nyuuzyou/wikis huggingface dataset:
Semidirect Product of Groups is Group

== Theorem ==
Let $H$ and $N$ be groups.
Let $\Aut N$ denote the automorphism group of $N$.
Let $\phi : H\to \Aut N$ be a group homomorphism, that is, let $H$ act on $N$.
Let $N \rtimes_\phi H$ be the semidirect product of $N$ and $H$ with respect to $\phi$, that is:
:$N \rtimes_\phi H = (N \times H, \circ)$ where
:$(n_1, h_1) \circ (n_2, h_2) = (n_1\cdot \phi(h_1)(n_2), h_1\cdot h_2)$
Then $N\rtimes_\phi H$ is a group.
== Proof ==
=== Associativity ===
Let $(n_1,h_1),(n_2,h_2),(n_3,h_3)\in N\times H$. Then
=== Identity Element ===
Let $e = (e_N,e_H)$ and $(n,h)\in N\times H$.
Then 
and
Thus $e$ is an identity element for $\circ$.
=== Existence of inverse element ===
Let $(n,h)\in N\times H$.
We are looking for $(m,g)$ with $(n,h)\circ(m,g) = (e_N,e_H) = (m,g)\circ(n,h)$, that is:
:$\begin{eqnarray}
 n \phi_h(m) &= e_N \\
 hg &= e_H \\
 m \phi_g(n) &= e_N \\
 gh &= e_H
\end{eqnarray}$
Thus we take $g=h^{-1}$.
From the first equation we have $m=\phi_h^{-1}(n^{-1})$; from the third $m=\phi_{h^{-1}}(n)^{-1}$.
These are equal because $H$ acts by automorphisms.
The inverse of $(n,h)$ is thus $(\phi_{h^{-1}}(n^{-1}),h^{-1})$.
Category:Semidirect Products
Semidirect Product with Trivial Action is Direct Product

== Theorem ==
Let $H$ and $N$ be groups.
Let $\Aut N$ denote the automorphism group of $N$.
Let $\phi: H \to \Aut N$ be defined as:
:$\forall h \in H: \map \phi h = I_N$ for all $h \in H$
where $I_N$ denotes the identity mapping on $N$.
Let $N \rtimes_\phi H$ be the corresponding semidirect product.
Then $N \rtimes_\phi H$ is the direct product of $N$ and $H$.
== Proof ==
Pick arbitrary $\tuple {n_1, h_1}, \tuple {n_2, h_2} \in N \rtimes_\phi H$.
which meets the definition of direct product.
Category:Semidirect Products
Semigroup/Examples/Order 2

== Examples of Semigroups ==
The Cayley tables for the complete set of semigroups of order $2$ are listed below.
The underlying set in all cases is $\set {a, b}$.
:$\begin{array}{r|rr}
 & a & b \\
\hline
a & a & a \\
b & a & a \\
\end{array}
\qquad
\begin{array}{r|rr}
 & a & b \\
\hline
a & a & a \\
b & a & b \\
\end{array}
\qquad
\begin{array}{r|rr}
 & a & b \\
\hline
a & a & a \\
b & b & b \\
\end{array}$
:$\begin{array}{r|rr}
 & a & b \\
\hline
a & a & b \\
b & a & b \\
\end{array}
\qquad
\begin{array}{r|rr}
 & a & b \\
\hline
a & a & b \\
b & b & a \\
\end{array}
\qquad
\begin{array}{r|rr}
 & a & b \\
\hline
a & a & b \\
b & b & b \\
\end{array}$
:$\begin{array}{r|rr}
 & a & b \\
\hline
a & b & a \\
b & a & b \\
\end{array}$
:$\begin{array}{r|rr}
 & a & b \\
\hline
a & b & b \\
b & b & b \\
\end{array}$
== Sources ==
* : Chapter $4$. Groups: Exercise $1 \ \text{(ii)}$
Category:Examples of Semigroups
Semigroup/Examples/x+y+xy on Positive Integers

