avewright/proofwiki-math huggingface dataset:
Theorem: Non-Palindromes in Base 2 by Reverse-and-Add Process

Let the number $22$ be expressed in binary: $10110_2$.
When the reverse-and-add process is performed on it repeatedly, it never becomes a palindromic number.

Proof:
{{begin-eqn}}
{{eqn | l = 10110_2 + 01101_2
      | r = 100011_2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 100011_2 + 110001_2
      | r = 1010100_2
      | c = 
}}
{{end-eqn}}
It remains to be shown that a binary number of this form does not become a palindromic number.
Let $d_n$ denote $n$ repetitions of a binary digit $d$ in a number.
Thus:
:$10111010000$
can be expressed as:
:$101_3010_4$
'''Beware''' that the subscript, from here on in, does not denote the number base.
It is to be shown that the reverse-and-add process applied to:
:$101_n010_{n + 1}$
leads after $4$ iterations to:
:$101_{n + 1}010_{n + 2}$
Thus:
{{begin-eqn}}
{{eqn | l = 101_n010_{n + 1} + 0_{n + 1}101_k01
      | r = 110_n101_{n - 1}01
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 110_n101_{n - 1}01 + 101_{n - 1}010_n11
      | r = 101_{n + 1}010_{n + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 101_{n + 1}010_{n + 1} + 0_{n + 1}101_{n + 1}01
      | r = 110_{n - 1}10001_{n - 1}01
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 110_{n - 1}10001_{n - 1}01 + 101_{n - 1}00010_{n - 1}11
      | r = 101_{n + 1}010_{n + 2}
      | c = 
}}
{{end-eqn}}
As neither $101_n010_{n + 1}$ nor $101_{n + 1}010_{n + 2}$ are palindromic numbers, nor are any of the intermediate results, the result follows.
{{qed}}
Theorem: Best Approximation from Below to 1 as Sum of Minimal Number of Unit Fractions

:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {43} = 1 -  \dfrac 1 {1806}$
This is the best approximation from below to $1$ as the minimal sum of unit fractions.

Proof:
{{ProofWanted|It is not clear exactly what is being stated here.}}
Theorem: Cuboid with Integer Edges and Face Diagonals

The smallest cuboid whose edges and the diagonals of whose faces are all integers has edge lengths $44$, $117$ and $240$.
Its space diagonal, however, is not an integer.
{{expand|Add the definition of Definition:Euler Brick, and rewrite and rename as appropriate.}}

Proof:
The edges are given as having lengths $44$, $117$ and $240$.
The faces are therefore:
:$44 \times 117$
:$44 \times 240$
:$117 \times 240$
The diagonals of these faces are given by Pythagoras's Theorem as follows:
{{begin-eqn}}
{{eqn | l = 44^2 + 117^2
      | r = 15 \, 625
      | c = 
}}
{{eqn | r = 125^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 44^2 + 240^2
      | r = 59 \, 536
      | c = 
}}
{{eqn | r = 244^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 117^2 + 240^2
      | r = 71 \, 289
      | c = 
}}
{{eqn | r = 267^2
      | c = 
}}
{{end-eqn}}
However, its space diagonal is calculated as:
{{begin-eqn}}
{{eqn | l = 44^2 + 117^2 + 240^2
      | r = 73 \, 225
      | c = 
}}
{{eqn | r = \paren {270 \cdotp 6012 \ldots}^2
      | c = 
}}
{{end-eqn}}
which, as can be seen, is not an integer.
{{ProofWanted|Still to be proved that this is the smallest.}}
Theorem: Largest n such that 1 to n can be Partitioned for no Element to be Sum of 2 Elements in Same Set

$44$ is the largest integer $n$ such that the set of integers from $1$ to $n$ can be partitioned into $4$ subsets such that no integer in any of these subsets is the sum of $2$ other integers in the same subset:
:$\set {1, 3, 5, 15, 17, 19, 26, 28, 40, 42, 44}$
:$\set {2, 7, 8, 18, 21, 24, 27, 33, 37, 38, 43}$
:$\set {4, 6, 13, 20, 22, 23, 25, 30, 32, 39, 41}$
:$\set {9, 10, 11, 12, 14, 16, 29, 31, 34, 35, 36}$

Proof:
{{ProofWanted|Outline: find all possible solutions of partitioning $44$ into sum-free sets, then show that $45$ cannot be inserted in any of the sets. The search is, of course, done by computer. Note that $x$ and $2 x$ cannot be in the same set.}}
Theorem: Sequences of 4 Consecutive Integers with Falling Divisor Sum

