nyuuzyou/wikis huggingface dataset:
Diophantus of Alexandria/Arithmetica/Book 1/Problem 17/Proof 1

== Problem ==
== Solution ==
== Proof ==
Let $x$ be the sum of all $4$ numbers.
Then the numbers individually are:
The result follows.
== Sources ==
* : Book $\text I$: Problem $17$
* : The Arithmetica: Book $\text {I}$: $17$
Category:Diophantus of Alexandria/Arithmetica/Book 1/Problem 17
Diophantus of Alexandria/Arithmetica/Book 1/Problem 17/Proof 2

== Problem ==
== Solution ==
== Proof 2 ==
Let $a$, $b$, $c$ and $d$ be the numbers requested.
Then we have:
We use the method of false position.
Set $c = 1$.
Then:
These sums are all $9$ less than what they should be.
So we have to add $3$ to each number, to make:
Category:Diophantus of Alexandria/Arithmetica/Book 1/Problem 17
Diophantus of Alexandria/Arithmetica/Book 1/Problem 17/Solution

== Problem ==
== Solution ==
== Sources ==
* : Book $\text I$: Problem $17$
* : The Arithmetica: Book $\text {I}$: $17$
* : The First Pure Number Puzzles: $25$
Category:Diophantus of Alexandria/Arithmetica/Book 1/Problem 17
Diophantus of Alexandria/Arithmetica/Book 1/Problem 22

== Example of Diophantine Problem ==
:To find $3$ numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal.
That is:
Let $\dfrac 1 p, \dfrac 1 q, \dfrac 1 r$ be given.
The exercise is to find a set of $3$ natural numbers $\set {x, y, z}$ such that:
where $m$ is a natural number.
=== Example: $\dfrac 1 3, \dfrac 1 4, \dfrac 1 5$ ===
== Sources ==
* : Book $\text I$: Problem $22$
* : The Arithmetica: Book $\text {I}$: $22$
* : The First Pure Number Puzzles: $26$
Category:Examples of Diophantine Problems
Diophantus of Alexandria/Arithmetica/Book 1/Problem 22/6, 4, 5

== Example of Diophantine Problem ==
:To find $3$ numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal.
Let:
:the first give $\dfrac 1 3$ of itself to the second
:the second give $\dfrac 1 4$ of itself to the third
:the third give $\dfrac 1 5$ of itself to the first.
What are the numbers?
== Solution ==
The solution given by is:
:$\set {6, 4, 5}$
but it is only the ratio which is important, so:
:$\set {12, 8, 10}$
is another solution.
== Proof ==
Let $m$ be the final number.
Let $x = 3 p$.
$y$ is assumed to be divisible by $4$.
Let $y = 4$ be chosen.
Thus after giving and taking to and from $y$, we have:
Thus $x$ must also become $p + 3$.
This $x$ must have taken $p + 3 - 2 p$, that is, $3 - p$.
Thus $3 - p$ must also be $\dfrac z 5$.
That is:
:$z = 15 - 5 p$
Thus:
Hence the solution:
== Sources ==
* : Book $\text I$: Problem $22$
* : The Arithmetica: Book $\text {I}$: $22$
* : The First Pure Number Puzzles: $26$
Category:Examples of Diophantine Problems
Diophantus of Alexandria/Arithmetica/Book 1/Problem 8

== Problem ==
:What number must be added to $100$ and to $20$ (the same number added to each) so that the sums are in the ratio $3 : 1$?
== Solution ==
:$20$
== Proof ==
Let $x$ be the number added.
Then we have:
== Sources ==
* : Book $\text I$: Problem $8$
* : The Arithmetica: Book $\text {I}$: $8$
* : The First Pure Number Puzzles: $23$
Category:Category:Diophantus of Alexandria/Arithmetica
Diophantus of Alexandria/Arithmetica/Book 3/Problem 12