== Example of Semigroup ==
Let $\circ: \Z_{\ge 0} \times \Z_{\ge 0}$ be the operation defined on the integers $\Z_{\ge 0}$ as:
:$\forall x, y \in \Z_{\ge 0}: x \circ y := x + y + x y$
Then $\struct {\Z_{\ge 0}, \circ}$ is a semigroup.
== Proof ==
We have that:
:$\forall x, y \in \Z_{\ge 0}: x + y + x y \in \Z_{\ge 0}$
and so $\struct {\Z_{\ge 0}, \circ}$ is closed.
Now let $x, y, z \in \Z_{\ge 0}$.
We have:
and:
As can be seen by inspection:
:$x \circ \paren {y \circ z} = \paren {x \circ y} \circ z$
and so $\circ$ is associative.
The result follows by definition of semigroup.
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $3$
Category:Examples of Semigroups
Semigroup/Examples/x+y-xy on Integers

== Example of Semigroup ==
Let $\circ: \Z \times \Z$ be the operation defined on the integers $\Z$ as:
:$\forall x, y \in \Z: x \circ y := x + y - x y$
Then $\struct {\Z, \circ}$ is a semigroup.
== Proof ==
We have that:
:$\forall x, y \in \Z: x \circ y \in \Z$
and so $\struct {\Z, \circ}$ is closed.
Now let $x, y, z \in \Z$.
We have:
and:
As can be seen by inspection:
:$x \circ \paren {y \circ z} = \paren {x \circ y} \circ z$
and so $\circ$ is associative.
The result follows by definition of semigroup.
== Also see ==
* Inclusion-Exclusion Principle (think about why)
== Sources ==
* : Chapter $5$: Semigroups: Exercise $1$
Category:Examples of Semigroups
Semigroup Isomorphism/Examples/Structure with Two Operations

== Examples of Semigroup Isomorphisms ==
Let $\struct {S_1, \circ_1, *_1}$ and $\struct {S_2, \circ_2, *_2}$ be algebraic structures such that:
:$\struct {S_1, \circ_1}$ is isomorphic to $\struct {S_2, \circ_2}$
:$\struct {S_1, *_1}$ is isomorphic to $\struct {S_2, *_2}$
Then it is not necessarily the case that $\struct {S_1, \circ_1, *_1}$ is isomorphic to $\struct {S_2, \circ_2, *_2}$.
== Proof ==
;Proof by Counterexample:
Let $\R$ denote the set of real numbers.
Let $\vee$ and $\wedge$ denote the max operation and min operation respectively.
Let $\struct {\R, \vee}$ and $\struct {\R, \wedge}$ denote the algebraic structures formed from the above.
From Max and Min Operations on Real Numbers are Isomorphic, $\struct {\R, \vee}$ is isomorphic to $\struct {\R, \wedge}$.
Let $\struct {\R, \times}$ denote the algebraic structure formed from $\R$ under multiplication .
From Real Numbers under Multiplication form Monoid, $\struct {\R, \times}$ is a monoid and hence a semigroup.
Let $I_\R: \R \to \R$ denote the identity mapping on $\R$.
From Identity Mapping is Semigroup Automorphism we have that $\struct {\R, \times}$ is an automorphism and hence a fortiori an isomorphism.
Now consider the algebraic structures $\struct {\R, \vee, \times}$ and $\struct {\R, \wedge, \times}$.
We have from above that
We also have that:
:$\struct {\R, \vee}$ is isomorphic to $\struct {\R, \wedge}$
:$\struct {\R, \times}$ is isomorphic to $\struct {\R, \times}$
 there exists an isomorphism $\phi$ from $\struct {\R, \vee, \times}$ to $\struct {\R, \wedge, \times}$.
Because $\phi$ is an isomorphism, it is by definition a bijection
Hence $\phi$ is both a surjection and an injection.
Then:
Then:
Thus we have that:
:$\map \phi 0 = 0$
and:
:$\map \phi 1 = 0$
and so $\phi$ is not an injection.
This contradicts our assertion that isomorphism.
Hence by Proof by Contradiction no such isomorphism exists.
Hence $\struct {\R, \vee, \times}$ and $\struct {\R, \wedge, \times}$ are not isomorphic.
The result follows.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.6 \ \text {(b)}$
Category:Examples of Semigroup Isomorphisms
Semigroup is Group Iff Latin Square Property Holds

