nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Non-Unity Variant of Echelon Form

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
For ongoing brevity in this proof, the term non-unity echelon matrix will be used to refer to this variant echelon matrix in which the leading coefficients are not necessarily equal to $1$.
=== $(1)$ implies $(2)$ ===
Let $\mathbf A$ be an non-unity echelon matrix by definition $1$.
Then, apart from zero rows, each row starts with strictly more zeroes than the one before it.
 there exist adjacent rows in $\mathbf A$ of the form:
:$\begin {pmatrix}
0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\
0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\
\end {pmatrix}$
where $y_1 \ne 0$.
If $x_1 \ne 0$, then the $2$nd of these rows starts with the same number of zeroes as the row before it.
If $x_1 = 0$, then the $2$nd of these rows starts with fewer zeroes as the row before it.
In both cases, this contradicts our definition of a non-unity echelon matrix.
That is, there there exist no adjacent rows in $\mathbf A$ of the form:
:$\begin {pmatrix}
0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\
0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\
\end {pmatrix}$
where $y_1 \ne 0$.
That is $\mathbf A$ is a non-unity echelon matrix by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\mathbf A$ be a non-unity echelon matrix by definition $2$.
Then by definition there exist no adjacent rows in $\mathbf A$ of the form:
:$\begin {pmatrix}
0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\
0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\
\end {pmatrix}$
where $y_1 \ne 0$.
Suppose row $k$ does not start with a sequence of zeroes.
Then it cannot start with strictly more zeroes than the previous row unless $k = 1$, in which case there is no previous row.
Hence criterion $1$ of definition $1$ is satisfied.
Suppose row $k$, where $k > 1$, is not a zero row.
Let its leading coefficient be in column $r$.
Then the leading coefficient of row $k - 1$ must be in column $s$ where $s < r$.
That is, row $k$ starts with strictly more zeroes than row $k - 1$.
Hence criterion $2$ of definition $1$ is satisfied.
Suppose row $k$ is a zero row such that $k < m$.
Then by definition row $k + 1$ cannot have a leading coefficient.
So row $k + 1$ and all rows following must be zero rows.
Hence criterion $3$ of definition $1$ is satisfied.
Thus $\mathbf A$ is a non-unity echelon matrix by definition $1$.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Echelon Matrices
Equivalence of Definitions of Norm of Linear Functional

== Theorem ==
Let $V$ be a normed vector space, and let $L$ be a bounded linear functional on $V$.
Define the following norms of $L$:
:$(1): \quad \norm L_1 = \sup \set {\size {L h}: \norm h \le 1}$
:$(2): \quad \norm L_2 = \sup \set {\size {L h}: \norm h = 1}$
:$(3): \quad \norm L_3 = \sup \set {\dfrac {\size {L h} } {\norm h}: h \in H \setminus \set {\mathbf 0} }$
:$(4): \quad \norm L_4 = \inf \set {c > 0: \forall h \in H: \size {L h} \le c \norm h}$
Then:
:$\norm L_1 = \norm L_2 = \norm L_3 = \norm L_4$
=== Corollary ===
== Proof ==
We have:
:$\set {v \in V : \norm v = 1} \subseteq \set {v \in V : \norm v \le 1} \subseteq V$
So it follows from the definition of the supremum that 
:$\norm L_2 \le \norm L_1 \le \norm L_3$
Next we show that $\norm L_2 = \norm L_3$:
Therefore
:$\norm L_1 = \norm L_2 = \norm L_3$
Moreover, if $\size {L v} \le c \norm v$ for all $v \in V \setminus \set {\mathbf 0}$, then we have:
:$\norm L_3 = \sup \set {\dfrac {\size {L v} } {\norm v}: v \in V \setminus \set {\mathbf 0} } \le c$
Taking the infimum over all such $c$ this reads:
:$\norm L_3 \le \norm L_4$
 $c_0 := \norm L_4 > \norm L_3$.
Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $v \in V \setminus \set {\mathbf 0}$:
:$\dfrac {\size {L v} } {\norm v} + \epsilon \le c_0$
But this in turn implies that for every $v \in V \setminus \set {\mathbf 0}$:
:$\size {L v} \le c_0 \norm v - \epsilon \norm v = \paren {c_0 - \epsilon} \norm v$
This contradicts the fact that $c_0$ is the least such number satisfying this inequality.
Therefore:
:$\norm L_3 = \norm L_4$
and the proof is complete.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Proposition $3.3$
Category:Bounded Linear Functionals
Category:Definition Equivalences
Equivalence of Definitions of Norm of Linear Functional/Corollary

