nyuuzyou/wikis huggingface dataset:
Difference between Adjacent Convergents But One of Simple Continued Fraction

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $C_0, C_1, C_2, \ldots$ be the convergents of $\sqbrk {a_0, a_1, a_2, \ldots}$.
For $k \ge 2$:
:$p_k q_{k - 2} - p_{k - 2} q_k = \paren {-1}^k a_k$
That is:
:$C_k - C_{k-2} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 2} } {q_{k - 2} } = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} }$
== Proof ==
Let $k \ge 2$.
Category:Simple Continued Fractions
Difference between Adjacent Convergents of Simple Continued Fraction

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $C_0, C_1, C_2, \ldots$ be the convergents of $\sqbrk {a_0, a_1, a_2, \ldots}$.
Then for $k \ge 1$:
:$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$
That is:
:$C_k - C_{k - 1} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 1} } {q_{k - 1} } = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
== Proof ==
Proof by induction:
For all $n \in \Z: n \ge 2$, let $\map P n$ be the proposition:
:$p_n q_{n - 1} - p_{n - 1} q_n = \paren {-1}^{n + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$p_0 = a_0, p_1 = a_0 a_1 + 1, q_0 = 1, q_1 = a_1$
So:
:$p_1 q_0 - p_0 q_1 = \paren {a_0 a_1 + 1} \times 1 - a_0 a_1 = 1 = \paren {-1}^2$
So $P(2)$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$
Then we need to show:
:$p_{k + 1} q_k - p_k q_{k + 1} = \paren {-1}^{k + 2}$
=== Induction Step ===
This is our induction step:
Consider $\sqbrk {a_0, a_1, a_2, \ldots, a_k, a_{k + 1} }$.
Its final numerator and denominator are by definition:
:$p_{k + 1} = a_{k + 1} p_k + p_{k - 1}, q_{k + 1} = a_{k + 1} q_k + q_{k - 1}$
Therefore:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \ge 1: p_n q_{n-1} - p_{n-1} q_n = \paren {-1}^{n + 1}$
Category:Simple Continued Fractions
Difference between Adjacent Terms of Farey Sequence

== Theroem ==
Let $F_n$ be a Farey sequence of order $n$.
Let $\dfrac a b$ and $\dfrac c d$ be consecutive terms of $F_n$ such that $\dfrac a b < \dfrac c d$.
Then:
:$\dfrac c d - \dfrac a b = \dfrac 1 {b d}$
or equivalently:
:$b c - a d = 1$
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:Difference between Adjacent Terms of Farey Sequence
Category:Farey Sequences
Difference between Adjacent Terms of Farey Sequence/Examples/Arbitrary Example 1

== Example of Difference between Adjacent Terms of Farey Sequence ==
Consider the Farey sequence $F_5$ of order $5$:
:$\dfrac 0 1, \dfrac 1 5, \dfrac 1 4, \dfrac 1 3, \dfrac 2 5, \dfrac 1 2, \dfrac 3 5, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5, \dfrac 1 1$
Consider the consecutive terms $\dfrac 3 5$ and $\dfrac 2 3$.
We have:
== Sources ==
* 
* 
Category:Difference between Adjacent Terms of Farey Sequence
Difference between Distances from Point on Hyperbola to Foci is Constant

== Theorem ==
Let $K$ be a hyperbola.
Let $F_1$ and $F_2$ be the foci of $K$.
Let $P$ be an arbitrary point on $K$.
Then the distance from $P$ to $F_1$ minus the distance from $P$ to $F_2$ is constant for all $P$ on $K$.
== Proof ==
== Sources ==
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Hyperbolas
Difference between Kaprekar Number and Square is Multiple of Repunit

== Theorem ==
Let $n$ be a Kaprekar number of $m$ digits.
Then:
: $n^2 - n = k R_m$
where:
:$R_m$ is the $m$-digit repunit
:$k$ is an integer.
== Proof ==
Since $n$ is a Kaprekar number of $m$ digits:
:$\begin {cases} n^2 = a \times 10^m + b \\ n = a + b \end {cases}$
for some positive integers $a$ and $b$.
Hence:
== Sources ==
* : $297$
* : $297$
Category:Kaprekar Numbers
Difference between Odd Squares is Divisible by 8