== Theorem ==
Let $\left({S, \circ}\right)$ be a semigroup.
Then $\left({S, \circ}\right)$ is a group for all $a, b \in S$ the Latin square property holds in $S$:
:$a \circ x = b$
:$y \circ a = b$
for $x$ and $y$ each unique in $S$.
== Proof 1 ==
== Proof 2 ==
Category:Group Theory
Category:Semigroups
Category:Semigroup is Group Iff Latin Square Property Holds
Semigroup is Group Iff Latin Square Property Holds/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $\struct {S, \circ}$ be a group.
$\struct {S, \circ}$ is a semigroup by the definition of a group.
By Group has Latin Square Property, the Latin square property holds in $S$.
=== Sufficient Condition === 
Let $\struct {S, \circ}$ be a semigroup on which the Latin square property holds.
Taking the group axioms in turn:
==== ====
$\struct {S, \circ}$ is semigroup.
Hence $\struct {S, \circ}$ is closed by definition.
==== ====
$\struct {S, \circ}$ is semigroup.
Hence $\circ$ is associative by definition.
==== ====
Let $a \in S$.
Then there is an $x \in S$ such that:
Let $b \in S$.
Then there is a $y \in S$ such that:
Proving that $x$ is a right identity:
Let $c \in S$.
Then there is a $z \in S$ such that:
Proving that $x$ is a left identity and thus an identity:
Relabel $x$ as $e$.
==== ====
Let $x \in S$.
Then $x$ has a right and left inverse by the property as there exists a $y$ and $z$ in $S$ such that:
We shall prove that $y$ is also the left inverse of $x$:
Hence every element of $S$ has a inverse.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Proposition $1.4$
Category:Semigroup is Group Iff Latin Square Property Holds
Semigroup is Group Iff Latin Square Property Holds/Proof 2

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $\struct {S, \circ}$ be a group.
$\struct {S, \circ}$ is a semigroup by the definition of a group.
By Group has Latin Square Property, the Latin square property holds in $S$.
=== Sufficient Condition === 
Let $a \in G$.
We have :
:$\exists e \in S: a \circ e = a$
and such an $e$ is unique.
Let $b \in G$.
Also :
:$\exists x \in G: x \circ a = b$
and such an $x$ is unique.
Hence:
Also :
:$\exists a' \in G: a' \circ a = e$
Thus $\struct {S, \circ}$ satisfies the right-hand Group Axioms
Thus $\struct {S, \circ}$ is a group.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Theorem $1$
Category:Semigroup is Group Iff Latin Square Property Holds
Semigroup is Subsemigroup of Itself

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {S, \circ}$ is a subsemigroup of itself.
== Proof ==
For all sets $S$, $S \subseteq S$, that is, $S$ is a subset of itself.
Thus $\struct {S, \circ}$ is a semigroup which is a subset of $\struct {S, \circ}$, and therefore a subsemigroup of $\struct {S, \circ}$.
== Sources ==
* : $\S 32$ Identity element and inverses
Category:Subsemigroups
Category:Semigroups
Semigroup of Bounded Linear Operators Uniformly Continuous iff Continuous as Map from Non-Negative Reals to Bounded Linear Operators

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a semigroup of bounded linear operators. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Then $\family {\map T t}_{t \ge 0}$ is uniformly continuous :
:the mapping $T : \hointr 0 \infty \to \map B X$ is continuous.
== Proof ==
=== Necessary Condition ===
If $T$ is continuous, then in particular it is continuous at $0$.
Since $\map T 0 = I$, we therefore have: 
:$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$
So $\family {\map T t}_{t \ge 0}$ is uniformly continuous.
=== Sufficient Condition ===
Suppose that $\family {\map T t}_{t \mathop \ge 0}$ is uniformly continuous.
 
