M \ge 0$
So:
:$\alpha_{M - 1} \ge -c_M p^M \ge p^M$
This contradicts the coherent sequence condition:
:$\alpha_{M - 1} < p^M$
So:
:$\forall n \in \N_{>0} : c_n \ge 0$
{{qed}}
Category:P-adic Number Theory
plnsql96qh7zh843m4s7q123mpzlw5c
Theorem: Vandermonde Matrix Identity for Cauchy Matrix

Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct in matrix
{{begin-eqn}}
{{eqn
  | l = C 
  | r = <nowiki>\begin {pmatrix}
        \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
        \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
        \vdots               & \vdots               & \cdots & \vdots \\
        \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
\end {pmatrix}</nowiki>
  | c = Cauchy matrix of order $n$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn
  | l = C
  | r = -P V_x^{-1} V_y Q^{-1}
  | c = Vandermonde matrix identity for a Cauchy matrix 
}}
{{end-eqn}}

Proof:
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.
For $1 \le i \le n$ express polynomial $p_i$ as:
:$\ds \map {p_i} x = \sum_{k \mathop = 1}^n a_{i k} x^{k - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \paren {\map {p_i} {x_j} }
      | r = \paren {a_{i j} } V_x
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | l = P
      | r = \paren {a_{i j} } V_x
      | c = as $\map {p_i} {x_j} = 0$ for $i \ne j$.
}}
{{eqn | l = \paren {a_{i j} }
      | r = P V_x^{-1}
      | c = solving for matrix $paren {a_{i j} }$
}}
{{eqn | l = \paren {\map {p_i} {y_j} }
      | r = \paren {a_{i j} } V_y
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | l = \paren {\map {p_i} {y_j} }
      | r = P V_x^{-1} V_y
      | c = substituting $paren {a_{i j} } = P V_x^{-1}$
}}
{{end-eqn}}
Use second equation $\map {p_i} {y_j} = \dfrac {\map p {y_j} } {y_j - x_i}$:
{{begin-eqn}}
{{eqn | l = \paren {\map {p_i} {y_j} }
      | r = -C Q
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | l = -C Q
      | r = P V_x^{-1} V_y
      | c = equating competing equations for $\paren {\map {p_i} {y_j} }$
}}
{{eqn | l = C
      | r = -P V_x^{-1} V_y Q^{-1}
      | c = solving for $C$
}}
{{end-eqn}}
{{qed}}
Theorem: Permutation is Product of Transpositions

Let $S_n$ denote the symmetric group on $n$ letters.
Every element of $S_n$ can be expressed as a product of transpositions.

Proof:
Let $\pi \in S_n$.
From Existence and Uniqueness of Cycle Decomposition, $\pi$ can be uniquely expressed as a cycle decomposition, up to the order of factors.
From K-Cycle can be Factored into Transpositions, each one of the cyclic permutations that compose this cycle decomposition can be expressed as a product of transpositions.
The result follows.
{{qed}}
Theorem: Vandermonde Matrix Identity for Hilbert Matrix

Define polynomial root sets $\set {1, 2, \ldots, n}$ and $\set {0, -1, \ldots, -n + 1}$ for Definition:Cauchy Matrix.
Let $H$ be the Hilbert matrix of order $n$:
:<nowiki>$H = \begin {pmatrix}
        1 & \dfrac 1 2 & \cdots & \dfrac 1 n \\
        \dfrac 1 2 & \dfrac 1 3   & \cdots & \dfrac 1 {n + 1} \\
        \vdots     & \vdots       & \cdots & \vdots \\
        \dfrac 1 n & \dfrac 1 {n + 1} & \cdots & \dfrac 1 {2 n - 1} \end {pmatrix}$</nowiki>
Then from Vandermonde Matrix Identity for Cauchy Matrix and Hilbert Matrix is Cauchy Matrix:
:$H = -P V_x^{-1} V_y Q^{-1}$
where $V_x$, $V_y$ are Vandermonde matrices:
:<nowiki>$V_x = \begin {pmatrix}
1         & 1         & \cdots & 1 \\
1         & 2       & \cdots & n \\
\vdots    & \vdots    & \ddots & \vdots \\
1 & 2^{n - 1} & \cdots & n^{n -1 } \\
\end {pmatrix}, \quad 
V_y = \begin {pmatrix}
1       & 1         & \cdots & 1 \\
0       & -1        & \cdots & -n + 1 \\
\vdots  & \vdots    & \ddots & \vdots \\
0 & \paren {-1}^{n - 1} & \cdots & \paren {-n + 1}^{n - 1} \\
\end {pmatrix}$</nowiki>
and  $P$, $Q$ are diagonal matrices:
:<nowiki>$P = \begin {pmatrix}
\map {p_1} 1 &  \cdots & 0 \\
\vdots   & \ddots  & \vdots \\
0        & \cdots  & \map {p_n} n \\
\end {pmatrix}, \quad
Q = \begin {pmatrix}
\map p 0 & \cdots  & 0 \\
\vdots  & \ddots  & \vdots \\
0       & \cdots  & \map p {-n + 1} \\
\end {pmatrix}$</nowiki>
Definitions of polynomials $p$, $p_1$, $\ldots$, $p_n$:
:$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - i}$
:$\ds \map {p_k} x = \dfrac {\map p x} {x - k} = \prod_{i \mathop = 1, i \mathop \ne k}^n \, \paren {x - i}$, $1 \mathop \le k \mathop \le n$