The following ordered quadruple of consecutive integers have divisor sums which are strictly decreasing:
:$44, 45, 46, 47$
:$104, 105, 106, 107$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {44}
      | r = 84
      | c = {{DSFLink|44}}
}}
{{eqn | l = \map {\sigma_1} {45}
      | r = 78
      | c = {{DSFLink|45}}
}}
{{eqn | l = \map {\sigma_1} {46}
      | r = 72
      | c = {{DSFLink|46}}
}}
{{eqn | l = \map {\sigma_1} {47}
      | r = 48
      | c = Divisor Sum of Prime Number: $47$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {104}
      | r = 210
      | c = {{DSFLink|104}}
}}
{{eqn | l = \map {\sigma_1} {105}
      | r = 192
      | c = {{DSFLink|105}}
}}
{{eqn | l = \map {\sigma_1} {106}
      | r = 162
      | c = {{DSFLink|106}}
}}
{{eqn | l = \map {\sigma_1} {107}
      | r = 108
      | c = Divisor Sum of Prime Number: $107$ is prime
}}
{{end-eqn}}
{{qed}}
Theorem: Pairs of Consecutive Integers with 6 Divisors

The following sequence of integers are those $n$  which fulfil the equation:
:$\map {\sigma_0} n = \map {\sigma_0} {n + 1} = 6$
where $\map {\sigma_0} n$ denotes the divisor count function.
That is, they are the first of pairs of consecutive integers which each have $6$ divisors:
:$44, 75, 98, 116, 147, 171, 242, 243, 244, 332, \ldots$
{{OEIS|A049103}}

Proof:
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {44}
      | r = 6
      | c = {{DCFLink|44}}
}}
{{eqn | l = \map {\sigma_0} {45}
      | r = 6
      | c = {{DCFLink|45}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {75}
      | r = 6
      | c = {{DCFLink|75}}
}}
{{eqn | l = \map {\sigma_0} {76}
      | r = 6
      | c = {{DCFLink|76}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {98}
      | r = 6
      | c = {{DCFLink|98}}
}}
{{eqn | l = \map {\sigma_0} {99}
      | r = 6
      | c = {{DCFLink|99}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {116}
      | r = 6
      | c = {{DCFLink|116}}
}}
{{eqn | l = \map {\sigma_0} {117}
      | r = 6
      | c = {{DCFLink|117}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {147}
      | r = 6
      | c = {{DCFLink|147}}
}}
{{eqn | l = \map {\sigma_0} {148}
      | r = 6
      | c = {{DCFLink|148}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {171}
      | r = 6
      | c = {{DCFLink|171}}
}}
{{eqn | l = \map {\sigma_0} {172}
      | r = 6
      | c = {{DCFLink|172}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {242}
      | r = 6
      | c = {{DCFLink|242}}
}}
{{eqn | l = \map {\sigma_0} {243}
      | r = 6
      | c = {{DCFLink|243}}
}}
{{eqn | l = \map {\sigma_0} {244}
      | r = 6
      | c = {{DCFLink|244}}
}}
{{eqn | l = \map {\sigma_0} {245}
      | r = 6
      | c = {{DCFLink|245}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {332}
      | r = 6
      | c = {{DCFLink|332}}
}}
{{eqn | l = \map {\sigma_0} {333}
      | r = 6
      | c = {{DCFLink|333}}
}}
{{end-eqn}}
{{qed}}
Theorem: Hermitian Conjugate is Involution

Let $\mathbf A$ be a complex-valued matrix.
Let $\mathbf A^*$ denote the Hermitian conjugate of $\mathbf A$.
Then the operation of Hermitian conjugate is an involution:
:$\paren {\mathbf A^*}^* = \mathbf A$

Proof:
{{begin-eqn}}
{{eqn | l = \sqbrk {\paren {\mathbf A^*}^* }_{i j}
      | r = \overline {\sqbrk {\mathbf A^*}_{j i} }
      | c = {{Defof|Hermitian Conjugate}}
}}
{{eqn | r = \overline {\paren {\overline {\sqbrk {\mathbf A}_{i j} } } }
      | c = {{Defof|Hermitian Conjugate}}
}}
{{eqn | r = \sqbrk {\mathbf A}_{i j}
      | c = Complex Conjugation is Involution
}}
{{end-eqn}}
So:
:$\paren {\mathbf A^*}^* = \mathbf A$
{{qed}}
Category:Hermitian Conjugates
Category:Involutions
bo7vtsqugxw9q7iptw324puiy4t54vb
Theorem: Number as Sum of Distinct Primes greater than 11

Every number greater than $45$ can be expressed as the sum of distinct primes greater than $11$.