== Theorem ==
Let $V$ be a normed vector space, and let $L$ be a bounded linear functional on $V$.
For all $v \in V$, the following inequality holds:
:$\size {L v} \le \norm L \norm v$
== Proof ==
If $v = \mathbf 0$ there is nothing to prove.
Let $v \ne \mathbf 0$.
By the definition of the supremum:
:$\dfrac {\size {L v} } {\norm v} \le \norm L_3 = \norm L$
whence:
:$\size {L v} \le \norm L \norm v$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Proposition $3.3$
Category:Bounded Linear Functionals
Equivalence of Definitions of Norm of Linear Transformation

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
Let:
From Norm on Bounded Linear Transformation is Finite:
:$\lambda_4 < \infty$
We will show that:
:$\lambda_4 \ge \lambda_2 \ge \lambda_1 \ge \lambda_3 \ge \lambda_4$
=== Lemma ===
=== Inequality: $\lambda_4 \ge \lambda_2$ ===
=== Inequality: $\lambda_2 \ge \lambda_1$ ===
=== Inequality: $\lambda_1 \ge \lambda_3$ ===
=== Inequality: $\lambda_3 \ge \lambda_4$ ===
It follows that the definitions are all equivalent. 
== Sources ==
* $\S \text {II}.1$
Category:Linear Transformations on Hilbert Spaces
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Norm of Linear Transformation/Definition 1 Greater or Equal Definition 3

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let:
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$
and
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$
Then:
:$\lambda_1 \ge \lambda_3$
== Proof ==
By definition of the supremum:
:$\forall h \in H, \norm h_H \le 1 : \norm{A h}_K \le \lambda_1$
In particular:
:$\forall h \in H, \norm h_H = 1 : \norm{A h}_K \le \lambda_1$
From Continuum Property:
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$ exists
By definition of the supremum:
:$\lambda_3 \le \lambda_1$
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Norm of Linear Transformation/Definition 2 Greater or Equal Definition 1

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let: 
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$
and
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$
Then:
:$\lambda_2 \ge \lambda_1$
== Proof ==
=== Lemma ===
By definition of the supremum:
:$\forall h \in H, h \ne \mathbf 0_H: \dfrac {\norm {A h}_K} {\norm h_H} \le \lambda_2$
Hence:
:$\forall h \in H, h \ne \mathbf 0_H: \norm {A h}_K \le \lambda_2 \norm h_H$
From Lemma:
:$\norm{A 0_H}_K = \lambda_2 \norm{0_H}_H$
Hence:
:$\forall h \in H: \norm {A h}_K \le \lambda_2 \norm h_H$
In particular:
From Continuum Property:
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$ exists
By definition of the supremum:
:$\lambda_1 \le \lambda_2$
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Norm of Linear Transformation/Definition 3 Greater or Equal Definition 4

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let:
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$
and
:$\lambda_4 = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$
Let:
:$\lambda_3 \ge \lambda_4$
== Proof ==
=== Lemma ===
Let $h \in H: h \ne 0_h$.
We have:
and
Hence:
:$\forall h \in H: h \ne 0_h: \norm{A h}_K \le \lambda_3 \norm h_H$
From Lemma:
:$\norm {A 0_H}_K = \lambda_3 \norm {0_H}_H$
Hence:
:$\forall h \in H: \norm{A h}_K \le \lambda_3 \norm h_H$
That is,
:$\lambda_3 \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$
By definition of the infimum:
:$\lambda_4 \le \lambda_3$
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Norm of Linear Transformation/Definition 4 Greater or Equal Definition 2

== Theorem ==
Let $\HH, \KK$ be Hilbert spaces.
Let $A: \HH \to \KK$ be a bounded linear transformation.
Let:
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_\KK} {\norm h_\HH}: h \in \HH, h \ne 0_\HH}$
and
:$\lambda_4 = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$
Then:
:$\lambda_4 \ge \lambda_2$
== Proof ==
From Norm on Bounded Linear Transformation is Finite:
:$\lambda_4 < \infty$
From Fundamental Property of Norm on Bounded Linear Transformation:
:$\forall h \in \HH : \norm{A h}_\KK \le \lambda_4 \norm h_\HH$
Hence:
:$\forall h \in H, h \ne 0_\HH : \dfrac {\norm{A h}_\KK} {\norm h_\HH} \le \lambda_4$
From Continuum Property:
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_\KK} {\norm h_\HH}: h \in \HH, h \ne 0_\HH}$ exists
By definition of the supremum:
:$\lambda_2 \le \lambda_4$
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Norm of Linear Transformation/Lemma

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a linear transformation.
Then:
:$\forall \lambda > 0 : \norm{A 0_H}_K = \lambda \norm{0_H}_H$
== Proof ==
Let $\lambda > 0$.
We have:
Category:Equivalence of Definitions of Norm of Linear Transformation
Equivalence of Definitions of Normal Extension