== Theorem ==
Let $a$ and $b$ be odd integers.
Then $a^2 - b^2$ is divisible by $8$.
== Proof ==
Let $a = 2 m + 1$, $b = 2 n + 1$.
Then:
Suppose $m - n$ is even such that $m - n = 2 k$.
Then:
:$a^2 - b^2 = 4 \paren {2 k} \paren {m + n + 1} = 8 k \paren {m + n + 1}$
and so is divisible by $8$.
Suppose $m - n$ is odd such that $m - n = 2 k + 1$.
Then:
:$m + n + 1 = m + \paren {2 k + 1 + m} + 1 = 2 m + 2 k = 2 \paren {m + k}$
and so:
:$a^2 - b^2 = 4 \paren {2 k + 1} 2 \paren {m + k} = 8 \paren {2 k + 1} \paren {m + k}$
and so is again divisible by $8$.
The result follows.
== Sources ==
* : $\text {2-1}$ Euclid's Division Lemma: Exercise $6$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $15 \ \text {(b)}$
Category:Odd Squares
Difference between Odd Squares is Divisible by 8/Solution/Mistake

== Source Work ==
:
:Chapter $2$: The Fundamental Theorem of Arithmetic
::$\text {2-1}$ Euclid's Division Lemma
:::Solution to Exercise $6$
== Mistake ==
:Since $a$ and $b$ are odd integers, $a = 2 r + 1$, and $b = 2 s + 1$. Thus
::$a^2 - b^2 = \paren {4 r^2 + 4 r + 1} - \paren {4 s^2 + 4 s + 1} = 4 \paren {r - s} \paren {r - s + 1}$.
:Now if $r - s$ is even, then $r - s = 2 m$ and $a^2 - b^2 = 8 m \paren {2 m + 1}$; if $r - s$ is odd, then $r - s = 2 n + 1$ and $a^2 - b^2 = 8 \paren {2 n + 1} \paren {n + 1}$. Thus in any case $a^2 - b^2$ is divisible by $8$ if $a$ and $b$ are odd integers.
== Correction ==
The last expression should be:
:$4 \paren {r - s} \paren {r + s + 1}$
The rest of the proof needs to be adjusted accordingly; for example, for odd $r - s$ we get this instead:
:$a^2 - b^2 = 8 \paren {2 n + 1} \paren {r + n}$
See Difference between Odd Squares is Divisible by 8 for a correct working.
== Sources ==
* : $\text {2-1}$ Euclid's Division Lemma: Exercise $6$
Category:Mistakes/Number Theory
Difference between Summation of Natural Logarithms and Summation of Harmonic Numbers

== Theorem ==
:$\ds \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!} \approx \gamma n + \dfrac {\ln n} 2 + 0 \cdotp 158$
where:
:$H_k$ denotes the $k$th harmonic number
:$n!$ denotes the $n$th factorial
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
From Summation over k to n of Natural Logarithm of k:
:$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$
Then:
Then we have that:
Then:
The result follows by ignoring the lower order terms in $n$.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $8$
Category:Harmonic Numbers
Category:Natural Logarithms
Difference between Terms of Wythoff Pair

== Theorem ==
Let $\tuple {\floor {k \phi}, \floor {k \phi^2} }$ be a Wythoff pair.
The difference between the coordinates of this Wythoff pair is $k$.
That is:
:$\floor {k \phi^2} - \floor {k \phi} = k$
== Proof ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Beatty Sequences
Difference between Two Squares equal to Repdigit

== Theorem ==
Some differences of two squares that each make a repdigit number include:
== Proof ==
Let $a, b$ be integers with $1 \le b < a \le 8$ and $a + b = 9$.
Then:
which is a repdigit number.
The examples above are instances with $\tuple {a, b} = \tuple {5, 4}, \tuple {6, 3}, \tuple {7, 2}, \tuple {8, 1}$.
== Sources ==
* : $1111$
* : $1111$
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit

== Theorem ==
The sequence of differences of two squares that each make a repunit begins:
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $x^2 - y^2 = R_n$ for some $n$, where $R_n$ denotes the $n$-digit repunit.
From Integer as Difference between Two Squares:
:$R_n$ has at least two distinct divisors of the same parity that multiply to $R_n$.
Then from Difference of Two Squares:
:$x = \dfrac {a + b} 2$
:$y = \dfrac {a - b} 2$
where:
:$R_n = a b$
for all $a, b$ where:
:$a b = R_n$
:$a$ and $b$ are of the same parity.
Here we have that $R_n$ is odd.
So both $a$ and $b$ are always odd and therefore always of the same parity.
It remains to perform the calculations and evaluate the examples.
== Examples ==
== Sources ==
* 
* : $111$
::Both of the above sources miss the obvious $56^2 - 55^2 = 111$.
Category:Square Numbers
Category:Repunits
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit/Corollary 1