We have that: 
:$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$
so $T$ is continuous at $0$.
Let $t > 0$ and $h > 0$. 
Then, we have:
so:
:$\ds \lim_{h \to 0^+} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$
Now let $-t < h < 0$. 
Then $h + t \in \closedint 0 t$ for all such $h$, so there exists $M > 0$:
:$\norm {\map T {t + h} }_{\map B X} \le M$
for all $-t < h < 0$, from Uniformly Continuous Semigroup Bounded on Compact Intervals.
Then we have: 
So we have: 
:$\ds \lim_{h \to 0^-} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$
and hence: 
:$\ds \lim_{h \to 0} \norm {\map T {t + h} - \map T t}_{\map B X} = 0$
for $t > 0$.
So $T : \hointr 0 \infty \to \map B X$ is continuous.
== Sources ==
* : $1.1$: Uniformly Continuous Semigroups of Bounded Linear Operators
Category:Uniformly Continuous Semigroups
Category:Semigroups of Bounded Linear Operators
Semigroup of Bounded Linear Operators is C0 iff Point Evaluations Continuous

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a semigroup of bounded linear operators. 
For each $x \in X$, define $x^\wedge : \hointr 0 \infty \to X$ by: 
:$\map {x^\wedge} t = \map T t x$
for each $t \in \hointr 0 \infty$. 
Then $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup $x^\wedge$ is continuous for each $x \in X$. 
 
== Proof ==
=== Sufficient Condition ===
Suppose that $x^\wedge$ is continuous for each $x \in X$.
In particular, for each $x \in X$ we have that $x^\wedge$ is continuous at $0$. 
That is: 
:$\ds \lim_{t \mathop \to 0^+} \map T t x = \map T 0 x = x$ for each $x \in X$.
So $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup.
=== Necessary Condition ===
Suppose that $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup.
By Bound on C0 Semigroup, there exists $M \ge 1$ and $\omega \ge 0$ such that: 
:$\norm {\map T t}_{\map B X} \le M e^{\omega t}$
for each $t \in \hointr 0 \infty$. 
Let $x \in X$.
We show that $x^\wedge$ is continuous.
We have that: 
:$\ds \lim_{t \mathop \to 0^+} \map T t x = x = \map T 0 x$
so $x^\wedge$ is continuous at $0$.
Let $t > 0$ and let $h > 0$.
Then:
That is: 
:$\ds \lim_{h \mathop \to 0^+} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$ 
Now let $0 \ge h > -t$.
We have: 
We have: 
:$\ds \lim_{h \mathop \to 0^-} M e^{\omega \paren {t + h} } = M e^{\omega t}$
Also:
:$\ds \lim_{h \mathop \to 0^-} \norm {\map T {-h} x - x} = 0$ 
from the definition of a $C_0$ semigroup.
From Product Rule for Limits of Functions and Squeeze Theorem, we have: 
:$\ds \lim_{h \mathop \to 0^-} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$
So:
:$\ds \lim_{h \mathop \to 0} \norm {\map {x^\wedge} {t + h} - \map {x^\wedge} t} = 0$
So $x^\wedge$ is continuous for each $x \in X$.
== Sources ==
* : $1.2$: Strongly Continuous Semigroups of Bounded Linear Operators
Category:C0 Semigroups
Category:Semigroups of Bounded Linear Operators
Semilattice Homomorphism is Order-Preserving

