nyuuzyou/wikis huggingface dataset:
Order of Subgroup Product/Proof 1

== Theorem ==
== Proof ==
From Intersection of Subgroups is Subgroup, we have that $H \cap K \le H$.
Let the number of left cosets of $H \cap K$ in $H$ be $r$.
Then the left coset space of $H \cap K$ in $H$ is:
: $\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \paren {H \cap K} }$
So each element of $H$ is in $x_i \paren {H \cap K}$ for some $1 \le i \le r$.
Also, if $i \ne j$, we have:
:$x_j^{-1} x_i \notin H \cap K$
Let $h k \in H K$.
We can write $h = x_i g$ for some $1 \le i \le r$ and some $g \in K$.
Thus:
:$h k = x_i \paren {g k}$
Since $g, k \in K$, this shows $h k \in x_i K$.
 the left cosets $x_i K$ are not all disjoint.
Then by Left Coset Space forms Partition:
:$x_i K = x_j K$ for some $i, j$.
So by Left Congruence Class Modulo Subgroup is Left Coset:
:$x_j^{-1} x_i \in K$
Since $x_i, x_j \in H$, we have:
:$x_j^{-1} x_i \in H \cap K$
which contradicts the definition.
Therefore the left cosets $x_i K$ are disjoint for $1 \le i \le r$.
This leads us to:
:$\dfrac {\order H} {\order {H \cap K} } = \dfrac {\order {H K} } {\order K} = r$
whence the result.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.18$
Category:Order of Subgroup Product
Order of Subgroup Product/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
We have that $H K$ is the union of all left cosets $h K$ with $h \in H$:
:$\ds H K = \bigcup_{h \mathop \in H} h K$
From Left Coset Space forms Partition, unequal $h K$ are disjoint.
From Cosets are Equivalent, each $h K$ contains $\order K$ elements.
From the Lemma, the number of different such left cosets is:
:$\index H {H \cap K}$
where $\index H {H \cap K}$ denotes the index of $H \cap K$ in $H$.
First, let $\order H < + \infty$.
Then, from Lagrange's Theorem:
$\index H {H \cap K} = \dfrac {\order H} {\order {H \cap K} }$
Hence:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
Finally, let $\order H = + \infty$.
Recall that $G$, $H$ and $K$ have the same identity element $e$ by Identity of Subgroup.
By :
:$H = H \set e \subseteq H K$
In particular, $\order {H K} = + \infty$,
Hence:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} } = + \infty$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $10$
Category:Order of Subgroup Product
Order of Subgroup Product/Proof 3

== Theorem ==
== Proof ==
The number of product elements $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:
:$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.
So, consider the Cartesian product $H \times K$.
From Cardinality of Cartesian Product of Finite Sets:
:$\size {H \times K} = \order H \times \order K$
Let us define a relation $\sim$ on $H \times K$ as:
:$\tuple {h_1, k_1} \sim \tuple {h_2, k_2} \iff h_1 k_1 = h_2 k_2$
As $\sim$ is based on the equality relation it is seen that $\sim$ is an equivalence relation:
:Reflexivity: $h_1 k_1 = h_1 k_1$
:Symmetry: $h_1 k_1 = h_2 k_2 \implies h_2 k_2 = h_1 k_1$
:Transitivity: $h_1 k_1 = h_2 k_2, h_2 k_2 = h_3 k_3 \implies h_1 k_1 = h_3 k_3$
Each equivalence class of $\sim$ corresponds to a particular element of $H K$.
Hence $\size {H K}$ is the number of equivalence classes of $\sim$.
It remains to be shown that each of these equivalence classes contains exactly $\order {H \cap K}$ elements.
Let $E$ be the equivalence class of $\tuple {h k}$.
We aim to prove that:
:$(1): \quad E = \set {\tuple {h x^{-1}, x k}: x \in H \cap K}$
Let $x \in H \cap K$.
Then:
:$a x^{-1} \in H$
and:
:$x k \in K$
so:
:$\tuple {h x^{-1}, x k} \in H \times K$
Conversely:
:$\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$
and so:
:$x = h_1^{-1} h = k_1 k^{-1} \in A \cap B$
Thus:
:$h_1 = h x^{-1}$
and:
:$k_1 = x k$
and $(1)$ is seen to hold.
Finally:
:$\tuple {h x^{-1}, x k} = \tuple {h y^{-1}, y k} \implies x = y$
Thus $E$ has exactly $\order {H \cap K}$ elements and the proof is complete.
== Sources ==
* : $\S 1.9$: Theorem $21$
Category:Order of Subgroup Product
Order of Subgroup of Cyclic Group