== Example of Diophantine Problem ==
To find $3$ numbers such that the product of any $2$ of them added to the $3$rd gives a square.
That is, let $\set {p, q, r}$ be a set of $3$ natural numbers such that:
:$p q + r$ is square
:$q r + p$ is square
:$r p + q$ is square
What are those $3$ numbers?
== Solution ==
The solution given by is:
:$\set {1, 7, 9}$
As can be seem:
== Proof ==
Take a square and subtract part of it for $r$.
Let $x^2 + 6 x + 9$ be $p q + r$, and let $r = 9$.
Hence:
:$p q = x^2 + 6 x = x \paren {x + 6}$
Let $p = x$ so that $q = x + 6$.
By the two remaining conditions:
:$10 x + 54$
:$10 x + 6$
are both square.
Thus we need to find $2$ squares whose difference is $48$.
In the words of :
:This is easy and can be done in an infinite number of ways.
This of course is not true, but the squares $16$ and $64$ come directly to mind.
Equating these to the given expressions:
and also:
and so $p = 1$.
The final number $q = x + 6$ yields that $q = 7$.
== Sources ==
* : Book $\text {III}$: Problem $12$
* : The Arithmetica: Book $\text {III}$: $12$
* : Square Problems: $27$
Category:Examples of Diophantine Problems
Diophantus of Alexandria/Arithmetica/Book 3/Problem 6

== Example of Diophantine Problem ==
To find $3$ numbers such that their sum is a square and the sum of any pair of them is a square.
That is, let $\set {p, q, r}$ be a set of $3$ natural numbers such that:
:$p + q + r$ is square
:$p + q$ is square
:$q + r$ is square
:$r + p$ is square.
What are those $3$ numbers?
== Solution ==
The solution given by is:
:$\set {41, 80, 320}$
As can be seem:
== Proof ==
Let $p + q + r$ be $x^2 + 2 x + 1$.
Let:
:$p + q = x^2$
and so:
:$r = 2 x + 1$
Let $q + r = \paren {x - 1}^2$.
Therefore:
But we have that $p + r$ is a square.
That is:
:$6 x + 1 = m^2$
which is satisfied by $m^2 = 121$.
This gives us:
:$x = 20$
and so the numbers are:
== Sources ==
* : Book $\text {III}$: Problem $6$
* : The Arithmetica: Book $\text {III}$: $6$
* : Square Problems: $28$
* : Chapter $7$: Patterns in Numbers: Diophantus
Category:Examples of Diophantine Problems
Diophantus of Alexandria/Arithmetica/Book 5/Problem 30

== Example of Diophantine Problem ==
:A man buys a certain number of measures of wine, some at $8$ drachmas, some at $5$ drachmas each.
:He pays for them a square number of drachmas;
:and if we add $60$ to this number, the result is a square, the side of which is equal to the whole number of measures.
:Find how many he bought at each price.
== Solution ==
The solution given by is:
:The number of $5$-drachma measures is $\dfrac {79} {12}$.
:The number of $8$-drachma measures is $\dfrac {59} {12}$.
== Proof ==
Let $x$ be the total number of measures bought.
Then $x^2 - 60$ is the price paid, which is itself a square number: $\paren {x - m}^2$, say.
So, $\dfrac 1 5$ of the price of the $5$-drachma measures plus $\dfrac 1 8$ of the price of the $8$-drachma measures equals $x$.
Hence the total price $x^2 - 60$ has to be divided into $2$ parts such that $\dfrac 1 5$ of one plus $\dfrac 1 8$ of the other equals $x$.
This has no real solution unless:
Thus we have:
:$x^2 > 5 x + 60$
and so $x^2$ equals $5 x$ plus a number greater than $60$.
Hence $x$ is not less than $11$.
Similarly, we have:
:$x^2 < 8 x + 60$
and so $x^2$ equals $8 x$ minus a number less than 60.
Hence $x$ is not greater than $12$.
That is:
:$11 < x < 12$
From above, we deduce:
:$x = \dfrac {m^2 + 60} {2 m}$
and so:
:$22 m < m^2 + 60 < 24 m$
So:
:$22 m$ equals $m^2$ plus some number less than $60$
and so:
:$m^2$ is not less than $19$.
Also:
:$24 m$ equals $m^2$ minus some number greater than $60$
and so:
:$m^2$ is less than $21$.
So we can put $m = 20$, and so:
:$x^2 - 60 = \paren {x - 20}^2$
which leads to:
It remains to divide $72 \frac 1 4$ into $2$ parts such that $\dfrac 1 5$ of one part plus $\dfrac 1 8$ of the other part equals $11 \frac 1 2$.
Let the first part be $5 z$.
Therefore $\dfrac 1 8$ times the second half equals $11 \frac 1 2 + z$
That is, the second half equals $92 - 8 z$m which leads to:
:$5 z + 92 - 8 z = 72 \frac 1 4$
and so:
:$z = \dfrac {79} {12}$
Hence the result.
== Sources ==
* : Book $\text {V}$: Problem $30$
* : The Arithmetica: Book $\text {V}$: $30$
* 
* : Square Problems: $29$
Category:Examples of Diophantine Problems
Dipper Operation/Examples