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semilattices.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a semilattice homomorphism.
Let $\preceq_1$ be the ordering on $S$ defined by:
:$a \preceq_1 b \iff \paren {a \circ b} = b$
Let $\preceq_2$ be the ordering on $T$ defined by:
:$x \preceq_2 y \iff \paren {x * y} = y$
Then:
:$\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is order-preserving.
== Proof ==
== Sources ==
* : Chapter $\text I$: Preliminaries, Definition $1.3$
Category:Semilattice Homomorphisms
Category:Increasing Mappings
Semilattice Induces Ordering

== Theorem ==
Let $\struct {S, \circ}$ be a semilattice.
Let $\RR$ be the relation on $S$ defined by, for all $a, b \in S$:
:$a \mathrel \RR b$ $a \circ b = b$
Then $\RR$ is an ordering.
== Proof ==
Let us verify that $\RR$ satisfies the three conditions for an ordering.
=== Reflexivity ===
From :
:$\forall a \in S: a \circ a = a$
Hence by definition of $\RR$:
:$\forall a \in S: a \mathrel \RR a = a$
So $\RR$ has been shown to be reflexive.
=== Antisymmetry ===
Let $a, b \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR a$.
So $\RR$ has been shown to be antisymmetric.
=== Transitivity ===
Let $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$.
Thus:
So $\RR$ has been shown to be transitive.
Having verified all three conditions, it follows that $\RR$ is an ordering.
== Sources ==
* : Chapter $\text I$: Preliminaries, Definition $1.3$
Category:Semilattices
Semilattice has Unique Ordering such that Operation is Supremum

== Theorem ==
Let $\struct {S, \vee}$ be a semilattice.
Then there exists a unique ordering $\preccurlyeq$ on $S$ such that:
:$x \vee y = \sup \set {x, y}$
where $\sup \set {x, y}$ is the supremum of $\set {x, y}$ $\preccurlyeq$.
== Proof ==
Let us define a relation $\RR$ on $S$ as:
:$\forall x, y \in S: x \mathrel \RR y \iff x \vee y = y$
From Semilattice Induces Ordering, we have that $\RR$ is an ordering.
Let us denote this ordering by $\preccurlyeq$, and recall its definition:
:$x \vee y = y \iff x \preccurlyeq y$
=== Supremum ===
Let $x, y \in S$ be arbitrary.
Let $x \vee y = z$.
We have:
Hence we see that $x \vee y$ is an upper bound of $\set {x, y}$ $\preccurlyeq$.
Now let $w$ be an arbitrary upper bound of $\set {x, y}$ $\preccurlyeq$.
That is:
:$x \preccurlyeq w$ and $y \preccurlyeq w$
Then:
So:
:$x \vee y$ is an upper bound of $\set {x, y}$ $\preccurlyeq$
:for every upper bound $w$ of $\set {x, y}$ $\preccurlyeq$, we have that $x \vee y \preccurlyeq w$
Hence, by definition, $x \vee y$ is the supremum of $\set {x, y}$ $\preccurlyeq$.
=== Uniqueness ===
It remains to be proved that the ordering $\preccurlyeq$, with the property that:
:$x \vee y = \sup \set {x, y}_\preccurlyeq$
is unique.
Indeed, suppose there exists another ordering $\preccurlyeq'$ such that:
:$x \vee y = \sup \set {x, y}_{\preccurlyeq'}$
We have:
The proof is complete.
== Also see ==
* Definition:Join (Order Theory)
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.22 \ \text {(b)}$
Category:Semilattices
Category:Suprema
Category:Orderings
Seminorm Maps Zero Vector to Zero

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_K$.
Let $X$ be a vector space over $\struct {K, \norm {\,\cdot\,}_K}$.
Let $\mathbf 0_X$ be the zero vector of $X$.
Let $p$ be a seminorm on $X$.
Then $\map p {\mathbf 0_X} = 0$.
== Proof ==
We have:
Category:Seminorms
Category:Zero Vectors
Seminorm is Sublinear Functional