== Theorem ==
Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$.
Let $a \in C_n: a = g^i$.
Let $H = \gen a$.
Then:
:$\order H = \dfrac n {\gcd \set {n, i} }$
where:
:$\order H$ denotes the order of $H$
:$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$.
== Proof ==
The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic.
We need to show that $H$ has $\dfrac n d$ elements.
Let $\order H = k$.
By Non-Trivial Group has Non-Trivial Cyclic Subgroup:
:$k = \order a$
where $\order a$ denotes the order of $a$.
That is:
:$a^k = e$
We have that $a = g^i$.
So:
We now need to calculate the smallest $k$ such that:
: $n \divides i k$
where $\divides$ denotes divisibility.
That is, the smallest $t \in \N$ such that $n t = i k$.
Let $d = \gcd \set {n, i}$.
Thus:
:$t = \dfrac {k \paren {i / d} } {n / d}$
From Integers Divided by GCD are Coprime, $\dfrac n d$ and $\dfrac i d$ are coprime.
Thus from Euclid's Lemma:
:$\dfrac n d \divides k$
As $a \divides b \implies a \le b$, the smallest value of $k$ such that $\dfrac k {\paren {n / d} } \in \Z$ is $\dfrac n d$.
Hence the result.
== Sources ==
* : Chapter $4$: Subgroups: Proposition $4.15$
Category:Cyclic Groups
Order of Subset Product with Singleton

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a singleton:
:$X = \set x$
Then:
:$\order {X \circ Y} = \order Y = \order {Y \circ X}$
where $\order S$ is defined as the order of $S$.
== Proof 1 ==
== Proof 2 ==
Category:Subset Products
Category:Singletons
Category:Order of Subset Product with Singleton
Order of Subset Product with Singleton/Proof 1

== Theorem ==
== Proof ==
From Regular Representations of Subset Product, we have that the left regular representation of $\struct {S, \circ}$ with respect to $a$ is:
:$\lambda_x \sqbrk S = \set x \circ S = x \circ S$
The result then follows directly from Regular Representation of Invertible Element is Permutation.
Category:Order of Subset Product with Singleton
Order of Subset Product with Singleton/Proof 2

== Theorem ==
== Proof ==
Let $\order Y = k$.
We define the mapping $\phi: Y \to X \circ Y$ such that:
:$\forall y \in Y: \map \phi y = x \circ y$
First we show that $\phi$ is injective.
Let $y_1, y_2 \in Y$.
Let $\map \phi {y_1} = \map \phi {y_2}$.
Hence $\phi$ is injective.
The fact that $\phi$ is surjective follows from the definition of $X \circ Y$.
Every element of $X \circ Y$ is of the form $x \circ y$ for some $y \in Y$.
Thus $\phi$ is surjective.
The other half of the result follows identically, by defining a similar function for $Y \circ X$.
Category:Order of Subset Product with Singleton
Order of Sum of Entire Functions

== Theorem ==
Let $f, g: \C \to \C$ be entire functions of order $\alpha$ and $\beta$.
Then $f + g$ has order at most $\map \max {\alpha, \beta}$, with equality if $\alpha \ne \beta$.
== Proof ==
Let $\map f z = \map \OO {e^{\cmod z^m} }$ and $\map g z = \map \OO {e^{\cmod z^k} }$.
, let $m \ge k$.
Then it follows that $\map f z + \map g z = \map \OO {e^{\cmod z^m} }$.
Then the order of $f + g$ is at most $m$.
Since $\alpha, \beta$ are the infima of $m, k$, it follows that the order of $f + g$ is at most $\alpha$.
Let $\alpha > \beta$ and $\map f z = \map \OO {e^{\cmod z^m} }$.
Then:
:$\map g z = \map {\mathcal o} {e^{\cmod z^m} }$
So $\dfrac {\map g z} {e^{\cmod z^m} } \to 0$ as $\cmod z \to \infty$.
In particular:
:$\dfrac {\map f z + \map g z} {e^{\cmod z^m} } \sim \dfrac {\map f z} {e^{\cmod z^m} } > 0$
and minimizing $m$ we see that the order of $f + g$ is at least $\alpha$.
== Also see ==
* Order of Product of Entire Functions
* Order of Product of Entire Function with Polynomial
Category:Entire Functions
Order of Sum of Reciprocal of Primes