== Examples of Dipper Operations ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
=== Example: $m = 0$ ===
=== Example: $n = 1$ ===
=== Example: $+_{3, 4}$ ===
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)

== Examples of Dipper Operation ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $m = 3$ and $n = 4$.
The Cayley table for $+_{3, 4}$ can be presented as follows:
== Examples ==
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)/Cayley Table

== Cayley Table for the Dipper Operation ==
The dipper operation $+_{3, 4}$ can be described by showing its Cayley table.
$\quad \begin{array}{r|rrrrrrr}
+_m & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6
\\
1 & 1 & 2 & 3 & 4 & 5 & 6 & 3
\\
2 & 2 & 3 & 4 & 5 & 6 & 3 & 4
\\
3 & 3 & 4 & 5 & 6 & 3 & 4 & 5
\\
4 & 4 & 5 & 6 & 3 & 4 & 5 & 6
\\
5 & 5 & 6 & 3 & 4 & 5 & 6 & 3
\\
6 & 6 & 3 & 4 & 5 & 6 & 3 & 4
\\
\end{array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(a)}$
Category:Examples of Cayley Tables
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)/Examples

== Examples of Equations on Dipper Operation $+_{3, 4}$ ==
Let $\N_{<7}$ denote the initial segment of the natural numbers:
:$\N_{<7} := \set {0, 1, \ldots, 6}$
Let $+_{3, 4}$ be the dipper operation on $\N_{<7}$ defined as:
:$\forall a, b \in \N_{<7}: a +_{3, 4} b = \begin{cases} a + b & : a + b < 3 \\ a + b - 4 k & : a + b \ge 3 \end{cases}$
where $k$ is the largest integer satisfying:
:$3 + 4 k \le a + b$
=== Example: $x +_{3, 4} 2 = 3$ ===
=== Example: $x +_{3, 4} x = 4$ ===
=== Example: $x +_{3, 4} x = x$ ===
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)/Examples/x +(3,4) 2 = 3

== Examples of Equations on Dipper Operation $+_{3, 4}$ ==
Let $\N_{<7}$ denote the initial segment of the natural numbers:
:$\N_{<7} := \set {0, 1, \ldots, 6}$
Let $+_{3, 4}$ be the dipper operation on $\N_{<7}$ defined as:
:$\forall a, b \in \N_{<7}: a +_{3, 4} b = \begin{cases} a + b & : a + b < 3 \\ a + b - 4 k & : a + b \ge 3 \end{cases}$
where $k$ is the largest integer satisfying:
:$3 + 4 k \le a + b$
Consider the equation:
:$x +_{3, 4} 2 = 3$
This has the solutions:
:$x = 1$
:$x = 5$
== Proof ==
Apparent from direct inspection of the Cayley table:
In the column headed by $2$ we see that $x +_{3, 4} 2 = 3$ when $x = 1$ and $x = 5$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(c)}$
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)/Examples/x +(3,4) x = 4

== Examples of Equations on Dipper Operation $+_{3, 4}$ ==
Let $\N_{<7}$ denote the initial segment of the natural numbers:
:$\N_{<7} := \set {0, 1, \ldots, 6}$
Let $+_{3, 4}$ be the dipper operation on $\N_{<7}$ defined as:
:$\forall a, b \in \N_{<7}: a +_{3, 4} b = \begin{cases} a + b & : a + b < 3 \\ a + b - 4 k & : a + b \ge 3 \end{cases}$
where $k$ is the largest integer satisfying:
:$3 + 4 k \le a + b$
Consider the equation:
:$x +_{3, 4} x = 4$
This has the solutions:
:$x = 2$
:$x = 4$
:$x = 6$
== Proof ==
Apparent from direct inspection of the Cayley table:
The main diagonal contains the elements $x$ of $\N_{<7}$ such that $x +_{3, 4} x$.
As can be seen, $x +_{3, 4} x = 4$ when $x = 2$, $x = 4$ and $x = 6$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(c)}$
Category:Examples of Dipper Operations
Dipper Operation/Examples/(m, n) = (3, 4)/Examples/x +(3,4) x = x