Proof:
Apply Vandermonde Matrix Identity for Cauchy Matrix and Hilbert Matrix is Cauchy Matrix.
Matrices $V_x$ and $V_y$ are invertible by Inverse of Vandermonde Matrix.
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.
{{qed}}
Theorem: Number of k-Cycles on Set of k Elements

Let $k \in \N$ be a natural number.
Let $S_k$ denote the symmetric group on $k$ letters.
The number of elements of $S_k$ which are $k$-cycles is $\paren {k - 1}!$.

Proof:
Let $\N_k$ denote the set $\set {1, 2, \ldots, k}$.
Let $\pi$ be a $k$-cycle in $S_k$.
By definition, $\pi$ moves all $k$ elements of $\set {1, 2, \ldots, k}$.
Let $j \in \N_k$ be an arbitrary element of $\N_k$.
By definition of $k$-cycle, $k$ is the smallest integer such that $\map {\pi^k} j = j$.
Thus for each $r \in \set {1, 2, \ldots, k}$, $\map {\pi^r} j$ is a different element of $\N_k$.
That is:
:$\set {\map {\pi^r} j: r \in \N_k} = \N_k$
Let $\pi$ and $\rho$ be $k$-cycles in $S_k$ such that $\pi \ne \rho$.
From the above, the distinguishing feature of $\rho$ and $\pi$ is the order of elements of $\N_k \setminus \set j$ that $\map {\pi^r} j$ and $\map {\rho^r} j$ move $j$ to for $1 \le r < k$.
Each different ordered $\paren {k - 1}$-tuple of $\N_k \setminus \set j$ corresponds to a different $k$-cycle of $S_k$
Thus there are as many $k$-cycles of $S_k$ as there are different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$.
From Number of Permutations, there are $\paren {k - 1}!$ different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$.
Hence the result.
{{qed}}
Category:Cyclic Permutations
pel2t5p5t2po3p461x98ymtwddg428n
Theorem: Sum of Elements of Inverse of Matrix with Column of Ones

Let $\mathbf B = \sqbrk b_n$ denote the inverse of a square matrix $\mathbf A$ of order $n$.
Let $\mathbf A$ be such that it has a row or column of all ones.
Then the sum of elements in $\mathbf B$ is one:
:$\ds \sum_{i \mathop  = 1}^n \sum_{j \mathop = 1}^n b_{ij}  = 1$

Proof:
If ones appear in a row of $\mathbf A$, then replace $\mathbf A$ by $\mathbf A^T$ and $\mathbf B$ by $\mathbf B^T$.
Assume $\mathbf A$ has a column of ones.
Apply Sum of Elements of Nonsingular Matrix to the inverse $\mathbf B = \mathbf A^{-1}$:
:$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$
where $\mathbf J_n$ denotes the square ones matrix of order $n$.
If $\mathbf A = \mathbf B^{-1}$ has a column of ones, then $\mathbf B^{-1} - \mathbf J_n$ has a column of zeros, implying determinant zero.
Substitute $\map \det {\mathbf B^{-1} - \mathbf J_n} = 0$ in Sum of Elements of Nonsingular Matrix:
:$\ds \sum_{i \mathop  = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - 0$
which implies the statement.
{{qed}}
Theorem: Coherent Sequence is Partial Sum of P-adic Expansion/Lemma

:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n b_{i, i} p^i$

Proof:
The theorem is proved by induction:
Theorem: Numbers Equal to Sum of Squares of Digits

There are exactly $2$ integers which are equal to the sum of the squares of their digits when expressed in base $10$:
:$0 = 0^2$
:$1 = 1^2$