== Theorem ==
Let $x \ge 2$ be a real number. 
We have: 
:$\ds \sum_{p \mathop \le x} \frac 1 p = \map \ln {\ln x} + \map \OO 1$
where:
:$\ds \sum_{p \mathop \le x}$ sums over the primes less than or equal to $x$
:$\OO$ is big-$\OO$ notation.
== Proof ==
We have: 
So we can write: 
We look to interchange summation and integral in our first term. 
=== Lemma ===
From Order of Sum over Primes of $\dfrac {\ln p} p$, there exists a real function $R : \hointr 2 \infty \to \R$ such that: 
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \ln x + \map R x$
with $\map R x = \map \OO 1$.
We then have: 
:$\ds \int_2^x \frac 1 {t \ln^2 t} \paren {\sum_{p \le t} \frac {\ln p} p} \rd t + \frac 1 {\ln x} \sum_{p \mathop \le x} \frac {\ln p} p = \int_2^x \frac {\ln t + \map R t} {t \ln^2 t} \rd t + 1 + \frac {\map R x} {\ln x}$
We have, by Linear Combination of Definite Integrals: 
:$\ds \int_2^x \frac {\ln t + \map R t} {t \ln^2 t} \rd t = \int_2^x \frac 1 {t \ln t} \rd t + \int_2^x \frac {\map R t} {t \ln^2 t} \rd t$
Evaluating the first term: 
We now have: 
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \map \ln {\ln x} + \paren {\int_2^x \frac {\map R t} {t \ln^2 t} \rd t + 1 + \frac {\map R x} {\ln x} - \map \ln {\ln 2} }$
We aim to show that the bracketed term is $\map \OO 1$. 
As shown in Order of Sum over Primes of $\dfrac {\ln p} p$, there exists a real constant $C > 0$ such that: 
:$\size {\map R t} \le C$
for $t \ge 2$. 
Then: 
So:
:$\ds \int_2^x \frac {\map R t} {t \ln^2 t} \rd t = \map \OO 1$
We also have: 
:$\ds \size {\frac {\map R x} {\ln x} } \le \frac C {\ln x}$
From Logarithm is Strictly Increasing, we then have:
:$\ds \frac 1 {\ln x} \le \frac 1 {\ln 2}$
for $x \ge 2$, and so:
:$\ds \size {\frac {\map R x} {\ln x} } \le \frac C {\ln 2}$ 
for $x \ge 2$. 
So:
:$\ds \frac {\map R x} {\ln x} = \map \OO 1$ 
Clearly:
:$1 - \map \ln {\ln 2} = \map \OO 1$ 
So from Sum of Big-$\OO$ Estimates, we have: 
:$\ds \int_2^x \frac {\map R t} {t \ln^2 t} \rd t + 1 + \frac {\map R x} {\ln x} - \map \ln {\ln 2} = \map \OO 1$ 
giving:
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \map \ln {\ln x} + \map \OO 1$
Category:Order of Sum of Reciprocal of Primes
Category:Analytic Number Theory
Order of Sum of Reciprocal of Primes/Lemma