== Theorem ==
The sequence of differences of two squares that each make a repunit begins:
and in general for integer $n$:
:$R_{2 n} = {\underbrace {55 \ldots 56}_{\text {$n - 1$ $5$'s} } }^2 - {\underbrace {44 \ldots 45}_{\text {$n - 1$ $4$'s} } }^2$
that is:
:$\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \paren {\sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6}^2 - \paren {\sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5}^2$
== Proof ==
From Difference between Two Squares equal to Repunit, $R_{2 n} = x^2 - y^2$ exactly when $R_{2 n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.
By the Basis Representation Theorem
 
 
Thus, let:
:$a = \paren {10^n + 1}$
:$b = \ds \sum_{k \mathop = 0}^{n - 1}$
So:
Similarly:
Hence the result.
== Sources ==
* : $1111$
* : $1111$
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit/Corollary 2

== Theorem ==
The sequence of differences of two squares that each make a repunit begins:
and in general for integer $n$:
:$R_{2 n} = {\underbrace{5050 \ldots 56}_{n - 1 \ 5 \text{'s} } }^2 - {\underbrace{5050 \ldots 45}_{n - 1 \ 5 \text{'s} } }^2$
that is:
:$\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6}\right)^2 - \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5}\right)^2$
== Proof ==
From Difference between Two Squares equal to Repunit, $R_{2n} = x^2 - y^2$ exactly when $R_{2n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.
By the Basis Representation Theorem
 
Thus, let:
:$a = \ds \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k}$
:$b = 11$
So:
Similarly:
Hence the result.
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit/Examples

== Examples of Difference between Two Squares equal to Repunit ==
=== Repunit $R_1$ ===
=== Repunit $R_2$ ===
=== Repunit $R_3$ ===
=== Repunit $R_4$ ===
=== Repunit $R_5$ ===
=== Repunit $R_6$ ===
=== Repunit $R_7$ ===
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit/Examples/R 6

== Example of Difference between Two Squares equal to Repunit ==
We have that:
:$111 \, 111 = 3 \times 7 \times 11 \times 13 \times 37$
So, from , there are $32$ divisors of $111 \, 111$, which can be grouped in $16$ pairs.
Each of these will generate a Difference between Two Squares equal to $111 \, 111$.
Hence:
== Sources ==
* 
Category:Difference between Two Squares equal to Repunit
Difference between Two Squares equal to Repunit/Mistake

== Source Work ==
:
:The Dictionary
::$111$
== Mistake ==
:$111 = 20^2 - 17^2$, the third difference of $2$ squares equal to a repunit. The sequence of such squares starts $1, 0$; $6, 5$; $20, 17$; $56, 45$; $156, 115$; $344, 85$; $356, 125 \ldots$
== Correction ==
The instances:
have been omitted.
This omission does not originate from -- but it does not appear (for reasons to be explained) in the list generated by and .
It is necessary to point out that the work by and started as an extension of the question:
:Find all two- and three-digit squares such that when one is subtracted from each digit, the resulting number is a square.
This was apparently a grade-school exercise set by .
 and expanded the question to extend to $n$-digit squares. The question they formulated was:
:For each $n$, find all positive integers $x$ and $y$ such that $x^2 - y^2 = 11 \cdots 1$ ($n$ ones) and $x^2$ contains no zero digits.
As such, this question is a flawed implementation of the problem in question, for the instance:
: $56^2 - 55^2 = 111$
while it satisfies this condition, it is not the case that by subtracting $1$ from each digit of $56^2$ you get $55^2$, but you do get a repunit by subtracting $55^2$ from $56^2$.
So, on analysis of the paper by and , we find that while they have indeed found:
:all the squares such that when one is subtracted from each digit, the resulting number is another square
they have not found:
:all positive integers $x$ and $y$ such that $x^2 - y^2 = 11 \cdots 1$ ($n$ ones) and $x^2$ contains no zero digits.
In particular, they have glossed over those squares of the form $56^2 - 55^2 = 111$:
and the like, as well as specifically considering and excluding those of the form:
, however, includes $20^2 - 17^2 = 400 - 289 = 111$ in his list, suggesting that he is not deliberately excluding such where $x^2$ has a zero digit.
Consequently it appears that the incompleteness of his list is the result of not having completely worked through the content of the and paper.
== Sources ==
* 
* : $111$
Category:Mistakes/Curious and Interesting Numbers
Difference in Naturally Ordered Semigroup is Unique