== Theorem ==
Let $L / K$ be an algebraic field extension.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition $1$ implies Definition $2$ ===
Let $\alpha \in L$ be an arbitrary element.
Let $\sigma: L \mapsto \overline K$ be an arbitrary embedding of $L$ fixing $K$.
We wish to show that $\map \sigma \alpha \in L$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$, which exists because $L / K$ is algebraic.
Since $\sigma$ fixes $K$, $\map \sigma \alpha$ must also be a root of $m_\alpha$.
By our assumption, $\alpha \in L$ implies that all roots of $m_\alpha$ are in $L$ and consequently $\map \sigma \alpha \in L$.
=== Definition $2$ implies Definition $1$ ===
Again, let $\alpha \in L$ and let $m_\alpha \in K \sqbrk x$ be its minimal polynomial over $K$.
We must show that for every root $\beta$ of $m_\alpha$, there exists an embedding $\sigma_\beta$, of $L$ in $\overline K$ such that $\map {\sigma_\beta} \alpha = \beta$. 
Consider the intermediate field $K \sqbrk \alpha \subset L$.
By Abstract Model of Algebraic Extensions, we have an automorphism $\tau_\beta$ for each root $\beta$ of $m_\alpha$ such that $\map {\tau_\beta} \alpha = \beta$ and $\tau_\beta$ fixes $K$
By Extension of Isomorphisms, each $\tau_\beta$ can be extended to an embedding $\sigma_\beta$ of $L$ in $\overline K$ such that:
:$\sigma_\beta \restriction_{K \sqbrk \alpha} = \tau_\beta$
By our assumption, $\sigma_\beta \sqbrk L = L$ for each $\beta$.
Consequently, every root of $m_\alpha$ is in $L$.
Category:Field Extensions
Equivalence of Definitions of Normal Subgroup

== Theorem ==
Let $G$ be a group.
Let $N$ be a subgroup of $G$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
=== Definition 6 ===
=== Definition 7 ===
== Proof ==
== Definition 1 iff Definition 2: Subgroup is Normal iff Left Cosets are Right Cosets ==
== Definition 1 iff Definition 3: Subgroup is Superset of Conjugate iff Normal ==
== Definition 1 iff Definition 4: Subgroup is Subset of Conjugate iff Normal ==
== Definitions 3 and 4 iff Definition 5: Subgroup equals Conjugate iff Normal ==
== Definition 1 iff Definition 6: Subgroup is Normal iff Contains Conjugate Elements ==
== Definition 1 iff Definition 7: Subgroup is Normal iff Normal Subset ==
Category:Normal Subgroups
Equivalence of Definitions of Normal Subset

== Theorem ==
=== Definition 1===
=== Definition 2===
=== Definition 3===
=== Definition 4===
=== Definition 5 ===
=== Definition 6 ===
=== Definition 7 ===
== Proof ==
=== Definition 1 is Equivalent to Definition 2 ===
=== Definition 2 Implies Definition 3 ===
=== Definition 3 is Equivalent to Definition 4 ===
=== Definitions 3 and 4 imply Definition 2 ===
=== Definition 3 is Equivalent to Definition 5 ===
Category:Group Theory
Category:Normal Subsets
Equivalence of Definitions of Normal Subset/1 iff 2

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ be a subset of $G$.
Then Normal Subset/Definition 1 is equivalent to Normal Subset/Definition 2.
That is, the following three statements are equivalent:
:$(1): \quad \forall g \in G: g \circ S = S \circ g$
:$(2): \quad \forall g \in G: g \circ S \circ g^{-1} = S$
:$(3): \quad \forall g \in G: g^{-1} \circ S \circ g = S$
== Proof ==
Let $e$ be the identity of $G$.
First note that:
:$(4): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} = S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g = S}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $S$ satisfies $(1)$.
Let $g \in G$.
Then:
=== Sufficient Condition ===
Suppose that $S$ satisfies $(2)$ or $(3)$.
By $(4)$, as long as one of these statements holds, the other one holds as well.
Let $g \in G$.
Then:
Category:Normal Subsets
Equivalence of Definitions of Normal Subset/2 implies 3

== Theorem ==
Let $\left({G, \circ}\right)$ be a group.
Let $S \subseteq G$.
Let $S$ be a normal subset of $G$ by Definition 2.
Then $S$ is a normal subset of $G$ by Definition 3.
That is, if:
:$\forall g \in G: g \circ S \circ g^{-1} = S$
or:
:$\forall g \in G: g^{-1} \circ S \circ g = S$
then:
:$\forall g \in G: g \circ S \circ g^{-1} \subseteq S$
and:
:$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$
== Proof ==
We have that:
:$\left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$
The result follows by definition of set equality.
Category:Normal Subsets
Equivalence of Definitions of Normal Subset/3 and 4 imply 2