== Theorem ==
Let $X$ be a vector space over $\R$.
Let $p : X \to \R$ be a seminorm on $X$. 
Then: 
:$p$ is a sublinear functional.
== Proof ==
Since $p$ is a seminorm, we have: 
:$\map p {x + y} \le \map p x + \map p y$ for each $x, y \in X$
We also have:
:$\map p {\lambda x} = \cmod \lambda \map p x$ for each $\lambda \in \R$ and $x \in X$.
and in particular: 
:$\map p {\lambda x} = \lambda \map p x$ for each $\lambda \in \R_{\ge 0}$ and $x \in X$.
So:
:$p$ is a sublinear functional.
Category:Seminorms
Category:Sublinear Functionals
Seminorm on Vector Space induces Norm on Quotient

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let:
:$N = \set {x \in X : \map p x = 0}$
From Set of Points for which Seminorm is Zero is Vector Subspace, $N$ is a vector subspace.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Then the mapping $\norm {\, \cdot \,} : X/N \to \hointr 0 \infty$ defined by:
:$\norm {\map \pi x} = \map p x$
is well-defined and a norm on $X/N$.
== Proof ==
We first want to show that if $\map \pi x = \map \pi y$ for $x, y \in X$, then $\map p x = \map p y$. 
From Quotient Mapping is Linear Transformation and Kernel of Quotient Mapping, it is enough to show that:
:if $y - x \in N$ then $\map p x = \map p y$.
It is therefore enough to show that if $x \in X$ and $z \in N$, then:
:$\map p {x + z} = \map p x$
Let $x \in X$ and $z \in N$. 
Then $\map p z = 0$. 
From Reverse Triangle Inequality: Seminormed Vector Space, we have:
:$\map p x = \size {\map p x - \map p z} \le \map p {x + z}$
From , we also have:
:$\map p {x + z} \le \map p x + \map p z = \map p x$ 
Hence we have $\map p {x + z} = \map p x$. 
Hence $\norm {\, \cdot \,}$ is well-defined.
We now prove the norm axioms. 
=== Proof of ===
Let $x \in X$. 
We have $\norm {\map \pi x} = 0$ $\map p x = 0$. 
Hence by the definition of $N$, we have $\norm {\map \pi x} = 0$ $x \in N$. 
From Kernel of Quotient Mapping, we have $\map \pi x = {\mathbf 0}_{X/N}$.
So for $x \in X$ we have $\norm {\map \pi x} = 0$ $\map \pi x = {\mathbf 0}_{X/N}$.
Hence we have proven .
=== Proof of ===
Let $x \in X$ and $\lambda \in \GF$.
Then, we have:
So we have proven . 
=== Proof of ===
Let $x, y \in X$.
Then, we have:
Hence we have proven .
Category:Seminorms
Category:Quotient Vector Spaces
Semiperfect Number is not Deficient

== Theorem ==
Let $n \in \Z_{>0}$ be a semiperfect number.
Then $n$ is not deficient.
== Proof ==
Let $n$ be semiperfect.
Then by definition, the sum of the aliquot parts of $n$ is not less than $n$.
The result follows by definition of deficient.
Category:Semiperfect Numbers
Category:Deficient Numbers
Semiperimeter of Integer Heronian Triangle is Composite