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $n, m \in S$ such that $m \preceq n$.
Then there exists a unique difference $n - m$ of $m$ and $n$.
== Proof ==
Since $m \preceq n$, by :
:$\exists p \in S: m \circ p = n$
Now suppose that $p, q \in S$ are such that:
:$m \circ p = m \circ q = n$
Then it follows from that:
:$p = q$
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Naturally Ordered Semigroup
Difference is Rational is Equivalence Relation

== Theorem ==
Define $\sim$ as the relation on real numbers given by:
:$x \sim y \iff x - y \in \Q$
That is, that the difference between $x$ and $y$ is rational.
Then $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
So $\sim$ is reflexive.
=== Symmetric ===
So $\sim$ is symmetric.
=== Transitive ===
So $\sim$ is transitive.
Thus $\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
Category:Examples of Equivalence Relations
Difference of Abscissae of Convergence

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Suppose that $\map f s$ has finite Abscissa of Convergence $\sigma_c$.
Then the Abscissa of Absolute Convergence $\sigma_a$ is finite, and:
:$0 \le \sigma_a - \sigma_c \le 1$
== Proof ==
It is trivial that $\sigma_a \ge \sigma_c$.
Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ converges.
It is sufficient to show that $\map f s$ converges absolutely for all $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$.
Pick an upper bound $M$ for the real numbers $\size {a_n n^{-s_0} }$.
Then for $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$:
:$\size {\dfrac {a_n} {n^s} } = \size {\dfrac {a_n} {n^{s_0} n^{s - s_0} } } = \dfrac M {n^{\sigma - \sigma_0} }$
Then by the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ converges.
== Sources ==
* : $\S 11.6$: Theorem $11.10$
Category:Dirichlet Series
Difference of Absolutely Convergent Series

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$
== Proof ==
The series $\ds \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is absolutely convergent, as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.
Then:
== Sources ==
* : $\S 4.4$. Power Series
Category:Absolute Convergence
Difference of Arccotangents

== Theorem ==
:$\arccot a - \arccot b = \arccot \dfrac {a b + 1} {a - b}$
where $\arccot$ denotes the arccotangent.
== Proof ==
Let $x = \arccot a$ and $y = \arccot b$.
Then:
== Also see ==
* Sum of Arccotangents
Category:Arccotangent Function
Difference of Arctangents

== Theorem ==
Let $\arctan a - \arctan b \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$ 
Then:
:$\arctan a - \arctan b = \arctan \dfrac {a - b} {1 + a b}$
where $\arctan$ denotes the arctangent.
== Proof ==
== Also see ==
* Sum of Arctangents
== Sources ==
* : $\text V$. Trigonometry: Formulae $(40)$
Category:Difference of Arctangents
Category:Arctangent Function
Difference of Arctangents/Mistake

== Source Work ==
:
:$\text V$. Trigonometry: Formulae $(40)$
== Mistake ==
:$\tan^{-1} a - \tan^{-1} b = \tan^{-1} \dfrac {a - b} {1 + a b}$
== Correction ==
This is valid only if $-\dfrac \pi 2 < \tan^{-1} a - \tan^{-1} b < \dfrac \pi 2$.
Outside that range, either:
:$\size {\tan^{-1} a - \tan^{-1} b} = \dfrac \pi 2$ in which case $\tan^{-1} \dfrac {a + b} {1 - a b}$ is undefined
or:
:$\size {\tan^{-1} a - \tan^{-1} b} > \dfrac \pi 2$ in which case $\tan^{-1} \dfrac {a - b} {1 + a b}$ is either $\tan^{-1} a - \tan^{-1} b + \pi$ or $\tan^{-1} a - \tan^{-1} b - \pi$ according to whether positive or negative.
== Sources ==
* : $\text V$. Trigonometry: Formulae $(40)$
Category:Mistakes/A Note Book in Pure Mathematics
Difference of Complex Conjugates

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\overline z$ denote the complex conjugate of the complex number $z$.
Then:
:$\overline {z_1 - z_2} = \overline {z_1} - \overline {z_2}$
== Proof ==
Let $w = -z_2$.
Then:
== Sources ==
* 
Category:Complex Conjugates
Category:Complex Subtraction
Difference of Complex Number with Conjugate