== Lemma ==
Let $x \ge 2$ be a real number.
Then:
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t} = \int_2^x \frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} \rd t$
where: 
:$\ds \sum_{p \mathop \le x}$ sums over the primes less than or equal to $x$.
== Proof ==
Let $\set {p_1, p_2, \ldots, p_n}$ be the primes in the interval $\closedint 2 x$, labelled so that $p_i < p_{i + 1}$ for each $1 \le i \le n - 1$.
Note that $p_1 = 2$. 
Then, we can write: 
:$\ds \int_2^x \frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} \rd t = \sum_{i \mathop = 1}^{n - 1} \paren {\int_{p_i}^{p_{i + 1} } \paren {\frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} } \rd t} + \int_{p_n}^x \frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} \rd t$
Note that since there are no primes between $p_i$ and $p_{i + 1}$, the sum: 
:$\ds \sum_{p \mathop \le t} \frac {\ln p} p$
is constant for $t \in \hointr {p_i} {p_{i + 1} }$, for each $i$. 
So:
:$\ds \int_{p_i}^{p_{i + 1} } \paren {\frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} } \rd t = \paren {\sum_{j \mathop = 1}^i \frac {\ln p_j} {p_j} } \int_{p_i}^{p_{i + 1} } \frac 1 {t \ln^2 t} \rd t$
and:
:$\ds \int_{p_n}^x \frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} \rd t = \paren {\sum_{j \mathop = 1}^{n - 1} \frac {\ln p_j} {p_j} } \int_{p_n}^x \frac 1 {t \ln^2 t} \rd t$ 
We can now rewrite: 
:$\ds \sum_{i \mathop = 1}^{n - 1} \paren {\int_{p_i}^{p_{i + 1} } \paren {\frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} } \rd t} = \sum_{i \mathop = 1}^{n - 1} \paren {\sum_{j \mathop = 1}^i \frac {\ln p_j} {p_j} } \int_{p_i}^{p_{i + 1} } \frac 1 {t \ln^2 t} \rd t$
Clearly: 
:$\ds \sum_{j \mathop = 1}^i \frac {\ln p_j} {p_j}$
sums $p = p_j$ $j \le i$.
With this in mind, we can write:
:$\ds \sum_{i \mathop = 1}^{n - 1} \paren {\sum_{j \mathop = 1}^i \frac {\ln p_j} {p_j} } \int_{p_i}^{p_{i + 1} } \frac 1 {t \ln^2 t} \rd t = \sum_{j \mathop = 1}^{n - 1} \frac {\ln p_j} {p_j} \paren {\sum_{i \mathop = j}^{n - 1} \int_{p_i}^{p_{i + 1} } \frac 1 {t \ln^2 t} \rd t}$
Then, from Sum of Integrals on Adjacent Intervals for Integrable Functions: 
:$\ds \sum_{j \mathop = 1}^{n - 1} \frac {\ln p_j} {p_j} \paren {\sum_{i \mathop = j}^n \int_{p_i}^{p_{i + 1} } \frac 1 {t \ln^2 t} \rd t} = \sum_{j \mathop = 1}^{n - 1} \frac {\ln p_j} {p_j} \int_{p_j}^{p_n} \frac 1 {t \ln^2 t} \rd t$
Putting everything together, we have: 
Category:Order of Sum of Reciprocal of Primes
Order of Sum over Primes of Logarithm of p over p

== Theorem ==
We have: 
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \ln x + \map \OO 1$
where:
:$\ds \sum_{p \mathop \le x}$ sums over the primes less than or equal to $x$
:$\OO$ is big-$\OO$ notation.
== Proof ==
From the definition of the Von Mangoldt function, we have that: 
:$\ds \map \Lambda m = \begin {cases}\ln p & m = p^k \text { for some prime } p \text { and } k \in \N \\ 0 & \text {otherwise} \end {cases}$
so:
that is: 
:$\ds \sum_{m \mathop \le x} \frac {\map \Lambda m} m - \sum_{p \mathop \le x} \frac {\ln p} p = \sum_{k \mathop \ge 2} \sum_{p^k \mathop \le x} \frac {\ln p} {p^k}$
Consider the sum: 
:$\ds \sum_{k \mathop \ge 2} \sum_{p^k \mathop \le x} \frac {\ln p} {p^k}$ 
The sum runs over the pairs of natural numbers $\tuple {k, p}$ for which $p$ is a prime with $p^k \le x$ and $k \ge 2$.
Note that if $p^k \le x$, then $p \le x$. 
So, alternatively, for each prime number $p \le x$ we can sum over the natural numbers $k \ge 2$ for which $p^k \le x$. 
That is: 
:$\ds \sum_{k \mathop \ge 2} \sum_{p^k \le x} \frac {\ln p} {p^k} = \sum_{p \mathop \le x} \sum_{k \mathop \ge 2, \, p^k \mathop \le x} \frac {\ln p} {p^k}$ 
Since $p \ge 2$, we have $\ln p > 0$, so we only increase the sum by adding more terms.
So: 
We have: 
Note that for $p \ge 2$, we have: 
:$\dfrac 1 2 p \le p - 1$ 
giving:
:$\dfrac 1 {p \paren {p - 1} } \le \dfrac 2 {p^2}$ 
So:
:$\ds \sum_{p \mathop \le x} \ln p \paren {\sum_{k \mathop \ge 2} \frac 1 {p^k} } \le 2 \sum_{p \mathop \le x} \frac {\ln p} {p^2}$
From Order of Natural Logarithm Function, we have: 
:$\dfrac {\ln p} {p^2} \le \dfrac {2 \sqrt p} {p^2} = \dfrac 2 {p^{3/2} }$
for $p \ge 2$. 
So: 
:$\ds 2 \sum_{p \mathop \le x} \frac {\ln p} {p^2} \le 4 \sum_{p \mathop \le x} \frac 1 {p^{3/2} }$
Note that we have: 
:$\ds 4 \sum_{p \mathop \le x} \frac 1 {p^{3/2} } \le 4 \sum_{n \mathop \le x} \frac 1 {n^{3/2} }$
From Convergence of P-Series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$ converges.
So from Convergent Real Sequence is Bounded, there exists a real number $C > 0$ such that: 
:$\ds \sum_{n \mathop \le x} \frac 1 {n^{3/2} } \le 4 C$
for all $x$.
Then:
:$\ds 0 \le \sum_{p \mathop \le x} \sum_{k \mathop \ge 2, \, p^k \mathop \le x} \frac {\ln p} {p^k} \le 4 C$
for all $x$.
So:
:$\ds \sum_{p \mathop \le x} \sum_{k \mathop \ge 2, \, p^k \mathop \le x} \frac {\ln p} {p^k} = \map \OO 1$
That is: 
:$\ds \sum_{m \mathop \le x} \frac {\map \Lambda m} m - \sum_{p \mathop \le x} \frac {\ln p} p = \map \OO 1$
So that: 
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p - \sum_{m \mathop \le x} \frac {\map \Lambda m} m = \map \OO 1$
Therefore:
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \sum_{m \mathop \le x} \frac {\map \Lambda m} m + \map \OO 1$
From Order of Sum of $\dfrac {\map \Lambda n} n$, we have: 
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \ln x + \map \OO 1 + \map \OO 1$
so, from Sum of Big-O Estimates:
:$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \ln x + \map \OO 1$
Category:Analytic Number Theory
Order of Symmetric Group