Proof:
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes greater than $11$ ordered by size.
Then $S = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, \dots}$.
By Bertrand-Chebyshev Theorem:
:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every integer $n$ where $45 < n \le 45 + s_{11} = 92$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_{10}} = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43}$.
Hence the result by Richert's Theorem.
{{qed|lemma}}
Here is a demonstration of our claim:
{{begin-eqn}}
{{eqn | l = 46
      | r = 17 + 29
}}
{{eqn | l = 47
      | r = 11 + 17 + 19
}}
{{eqn | l = 48
      | r = 19 + 29
}}
{{eqn | l = 49
      | r = 13 + 17 + 19
}}
{{eqn | l = 50
      | r = 19 + 31
}}
{{eqn | l = 51
      | r = 11 + 17 + 23
}}
{{eqn | l = 52
      | r = 23 + 29
}}
{{eqn | l = 53
      | r = 13 + 17 + 23
}}
{{eqn | l = 54
      | r = 23 + 31
}}
{{eqn | l = 55
      | r = 13 + 19 + 23
}}
{{eqn | l = 56
      | r = 19 + 37
}}
{{eqn | l = 57
      | r = 11 + 17 + 29
}}
{{eqn | l = 58
      | r = 17 + 41
}}
{{eqn | l = 59
      | r = 13 + 17 + 29
}}
{{eqn | l = 60
      | r = 19 + 41
}}
{{eqn | l = 61
      | r = 13 + 19 + 29
}}
{{eqn | l = 62
      | r = 19 + 43
}}
{{eqn | l = 63
      | r = 13 + 19 + 31
}}
{{eqn | l = 64
      | r = 23 + 41
}}
{{eqn | l = 65
      | r = 11 + 17 + 37
}}
{{eqn | l = 66
      | r = 23 + 43
}}
{{eqn | l = 67
      | r = 13 + 17 + 37
}}
{{eqn | l = 68
      | r = 31 + 37
}}
{{eqn | l = 69
      | r = 13 + 19 + 37
}}
{{eqn | l = 70
      | r = 11 + 17 + 19 + 23
}}
{{eqn | l = 71
      | r = 17 + 23 + 31
}}
{{eqn | l = 72
      | r = 13 + 17 + 19 + 23
}}
{{eqn | l = 73
      | r = 19 + 23 + 31
}}
{{eqn | l = 74
      | r = 11 + 13 + 19 + 31
}}
{{eqn | l = 75
      | r = 13 + 19 + 43
}}
{{eqn | l = 76
      | r = 11 + 17 + 19 + 29
}}
{{eqn | l = 77
      | r = 17 + 19 + 41
}}
{{eqn | l = 78
      | r = 11 + 17 + 19 + 31
}}
{{eqn | l = 79
      | r = 17 + 19 + 43
}}
{{eqn | l = 80
      | r = 13 + 17 + 19 + 31
}}
{{eqn | l = 81
      | r = 17 + 23 + 41
}}
{{eqn | l = 82
      | r = 11 + 17 + 23 + 31
}}
{{eqn | l = 83
      | r = 19 + 23 + 41
}}
{{eqn | l = 84
      | r = 13 + 17 + 23 + 31
}}
{{eqn | l = 85
      | r = 19 + 23 + 43
}}
{{eqn | l = 86
      | r = 13 + 19 + 23 + 31
}}
{{eqn | l = 87
      | r = 17 + 29 + 41
}}
{{eqn | l = 88
      | r = 11 + 17 + 29 + 31
}}
{{eqn | l = 89
      | r = 19 + 29 + 41
}}
{{eqn | l = 90
      | r = 13 + 17 + 29 + 31
}}
{{eqn | l = 91
      | r = 19 + 31 + 41
}}
{{eqn | l = 92
      | r = 13 + 19 + 29 + 31
}}
{{end-eqn}}
{{qed}}
Theorem: Subclass of Set is Set

Let $A$ be a set.
Let $\map \phi x$ be a condition in which $x$ is taken to be a set.
Then there exists a set that consists of all of the elements of $A$ that satisfies this condition.
In ZF, this result is known as the Axiom of Specification.