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Let $\map \Im z$ be the imaginary part of $z$.
Then
:$z - \overline z = 2 i \, \map \Im z$
== Proof ==
Let $z = x + i y$.
Then:
== Also defined as ==
This result is also reported as:
:$\map \Im z = \dfrac {z - \overline z} {2 i}$
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $58 \ \text{(b)}$
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation: $(2)$
Category:Complex Conjugates
Difference of Consecutive terms of Coherent Sequence

== Theorem ==
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:
:$\forall n \in \N_{>0} : 0 \le c_n < p$
 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:
:$\forall n \in \N_{>0} : c_n < p$
 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:
:$\forall n \in \N_{>0} : c_n \ge 0$
Category:P-adic Number Theory
Difference of Even Powers of z + a and z - a

== Theorem ==
Let $m \in \Z$ be an integer such that $m > 1$.
Then for all complex number $z$:
:$\paren {z + a}^{2 m} - \paren {z - a}^{2 m} = 4 m a z \ds \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \dfrac {k \pi} {2 m} }$
== Proof ==
From Factors of Difference of Two Even Powers:
:$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \ds \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$
Substituting $z + a$ for $x$, $z - a$ for $y$, and $m$ for $n$ we get:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $146$
Category:Difference of Two Powers
Difference of Fourth Powers of Cosine and Sine

== Theorem ==
:$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$
where $\sin$ and $\cos$ denote sine and cosine respectively.
== Proof 1 ==
== Proof 2 ==
Category:Difference of Fourth Powers of Cosine and Sine
Category:Sine Function
Category:Cosine Function
Difference of Functions of Bounded Variation is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$. 
Then $f - g$ is of bounded variation with:
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
where $V_{f - g}$ denotes the total variation of $f - g$ on $\closedint a b$.
== Proof ==
By Multiple of Function of Bounded Variation is of Bounded Variation, we have that: 
:$-g$ is of bounded variation.
So, by Sum of Functions of Bounded Variation is of Bounded Variation, we have that: 
:$f + \paren {-g} = f - g$ is of bounded variation
with: 
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_{-g} } {\closedint a b}$
where $V_{-g}$ is the total variation of $-g$ on $\closedint a b$. 
We have, by Multiple of Function of Bounded Variation is of Bounded Variation:
so: 
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
== Sources ==
* : $\S 6.4$: Total Variation: Theorem $6.9$
Category:Total Variation of Real Function
Category:Bounded Variation
Difference of Images under Mapping not necessarily equal to Image of Difference

== Theorem ==
Let $f: S \to T$ be a mapping.
The image of the set difference of two subsets of $S$ is not necessarily equal to the set difference of the images.
That is:
Let $S_1$ and $S_2$ be subsets of $S$.
Then it is not always the case that:
:$f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$
where $\setminus$ denotes set difference.
== Proof ==
Note that from Image of Set Difference under Mapping:
:$f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$
By Proof by Counterexample it is demonstrated that the inclusion does not necessarily apply in the other direction.
Let:
:$S_1 = \set {x \in \Z: x \le 0}$
:$S_2 = \set {x \in \Z: x \ge 0}$
:$f: \Z \to \Z: \forall x \in \Z: \map f x = x^2$
We have:
:$S_1 \setminus S_2 = \set {-1, -2, -3, \ldots}$
:$f \sqbrk {S_1} = \set {0, 1, 4, 9, 16, \ldots} = f \sqbrk {S_2}$
Then from Set Difference with Self is Empty Set:
:$f \sqbrk {S_1} \setminus f \sqbrk {S_2} = \O$
but:
:$f \sqbrk {S_1 \setminus S_2} = f \sqbrk {\set {x \in \Z: x > 0} } = \set {1, 4, 9, 16, \ldots}$
== Sources ==
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $1$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$
* : $\S 2$
* : $\S 2$: Problem $4 \ \text{(i)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(h)}$
Category:Images
Category:Set Difference
Category:Image of Set Difference under Mapping
Difference of Logarithms

== Theorem ==
Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$.
Then:
:$\log_b x - \log_b y = \map {\log_b} {\dfrac x y}$
where $\log_b$ denotes the logarithm to base $b$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 7$: Laws of Logarithms: $7.11$
* : Chapter $1$: Computations Using Logarithms
Category:Logarithms
Category:Difference of Logarithms