== Theorem ==
The semiperimeter of an integer Heronian triangle is always a composite number.
== Proof ==
Let $a, b, c$ be the side lengths of an integer Heronian triangle.
By Heron's Formula, its area is given by:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } \in \N$
where the semiperimeter $s$ is given by:
:$s = \dfrac {a + b + c} 2$
First we prove that $s$ is indeed an integer.
 not.
Since $2 s = a + b + c \in \N$, $2 s$ must be odd.
Hence $2 s - 2 a, 2 s - 2 b, 2 s - 2 c$ are odd as well.
Thus:
Since $16 \AA^2$ is a product of odd numbers, it must be odd.
But then $\AA^2$ is not an integer, a contradiction.
Therefore $s \in \N$.
Now we show that $s$ is composite number.
 not.
Then $s$ is either $1$ or prime.
Since $a, b, c \ge 1$, $s \ge \dfrac 3 2 > 1$.
Hence $s$ is prime.
Since:
:$\AA^2 = s \paren {s - a} \paren {s - b} \paren {s - c}$
We have $s \divides \AA^2$.
By Prime Divides Power, $s^2 \divides \AA^2$.
Thus $s \divides \paren {s - a} \paren {s - b} \paren {s - c}$.
By Euclid's Lemma, $s$ divides some $s - x$.
However by Absolute Value of Integer is not less than Divisors:
:$s \le s - x$
which is a contradiction.
Therefore $s$ is composite.
Category:Heronian Triangles
Separability in Uncountable Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Then $H$ is not separable.
== Proof ==
By definition, $H$ is separable there exists a countable subset of $S$ which is everywhere dense in $T$.
Let $V \subseteq H$ where $V$ is countable.
$V$ is not open in $T$ as it does not contain $p$.
From Subset of Particular Point Space is either Open or Closed it follows that $V$ is closed.
From Closed Set Equals its Closure, $V^- = V$.
But $V^- \ne H$ as $V$ is countable and $H$ is uncountable.
So whatever $V$ is, if it is countable it is not everywhere dense.
The result follows from definition of separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $10$. Uncountable Particular Point Topology: $6$
Category:Separable Spaces
Category:Particular Point Topology
Separability is not Weakly Hereditary

== Theorem ==
The property of separability is not weakly hereditary.
== Proof ==
It needs to be demonstrated that there exists a separable topological space which has a subspace which is closed but not separable.
Consider an uncountable particular point space $T = \struct {S, \tau_p}$.
From Particular Point Space is Separable, $T$ is separable.
By definition, the particular point $p$ is an open point of $T$.
Thus the subset $S \setminus \set p$ is by definition closed in $T$.
But from Separability in Uncountable Particular Point Space, $S \setminus \set p$ is not separable.
Thus by Proof by Counterexample, separability is not weakly hereditary.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
Category:Separable Spaces
Category:Weakly Hereditary Properties
Separability of Normed Vector Space preserved under Isometric Isomorphism

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a separable normed vector space.
Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space that is isometrically isomorphic to $\struct {X, \norm \cdot_X}$.
Let $T : X \to Y$ be an isometric isomorphism between $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$.
Then $\struct {Y, \norm \cdot_Y}$ is separable.
== Proof ==
Let $\mathcal S = \set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$. 
We show that $\map T {\mathcal S} = \set {T x_n : n \in \N}$ is a countable everywhere dense subset of $Y$.
Let $y \in Y$ and $\epsilon > 0$. 
Since $T$ is a bijection, there exists $x \in X$ such that $y = T x$.
Since $\mathcal S$ is everywhere dense in $X$, there exists $j \in \N$ such that: 
:$\norm {x - x_j}_X < \epsilon$
Since $T$ is a linear isometry, we have: 
:$\norm {\map T {x - x_j} }_Y = \norm {x - x_j}_X$
and so: 
:$\norm {\map T {x - x_j} }_Y < \epsilon$
That is: 
:$\norm {T x - T x_j}_Y < \epsilon$
so:
:$\norm {y - T x_j}_Y < \epsilon$
Since $\epsilon > 0$ and $y \in Y$ were arbitrary, and $T x_j \in \map T {\mathcal S}$, we have that:
:$\map T {\mathcal S}$ is everywhere dense in $Y$.
From Image of Countable Set under Mapping is Countable, we have: 
:$\map T {\mathcal S}$ is countable.
So:
:$\struct {Y, \norm \cdot_Y}$ is separable.
Category:Isometric Isomorphisms (Normed Vector Spaces)
Category:Separable Spaces