Proof:
We see the cases $0$ and $1$ above hold.
Suppose $N > 1$ is equal to the sum of the squares of their digits when expressed in base $10$.
Since $N^2 > N$, $N$ cannot be a $1$-digit integer.
Suppose $N$ is a $2$-digit integer.
Write $N = \sqbrk {a b} = 10 a + b$.
Then we have $a^2 + b^2 = 10 a + b$.
This can be reduced to $b \paren {b - 1} = a \paren {10 - a}$.
Since $b \paren {b - 1}$ is even, $a$ must be even as well.
The list of possible values of $a \paren {10 - a}$ are:
:$2 \paren {10 - 2} = 8 \paren {10 - 8} = 16$
:$4 \paren {10 - 4} = 6 \paren {10 - 6} = 24$
The list of possible values of $b \paren {b - 1}$ are:
:$b \paren {b - 1} \le 4 \paren {4 - 1} = 12$ for $b \le 4$
:$5 \paren {5 - 1} = 20$
:$b \paren {b - 1} \ge 6 \paren {6 - 1} = 30$ for $b \ge 6$
We see that they do not coincide.
Thus $N$ cannot be a $2$-digit integer.
Suppose $100 \le N \le 199$.
Write $N = \sqbrk {1 a b} = 100 + 10 a + b$.
Then we have $1^2 + a^2 + b^2 = 100 + 10 a + b$.
This can be reduced to $b \paren {b - 1} = a \paren {10 - a} + 99$.
But we have $b \paren {b - 1} < 9 \times 8 = 72 < 99 \le a \paren {10 - a} + 99$.
So $N$ cannot be in this range.
Suppose $200 \le N \le 299$.
Then the sum of the squares of their digits cannot exceed $2^2 + 9^2 + 9^2 = 186 < 200$.
So $N$ cannot be in this range.
Suppose $300 \le N \le 999$.
Then the sum of the squares of their digits cannot exceed $9^2 + 9^2 + 9^2 = 243 < 300$.
So $N$ cannot be in this range.
Suppose $N$ is a $k$-digit integer with $k \ge 4$.
Then the sum of the squares of their digits cannot exceed $9^2 \times k$.
We have:
{{begin-eqn}}
{{eqn | l = N
      | o = \ge
      | r = 10^{k - 1}
      | c = The smallest $k$-digit number
}}
{{eqn | o = \ge
      | r = 100 \paren {1 + 9 \paren {k - 3} }
      | c = Bernoulli's Inequality
}}
{{eqn | o = >
      | r = 81 \paren {k + 8 k - 26}
}}
{{eqn | o = >
      | r = 9^2 \times k
      | c = From $k \ge 4$
}}
{{end-eqn}}
So $N$ cannot be greater than $1$.
{{qed}}
Theorem: Convergent Sequence in P-adic Numbers has Unique Limit

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence {x_n} $ be a sequence in $\Q_p$.
Then $\sequence {x_n}$ can have at most one limit.

Proof:
From P-adic Metric on P-adic Numbers is Non-Archimedean Metric the $p$-adic metric is a metric on $\Q_p$.
By definition, the sequence $\sequence {x_n}$ converges in $\Q_p$ {{iff}}:
:$\sequence {x_n}$ converges in the $p$-adic metric. 
The result then follows from Convergent Sequence in Metric Space has Unique Limit.
{{qed}}
Category:Sequences
Category:Convergence
Category:P-adic Number Theory
2yotou9ieejha9zlu7dw2i9a44161so
Theorem: Limit to Infinity of Summation of Euler Phi Function over Square