Proof:
By the axiom of class comprehension, let $B$ be the class defined as:
{{begin-eqn}}
{{eqn | l = B
      | r = \set {x: x \in A \land \map \phi x}
}}
{{eqn | r = \set {x \in A: \map \phi x}
      | c = Set-Builder Notation
}}
{{end-eqn}}
{{AimForCont}} that $B$ is not a set.
Then $B$ must be a proper class.
It is easily seen that $B \subseteq A$.
So by the Axiom of Powers:
:$B \in \powerset A$
where $\powerset A$ is denotes the power set of $A$.
But by Proper Class is not Element of Class, this is a contradiction.
Therefore by Proof by Contradiction it follows that $B$ is a set.
{{qed}}
Theorem: Hexagonal Number as 4 times Triangular Number plus n

Let $H_n$ be the $n$th hexagonal number.
Then:
:$H_n = 4 T_{n - 1} + n$
where $T_{n - 1}$ is the $n - 1$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = H_n
      | r = n \paren {2 n - 1}
      | c = Closed Form for Hexagonal Numbers
}}
{{eqn | r = \frac {2 n \paren {2 n - 1} } 2
      | c = 
}}
{{eqn | r = \frac {2 n \paren {2 n - 2 + 1} } 2
      | c = 
}}
{{eqn | r = 4 \frac {n \paren {n - 1} } 2 + \frac {2 n} 2
      | c = 
}}
{{eqn | r = 4 T_{n - 1} + n
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Dependent on N Functions

Let $K$ be a (real) functional, such that:
:$\ds K \sqbrk {\mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$
where:
:$\mathbf h$ is an $N$-dimensional vector
:$\mathbf Q$ is a $N \times N$ matrix
:$\mathbf P$ is a $N\times N$ symmetric positive definite matrix.
Let $\closedint a b$ be such that it does not contain a point conjugate to $a$.
Then:
:$\forall \mathbf h: \map {\mathbf h} a = \map {\mathbf h} b = 0: K \sqbrk {\mathbf h} > 0$
{{iff}} the above holds.
{{explain|"the above" -- reference the specific statement, there is plenty of "above" and it may not be immediately clear exactly what}}

Proof:
=== Necessary Condition ===
Let $\mathbf W$ be an arbitrary differentiable symmetric matrix.
Then
{{begin-eqn}}
{{eqn | l = 0
      | r = \int_a^b \map {\frac \d {\d x} } {\mathbf h \mathbf W \mathbf h} \rd x
}}
{{eqn | r = \int_a^b \mathbf h \mathbf W' \mathbf h \rd x + 2 \int_a^b \mathbf h' \mathbf W \mathbf h \rd x
}}
{{end-eqn}}
Suppose, $\mathbf W$ is such that:
:$\mathbf Q + \mathbf W' = \mathbf W \mathbf P^{-1} \mathbf W$
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x
      | r = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf Q \mathbf h + \mathbf h \mathbf W' \mathbf h} \rd x 
}}
{{eqn | r = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf W \mathbf P^{-1} \mathbf W \mathbf h} \rd x
}}
{{eqn | r = \int_a^b \paren {\mathbf P^{1/2} \mathbf h' + \mathbf P^{1/2} \mathbf h' \mathbf P^{-1/2} \mathbf W \mathbf h + \mathbf P^{-1/2} \mathbf W \mathbf h}^2 \rd x
}}
{{end-eqn}}
{{explain|Clarify the below -- what is being used to show what?}}
Note that:
:$\mathbf P^{1/2} \mathbf h' + \mathbf P^{-1/2} \mathbf W \mathbf h \ne 0$
unless:
:$\forall x\in\closedint a b: \map {\mathbf h} x = 0$
However, this contradicts the absence of conjugate points.
Hence:
:$K > 0$
{{qed|lemma}}
Theorem: Set of Upper Closures of Compact Elements is Basis implies Complete Scott Topological Lattice is Algebraic

Let $L = \struct {S, \preceq, \tau}$ be a complete Scott Definition:Topological Lattice.
Let $\BB = \set {x^\succeq: x \in \map K L}$ be a basis of $L$, where:
:$x^\succeq$ denotes the upper closure of $x$,
:$\map K L$ denotes the compact subset of $L$.
Then $L$ is algebraic.