== Theorem ==
Let $S$ be a finite set of cardinality $n$.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$.
Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see factorial).
== Proof ==
A direct application of Cardinality of Set of Bijections.
== Examples ==
== Sources ==
* : $\S 4.5$. Examples of groups: Example $79$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.5$
* : $\S 1.6$
* : Chapter $2$: Examples of Group Structure: $\S 30 \alpha$
* : Chapter $2$: Subgroups and Cosets: $\S 38$
* : Chapter $2$: The Symmetric Groups: $\S 78$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $5$
* 
Category:Symmetric Groups
Order of Symmetric Group/Examples/Degree 3

== Example of Order of Symmetric Group ==
Let $S = \set {1, 2, 3}$.
There are $3 \times 2 \times 1 = 6$ permutations on $S$:
:$\begin{bmatrix}
1 & 2 & 3 \\
1 & 2 & 3 
\end{bmatrix} \qquad
\begin{bmatrix}
1 & 2 & 3 \\
1 & 3 & 2 
\end{bmatrix} \qquad
\begin{bmatrix}
1 & 2 & 3 \\
2 & 1 & 3 
\end{bmatrix}$
:$\begin{bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 
\end{bmatrix} \qquad
\begin{bmatrix}
1 & 2 & 3 \\
3 & 1 & 2 
\end{bmatrix} \qquad
\begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 
\end{bmatrix}$
and so $\struct {\map \Gamma S, \circ}$ has $6$ elements.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Example $2.19$
Category:Examples of Order of Symmetric Group
Category:Symmetric Group on 3 Letters
Ordered Basis for Coordinate Plane

== Theorem ==
Let $a_1, a_2 \in \R^2$ such that $\set {a_1, a_2}$ forms a linearly independent set.
Then $\tuple {a_1, a_2}$ is an ordered basis for the $\R$-vector space $\R^2$.
Hence the points on the plane can be uniquely identified by means of linear combinations of $a_1$ and $a_2$.
== Proof ==
:
Let $P$ be any point in the plane for which we want to provide a linear combination of $a_1$ and $a_2$.
Let the distance from $O$ to the point determined by $a_1$ be defined as being $1$ unit of length on the line $L_1$.
Let the distance from $O$ to the point determined by $a_2$ be defined as being $1$ unit of length on the line $L_2$.
Draw lines parallel to $L_1$ and $L_2$ through $P$.
Then the coordinates $\lambda_1$ and $\lambda_2$ of $P$ are given by:
:$P = \lambda_1 a_1 + \lambda_2 a_2$
by the Parallelogram Law.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
Category:Coordinate Systems
Ordered Field of Rational Cuts is Isomorphic to Rational Numbers