== Examples of Equations on Dipper Operation $+_{3, 4}$ ==
Let $\N_{<7}$ denote the initial segment of the natural numbers:
:$\N_{<7} := \set {0, 1, \ldots, 6}$
Let $+_{3, 4}$ be the dipper operation on $\N_{<7}$ defined as:
:$\forall a, b \in \N_{<7}: a +_{3, 4} b = \begin{cases} a + b & : a + b < 3 \\ a + b - 4 k & : a + b \ge 3 \end{cases}$
where $k$ is the largest integer satisfying:
:$3 + 4 k \le a + b$
Consider the equation:
:$x +_{3, 4} x = x$
This has the solutions:
:$x = 0$
:$x = 4$
== Proof ==
Apparent from direct inspection of the Cayley table:
The main diagonal contains the elements $x$ of $\N_{<7}$ such that $x +_{3, 4} x$.
As can be seen, $x +_{3, 4} x = x$ when $x = 0$ and $x = 4$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(c)}$
Category:Examples of Dipper Operations
Dipper Operation/Examples/m = 0

== Examples of Dipper Operation ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $m = 0$.
Then $+_{m, n}$ degenerates to modulo addition modulo $n$ on $\N_{<n}$:
:$\forall a, b \in \N_{<n}: a +_n b = a + b - k n$
where $k$ is the largest integer satisfying:
:$k n \le a + b$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(a)}$
Category:Examples of Dipper Operations
Dipper Operation/Examples/n = 1

== Examples of Dipper Operation ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $n = 1$.
Then $+_{m, n}$ degenerates to the following operation on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, 1} b = \begin{cases} a + b & : a + b < m \\ m & : a + b \ge m \end{cases}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(a)}$
Category:Examples of Dipper Operations
Dipper Operation is Associative

== Theorem ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
The dipper operation on $\N_{< \paren {m \mathop + n} }$ is associative.
== Proof ==
Recall the definition of the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \Z_{>0}: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $a, b, c \in \N_{< \paren {m \mathop + n} }$ be arbitrary.
There are a number of cases to address.
By Dipper Operation is Commutative it is possible to arrange $a$, $b$ and $c$ into any order, and the result of $\paren {a +_{m, n} b} +_{m, n} c$ and $a +_{m, n} \paren {b +_{m, n} c}$ will be unaffected.
, therefore, it is sufficient to consider the scenario where $a \le b \le c$.
;Case 1: Let $a + b + c < m$.
Then:
Otherwise $a + b + c \ge m$.
;Case 2: Let $a + b < m$ and $b + c < m$.
Then:
;Case 3: Let $a + b < m$ and $b + c \ge m$.
Then:
and:
;Case 4: Let $a + b \ge m$ and $b + c \ge m$.
Then:
and:
So in all cases:
:$\paren {a +_{m, n} b} +_{m, n} c = a +_{m, n} \paren {b +_{m, n} c}$
and the result follows by definition of associative operation.
== Also see ==
* Dipper Operation is Commutative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(b)}$
Category:Dipper Operations
Category:Examples of Associative Operations
Dipper Operation is Closed on Initial Segment

== Theorem ==
Let $m \in \N$ be a natural number.
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ denote the dipper operation on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Then $+_{m, n}$ is closed on $\N_{< \paren {m \mathop + n} }$.
== Proof ==
Let $a + b < m$.
Then:
:$0 \le a +_{m, n} b < m$
and so:
:$a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$
Let $a + b > m$.
Then:
:$a +_{m, n} b = a + b - k n$
where $k$ is the largest such that $m + k n < a + b$
Hence:
That is:
:$a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$
That is, $+_{m, n}$ is closed on $\N_{< \paren {m \mathop + n} }$.
Category:Dipper Operations
Category:Algebraic Closure
Dipper Operation is Commutative

== Theorem ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
The dipper operation on $\N_{< \paren {m \mathop + n} }$ is commutative.
== Proof ==
Recall the definition of the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \Z_{>0}: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $a + b < m$.
Then:
Otherwise $a + b \ge m$.
Then:
So in both cases:
:$a +_{m, n} b = b +_{m, n} a$
and the result follows by definition of commutative operation.
== Also see ==
* Dipper Operation is Associative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(b)}$
Category:Dipper Operations
Category:Examples of Commutative Operations
Dipper Relation is Congruence for Addition