Proof:
Thus by Compact Closure is Directed:
:$\forall x \in S:x^{\mathrm {compact} }$ is directed
where $x^{\mathrm {compact} }$ denotes the compact closure of $x$.
Thus by definition of complete lattice:
:$L$ is up-complete.
Let $x \in S$.
By definition of lower closure of element:
:$x$ is upper closure for $x^\preceq$
By definition of supremum:
:$\map \sup {x^\preceq} \preceq x$
By Compact Closure is Intersection of Lower Closure and Compact Subset:
:$x^{\mathrm {compact} } = x^\preceq \cap \map K L$
By Intersection is Subset:
:$x^{\mathrm {compact} } \subseteq x^\preceq$
By Supremum of Subset:
:$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {x^\preceq}$
By definition of transitivity:
:$\map \sup {x^{\mathrm {compact} } } \preceq x$
{{AimForCont}}
:$x \ne \map \sup {x^{\mathrm {compact} } }$
We will prove that:
:$x \notin \paren {x^{\mathrm {compact} } }^\preceq$
{{AimForCont}}:
:$x \in \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
By definition of lower closure of element:
:$x \preceq \map \sup {x^{\mathrm {compact} } }$
Thus by definition of antisymmetry:
:this contradicts $x \ne \map \sup {x^{\mathrm {compact} } }$
{{qed|lemma}}
By definition of relative complement:
:$x \in \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By Complement of Lower Closure of Element is Open in Scott Topological Ordered Set:
:$\relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$ is open.
By definition of basis:
:$\relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} = \bigcup \set {G \in \BB: G \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} }$
By definition of union:
:$\exists X \in \set {G \in \BB: G \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} }: x \in X$
Then:
:$X \in \BB \land X \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By definition of $\BB$:
:$\exists k \in \map K L: X = k^\succeq$
By definition of compact subset:
:$k$ is compact.
By definition of upper closure of element:
:$k \preceq x$
By definition of compact closure:
:$k \in x^{\mathrm {compact} }$
By definitions of supremum and upper bound:
:$k \preceq \map \sup {x^{\mathrm {compact} } }$
By definition of upper closure of element:
:$\map \sup {x^{\mathrm {compact} } } \in X$
By gefinition of subset:
:$\map \sup {x^{\mathrm {compact} } } \in \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By definition of difference:
:$\map \sup {x^{\mathrm {compact} } } \notin \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
Thus by definition of lower closure of element:
:this contradicts $\map \sup {x^{\mathrm {compact} } } \in \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
{{qed}}
Theorem: Class is Proper iff Bijection from Class to Proper Class

Let $A$ be a class.
Let $\mathrm P$ be a proper class.
Then $A$ is proper {{iff}} there exists a bijection from $A$ to $\mathrm P$.

Proof:
{{NotZFC}}
Theorem: Numbers which Multiplied by 2 are the Reverse of when Added to 2

{{begin-eqn}}
{{eqn | l = 47 + 2
      | r = 49
}}
{{eqn | l = 47 \times 2
      | r = 94
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 497 + 2
      | r = 499
}}
{{eqn | l = 497 \times 2
      | r = 994
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 4997 + 2
      | r = 4999
}}
{{eqn | l = 4997 \times 2
      | r = 9994
}}
{{end-eqn}}
... and so on:

Proof:
We have that:
:$\ds \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} + 2 = 4 \times 10^n + \sum_{k \mathop = 0}^{n - 1} 9 \times 10^k$
using the Basis Representation Theorem.
It remains to be demonstrated that:
:$\ds 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} = \sum_{k \mathop = 1}^n 9 \times 10^k + 4$
again using the Basis Representation Theorem.
Thus:
{{begin-eqn}}
{{eqn | l = 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7}
      | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 2 \times 9 \times 10^k + 2 \times 7
}}
{{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 18 \times 10^k + 14
}}
{{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^{n - 1} 10 \times 10^k + 10
}}
{{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 0}^{n - 1} 10 \times 10^k
}}
{{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^n 10^k
}}
{{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + 10^n + \sum_{k \mathop = 1}^{n - 1} 10^k
}}
{{eqn | r = 9 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 4
}}
{{eqn | r = \sum_{k \mathop = 1}^n 9 \times 10^k + 4
}}
{{end-eqn}}
{{qed}}
Theorem: Infinite Product of Analytic Functions

Let $D \subset \C$ be an open connected set.
Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$ that are not identically zero.
Let $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converge locally uniformly absolutely on $D$.
Then:
:$(1): \quad f = \ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely on $D$
:$(2): \quad f$ is analytic
:$(3): \quad$ For each $z \in D$, $\map {f_n} z = 0$ for finitely many $n \in \N$
:$(4): \quad$ For each $z \in D$, $\map {\operatorname {mult}_z} f = \ds \sum_{n \mathop = 1}^\infty \map {\operatorname {mult}_z} {f_n}$
where $\operatorname {mult}$ denotes multiplicity.
{{explain|The definition of multiplicity is inadequate on that page.}}

Proof:
Because $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converges locally uniformly absolutely on $D$, the series converges locally uniformly absolutely.
Thus $f = \ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely.
By Uniform Limit of Analytic Functions is Analytic, $f$ is analytic
The last claim follows from Zeroes of Infinite Product of Analytic Functions
{{qed}}
Theorem: Prime between n and 9 n divided by 8

Let $n \in \Z$ be an integer such that $n \ge 48$.
Then there exists a prime number $p$ such that $n < p < \dfrac {9 n} 8$.