== Theorem ==
Let $\struct {\RR, +, \times, \le}$ denote the ordered field of rational cuts.
Let $\struct {\Q, +, \times, \le}$ denote the field of rational numbers.
Then $\struct {\RR, +, \times, \le}$ and $\struct {\Q, +, \times, \le}$ are isomorphic.
== Proof ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers
Category:Cuts
Category:Rational Numbers
Ordered Field with Archimedean Property on which Monotone Convergence Theorem Holds has Continuum Property

== Theorem ==
Let $\FF$ be an ordered field on which the Archimedean property holds.
Let $\FF$ be such that every sequence which is increasing and bounded above is convergent.
Then every non-empty subset of $\FF$ which is bounded above has a supremum.
== Proof ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.7$
Category:Ordered Fields
Ordered Group whose Doubletons with Identity admit Supremum is Lattice

== Theorem ==
Let $\struct {G, \odot, \preccurlyeq}$ be an ordered group whose identity element is $e$.
Let $\preccurlyeq$ be such that for all $x \in G$, $\set {x, e}$ admits a supremum.
Then $\preccurlyeq$ is a lattice ordering.
== Proof ==
Let $x, y \in G$ be arbitrary.
We have that $y = x \odot z$ for some $z \in G$.
From Suprema in Ordered Group:
:$\set {z, e}$ admits a supremum $\set {x \odot z, x \odot e}$ admits a supremum.
It follows that:
:$\set {x, y}$ admits a supremum.
As $x$ and $y$ are arbitrary, it follows that:
:$\forall x, y \in G: \set {x, y}$ admits a supremum.
Let $x = a^{-1}$ and $y = b^{-1}$.
That is:
:$a = x^{-1}$ and $b = y^{-1}$
Thus we have that:
:$\set {a, b}$ admits a supremum.
From Inverse of Supremum in Ordered Group is Infimum of Inverses it follows directly that:
:$\set {a^{-1}, b^{-1} }$ admits an infimum
from which:
:$\set {x, y}$ admits an infimum.
As $x$ and $y$ are arbitrary, we have that:
:$\forall x, y \in G: \set {x, y}$ admits both a supremum and an infimum,
The result follows by definition of lattice ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.12$
Category:Ordered Groups
Category:Lattice Theory
Category:Suprema
Ordered Integral Domain is Totally Ordered Ring

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
Then $\struct {D, +, \times, \le}$ is a totally ordered ring.
== Proof ==
By definition, $\struct {D, +, \times, \le}$ is an integral domain endowed with a strict positivity property.
From Strict Positivity Property induces Total Ordering, the ordering $\le$ on $\struct {D, +, \times, \le}$ is a total ordering.
Hence the result by definition of totally ordered ring.
Category:Ordered Integral Domains
Category:Totally Ordered Rings
Ordered Pair/Kuratowski Formalization/Motivation

== Theorem ==
The only reason for the Kuratowski formalization of ordered pairs:
:$\tuple {a, b} = \set {\set a, \set {a, b} }$
is so their existence can be justified in the strictures of the axiomatic set theory, in particular Zermelo-Fraenkel set theory (ZF).
Once that has been demonstrated, there is no need to invoke it again.
The fact that this formulation allows that:
:$\tuple {a, b} = \tuple {c, d} \iff a = c, b = d$
is its stated aim.
The fact that $\set {a, b} \in \tuple {a, b}$ is an unfortunate side-effect brought about by means of the definition.
It would be possible to add another axiom to ZF or ZFC specifically to allow for ordered pairs to be defined, and in some systems of axiomatic set theory this is what is done.
== Sources ==
* : $\S 6$: Ordered Pairs
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions
Category:Ordered Pairs
Ordered Pair/Kuratowski Formalization/Warning

== Warning ==
The weakness of the Kuratowski formalization of the ordered pair shows up when $a = b$:
Thus the ordered pair degenerates into the set $\set {\set a}$.
Most works on this subject gloss over this point, and indeed, completely fail to mention it.
== Sources ==
* : $\S 0.2$. Sets
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
Category:Ordered Pairs