== Theorem ==
Let $m \in \N$ be a natural number.
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\RR_{m, n}$ be the dipper relation on $\N$:
:$\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR_{m, n}$ is a congruence relation for addition.
== Proof ==
From Dipper Relation is Equivalence Relation we have that $\RR_{m, n}$ is an equivalence relation.
From Equivalence Relation is Congruence iff Compatible with Operation, it is sufficient to show that:
:$\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {x + z} \mathrel {\RR_{m, n} } \paren {y + z}$
and:
:$\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {z + x} \mathrel {\RR_{m, n} } \paren {z + y}$
Furthermore, because Natural Number Addition is Commutative, it is sufficient to demonstrate the first of these only.
So, let $x, y \in \N$ be such that:
:$x \mathrel {\RR_{m, n} } y$
First let $x = y$.
We have that:
:$x = y \implies x + z = y + z$
and so:
:$\paren {x + z} \mathrel {\RR_{m, n} } \paren {y + z}$
Otherwise, we have that:
:$x \ne y$ and $x, y \ge m$
, let $x < y$.
If not, then as $\RR_{m, n}$ is symmetric we can rename $x$ and $y$ as necessary.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.6 \ \text {(a)}$
Category:Dipper Relations
Category:Examples of Congruence Relations
Dipper Relation is Congruence for Multiplication

== Theorem ==
Let $m \in \N$ be a natural number.
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\RR_{m, n}$ be the dipper relation on $\N$:
:$\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR_{m, n}$ is a congruence relation for multiplication.
== Proof ==
From Dipper Relation is Equivalence Relation we have that $\RR_{m, n}$ is an equivalence relation.
From Equivalence Relation is Congruence iff Compatible with Operation, it is sufficient to show that:
:$\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$
and:
:$\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {z x} \mathrel {\RR_{m, n} } \paren {z y}$
Furthermore, because Natural Number Addition is Commutative, it is sufficient to demonstrate the first of these only.
First let $x = y$.
We have that:
:$x = y \implies x z = y z$
and so:
:$\paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$
Otherwise, we have that:
:$x \ne y$ and $x, y \ge m$
, let $x < y$.
If not, then as $\RR_{m, n}$ is symmetric we can rename $x$ and $y$ as necessary.
First note that if $z = 0$, then:
:$\paren {x z} = \paren {y z}= 0$
and so:
:$\paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$
Otherwise we have that $z \ge 1$ and so both $x z \ge m$ and $y z \ge m$.
Then:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.6 \ \text {(a)}$
Category:Dipper Relations
Category:Examples of Congruence Relations
Dipper Relation is Equivalence Relation

== Theorem ==
Let $m \in \N$ be a natural number.
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\RR_{m, n}$ be the dipper relation on $\N$:
:$\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR_{m, n}$ is an equivalence relation.
== Proof ==
First let it be noted that $\RR_{m, n}$ can be written as:
:$\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x, y \text { and } n \divides \size {x - y} \end {cases}$
where $\size {x - y}$ denotes the absolute value of $x - y$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
We have that:
:$\forall x \in \N: x = x$
and so:
:$x \mathrel {\RR_{m, n} } x$
Thus $\RR_{m, n}$ is seen to be reflexive.
=== Symmetry ===
Let $x, y \in \N$ such that $x \mathrel {\RR_{m, n} } y$.
If $x = y$ then trivially $y = x$ and so:
:$y \mathrel {\RR_{m, n} } x$
Otherwise note that:
:$n \divides \size {x - y} \iff n \divides \size {y - x}$
and it follows directly that:
:$y \mathrel {\RR_{m, n} } x$
Thus $\RR_{m, n}$ is seen to be symmetric.
=== Transitivity ===
Let $x, y, z \in \N$ such that $x \mathrel {\RR_{m, n} } y$ and $y \mathrel {\RR_{m, n} } z$.
First suppose $x = y$ or $y = z$ or $x = z$.
Then trivially $x \mathrel {\RR_{m, n} } z$.
Otherwise we have that:
:$x \ne y \ne z \ne x$
and:
:$x, y, z \ge m$
, let $x < y < z$.
If not, then as $\RR_{m, n}$ is symmetric we can rename $x$, $y$ and $z$ as necessary.
Then:
Thus $\RR_{m, n}$ is seen to be transitive.
Thus $\RR_{m, n}$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.6 \ \text {(a)}$
Category:Dipper Relations
Category:Examples of Equivalence Relations
Category:Dipper Relation is Equivalence Relation