:$\ds \lim_{n \mathop \to \infty} \dfrac {\map \Phi n} {n^2} = \dfrac 3 {\pi^2}$
where:
:$\map \Phi n = \ds \sum_{k \mathop = 1}^n \map \phi k$
:$\map \phi k$ is the Euler $\phi$ function of $k$.
Numerically, this evaluates to:
:$\dfrac 3 {\pi^2} \approx 0 \cdotp 30396 35509 \ldots$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Phi n
      | r = \sum_{k \mathop = 1}^n \map \phi k
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\sum_{d \mathop \divides k} \map \mu d \frac k d}
      | c = Euler Phi Function in terms of Möbius Function
}}
{{eqn | r = \sum_{d d' \mathop \le n} d' \map \mu d
      | c = taking $d d' = k$
}}
{{eqn | r = \sum_{d \mathop = 1}^n \map \mu d \sum_{d' \mathop = 1}^{\floor {n / d} } d'
      | c = $d'$ takes value from $1$ to $\floor {\frac n d}$ for each $d$
}}
{{eqn | r = \sum_{d \mathop = 1}^n \map \mu d \paren {\frac 1 2 \paren {\floor {\frac n d}^2 + \floor {\frac n d} } }
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac 1 2 \sum_{d \mathop = 1}^n \map \mu d \paren {\paren {\frac n d}^2 + \map \OO {\frac n d} }
      | c = 
}}
{{eqn | r = \frac 1 2 n^2 \sum_{d \mathop = 1}^n \frac {\map \mu d} {d^2} + \map \OO {\sum_{d \mathop = 1}^n \frac n d}
      | c = 
}}
{{eqn | r = \frac 1 2 n^2 \sum_{d \mathop = 1}^\infty \frac {\map \mu d} {d^2} + \map \OO {n^2 \sum_{d \mathop = n + 1}^\infty \frac 1 {d^2} } + \map \OO {n \ln n}
      | c = Approximate Size of Sum of Harmonic Series
}}
{{eqn | r = \frac 1 2 n^2 \times \frac 1 {\map \zeta 2}+ \map \OO {\frac {n^2} n} + \map \OO {n \ln n}
      | c = Reciprocal of Riemann Zeta Function
}}
{{eqn | r = \frac {3 n^2} {\pi^2} + \map \OO {n \ln n}
      | c = Basel Problem
}}
{{eqn | ll= \leadsto
      | l = \frac {\map \Phi n} {n^2}
      | r = \frac 3 {\pi^2} + \map \OO {n^{-1} {\ln n} }
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \to \infty} \frac {\map \Phi n} {n^2}
      | r = \frac 3 {\pi^2}
      | c = Powers Drown Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Complex Sine Function is Unbounded

Let $\sin: \C \to \C$ be the complex sine function. 
Then $\sin$ is unbounded.

Proof:
By Complex Sine Function is Entire, we have that $\sin$ is an entire function. 
{{AimForCont}} that $\sin$ was a bounded function.
Then, by Liouville's Theorem, we would have that $\sin$ is a constant function.
However we have, for instance, by Sine of Zero is Zero: 
:$\sin 0 = 0$
and by Sine of 90 Degrees: 
:$\sin \dfrac \pi 2 = 1$
Therefore, $\sin$ is clearly not a constant function, a contradiction.
We hence conclude, by Proof by Contradiction, that $\sin$ is unbounded.
{{qed}}
Theorem: Complex Sine Function is Entire

Let $\sin: \C \to \C$ be the complex sine function. 
Then $\sin$ is entire.

Proof:
By the definition of the complex sine function, $\sin$ admits a power series expansion about $0$: 
:$\ds \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$
By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\sin$ is entire it suffices to show that this series is everywhere convergent.
From Radius of Convergence from Limit of Sequence: Complex Case, it is sufficient to show that: 
:$\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 3}!} \times \frac {\paren {2 n + 1}!} {\paren {-1}^n} } = 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 3}!} \times \frac {\paren {2 n + 1}!} {\paren {-1}^n} }
	| r = \size {-1} \lim_{n \mathop \to \infty} \size {\frac {\paren {2 n + 1}!} {\paren {2 n + 3} \paren {2 n + 2} \paren {2 n + 1}!} }
	| c = {{Defof|Factorial}}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\frac 1 {\paren {2 n + 3} \paren {2 n + 2} } }
}}
{{eqn	| r = 0
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Complex Sine Function is Entire

Let $\sin: \C \to \C$ be the complex sine function. 
Then $\sin$ is entire.

Proof:
Let: 
{{begin-eqn}}
{{eqn | l = \map f z
      | r = \exp z
}}
{{eqn | l = \map g z
      | r = i z
}}
{{eqn | l = \map h z
      | r = -i z
}}
{{end-eqn}}
for all $z \in \C$. 
By Complex Exponential Function is Entire, we have that $f$ is entire. 
By Polynomial is Entire, we have that $g$ and $h$ are entire. 
Therefore, by Composition of Entire Functions is Entire, we have that both $f \circ g$ and $f \circ h$ are entire.
By Linear Combination of Entire Functions is Entire, we then have that: 
:$\dfrac 1 {2 i} \paren {f \circ g - f \circ h}$
is entire.
Note that: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2 i} \paren {\map {\paren {f \circ g} } z - \map {\paren {f \circ h} } z}
      | r = \frac 1 {2 i} \paren {\map \exp {i z} - \map \exp {-i z} }
}}
{{eqn | r = \sin z
      | c = Euler's Sine Identity
}}
{{end-eqn}}
Therefore, $\sin$ is an entire function.
{{qed}}