Proof:
Let $\map P n$ be the property:
:there exists a prime number $p$ such that $n \le p \le \dfrac {9 n} 8$.
First note that $\map P n$ does not hold for the following $n < 48$:
:$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 15, 19, 21, 23, 24, 25, 31, 32, 47$
Taking $47$ as an example:
{{begin-eqn}}
{{eqn | l = 48
      | r = 2^4 \times 3
      | c = is not prime
}}
{{eqn | l = 49
      | r = 7^2
      | c = is not prime
}}
{{eqn | l = 50
      | r = 2 \times 5^2
      | c = is not prime
}}
{{eqn | l = 51
      | r = 3 \times 17
      | c = is not prime
}}
{{eqn | l = 52
      | r = 2^2 \times 13
      | c = is not prime
}}
{{end-eqn}}
We have that:
: $\dfrac {9 \times 47} 8 = 52 \cdotp 875$
and so it is seen that $\map P {47}$ does not hold.
{{ProofWanted|It remains to be shown that $\map P n$ holds for all $n \ge 48$ as well.}}
Theorem: Uniformly Continuous Function Preserves Uniform Convergence

Let $X$ be a set.
Let $M = \struct {A_1, d_1}$ and $N = \struct {A_2, d_2}$ be metric spaces.
Let $\sequence {g_n}$ be a sequence of mappings $g_n: X \to A_1$.
Let $g_n$ converge to $g$ uniformly on $X$ as $n \to \infty$.
Let $f: A_1 \to A_2$ be uniformly continuous on $A_1$.
Then the sequence  $\sequence {f \circ g_n}$ converges to $g \circ f$ uniformly  on $X$ as $n \to \infty$.

Proof:
Let $\epsilon \in \R_{>0}$.
Because $f$ is uniformly continuous on $A_1$, there exists $\delta \in \R_{>0}$ such that:
:$\forall x, y \in A_1: \map {d_2} {\map f x, \map f y} < \epsilon$ for $\map {d_1} {x, y} < \delta$
Because $\sequence {g_n}$ converges to $g$ uniformly on $X$ as $n \to \infty$, there exists $N > 0$ such that:
:$\map {d_2} {\map {g_n} x, \map g x} < \epsilon$ 
for all $n > N$ and all $x \in X$.
Thus:
:$\map {d_2} {\map f {\map {g_n} x}, \map f {\map g x} } < \epsilon$
for all $n > N$ and all $x \in X$.
Hence the sequence  $\sequence {f \circ g_n}$ converges to $g \circ f$ uniformly  on $X$ as $n \to \infty$.
{{qed}}
{{proofread}}
Theorem: Complex Exponential is Uniformly Continuous on Half-Planes

Let $a \in \R$.
Then $\exp$ is uniformly continuous on the half-plane $\set {z \in \C : \map \Re z \le a}$.

Proof:
Let $\epsilon > 0$.
For $x, y \in \C$ with $\map \Re x, \map \Re y \le a$:
{{begin-eqn}}
{{eqn | l = \cmod {e^x - e^y}
      | r = \cmod {e^y} \cdot \cmod {e^{x - y} - 1}
}}
{{eqn | r = e^{\map \Re y} \cdot \cmod {e^{x - y} - 1}
      | c = Modulus of Exponential is Exponential of Real Part
}}
{{eqn | o = \le
      | r = e^a \cdot \cmod {e^{x - y} - 1}
      | c = Exponential is Strictly Increasing
}}
{{end-eqn}}
Because Exponential Function is Continuous, there exists $\delta > 0$ such that $\cmod {e^z - 1} < \epsilon$ for $\cmod z < \delta$.
Thus if $\cmod {x - y} < \delta$, $\cmod {e^x - e^y} < e^a \epsilon$.
Thus $\exp$ is uniformly continuous.
{{qed}}
Category:Exponential Function
Category:Uniform Continuity
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