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $S$ be a normal subset of $G$ by Definition 3 and Definition 4.
Then $S$ is a normal subset of $G$ by Definition 2.
== Proof ==
By Equivalence of Definitions of Normal Subset: 3 iff 4, $S$ being a normal subset of $G$ by Definition 3 and Definition 4 implies that the following hold:
:$(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
:$(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
:$(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$
:$(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$
By $(1)$ and $(3)$ and definition of set equality:
:$\forall g \in G: g \circ S \circ g^{-1} = S$
By $(2)$ and $(4)$ and definition of set equality:
:$\forall g \in G: g^{-1} \circ S \circ g = S$
Category:Normal Subsets
Equivalence of Definitions of Normal Subset/3 iff 4

== Theorem ==
Let $\struct {G,\circ}$ be a group.
Let $S \subseteq G$.
Then:
: $S$ is a normal subset of $G$ by Definition 3
:
: $S$ is a normal subset of $G$ by Definition 4.
That is, the following conditions are equivalent:
: $(1) \quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
: $(2) \quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
: $(3) \quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$
: $(4) \quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$
== Proof ==
First note that:
:$(5): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} \subseteq S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g \subseteq S}$
:$(6): \quad \paren {\forall g \in G: S \subseteq g \circ S \circ g^{-1}} \iff \paren {\forall g \in G: S \subseteq g^{-1} \circ S \circ g}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
Therefore:
: conditions $(1)$ and $(2)$ are equivalent
and:
: conditions $(3)$ and $(4)$ are equivalent.
It remains to be shown that condition $(1)$ is equivalent to condition $(3)$.
Suppose that $(1)$ holds.
Then:
Thus condition $(1)$ implies condition $(3)$.
The exact same argument, substituting $\supseteq$ for $\subseteq$ and using Superset Relation is Compatible with Subset Product instead of Subset Relation is Compatible with Subset Product proves that $(3)$ implies $(1)$.
Category:Normal Subsets
Equivalence of Definitions of Normal Subset/3 iff 5

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Then:
: $S$ is a normal subset of $G$ by Definition 3
:
: $S$ is a normal subset of $G$ by Definition 5.
== Proof ==
==== 3 implies 5 ====
Suppose that $S$ is a normal subset of $G$ by Definition 3.
That is:
:$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$.
Let $x, y \in G$ such that $x \circ y \in S$.
Then:
==== 5 implies 3 ====
Suppose that $S$ is a normal subset of $G$ by Definition 5.
That is:
:$\forall x, y \in G: x \circ y \in S \implies y \circ x \in S$
Let $g \in G$.
Then:
Category:Normal Subsets
Equivalence of Definitions of Nowhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $H$ be nowhere dense in $T$ by definition $1$.
Then by definition:
:$\paren {H^-}^\circ = \O$
Hence by definition of interior:
:the union of all subsets of $H$ which are open in $T$.
But this union is empty.
Hence all subsets of $H$ which are open in $T$ must themselves be empty.
Thus $H$ is nowhere dense in $T$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $H$ be nowhere dense in $T$ by definition $2$.
Then by definition:
:$H^-$ contains no open set of $T$ which is non-empty
where $H^-$ denotes the closure of $H$.
Hence the union of all subsets of $H$ which are open in $T$ must itself be empty.
Thus $H$ is nowhere dense in $T$ by definition $1$.
Category:Nowhere Dense
Equivalence of Definitions of Null Set in Euclidean Space

== Theorem ==
Let $\lambda^n$ be $n$-dimensional Lebesgue measure on $\R^n$.
Let $E \subseteq \R^n$.
Then the following are equivalent:
:$(1):\quad \exists B \in \map \BB {\R^n}: E \subseteq B, \map {\lambda^n} B = 0$
:$(2):\quad$ For every $\epsilon > 0$, there exists a countable cover $\family {J_i}_{i \mathop \in \N}$ of $E$ by open $n$-rectangles, such that:
::$\ds \sum_{i \mathop = 1}^\infty \map {\operatorname {vol} } {J_i} \le \epsilon$
== Proof ==
== Sources ==
* : $\S 6$: Problem $8$
Category:Measure Theory
Category:Definition Equivalences
Equivalence of Definitions of Octagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $O_n$ be an octagonal number by definition 1.
Let $n = 0$.
By definition:
:$O_0 = 0$
By vacuous summation:
:$\ds O_0 = \sum_{i \mathop = 1}^0 \paren {6 i - 5} = 0$
By definition of summation:
and so:
Thus $O_n$ is an octagonal number by definition 2.
=== Definition 2 implies Definition 1 ===
Let $O_n$ be an octagonal number by definition 2.
Then:
Then:
:$\ds O_0 = \sum_{i \mathop = 1}^0 \paren {6 i - 5}$
is a vacuous summation and so:
:$O_0 = 0$
Thus $O_n$ is an octagonal number by definition 1.
=== Definition 1 equivalent to Definition 3 ===
We have by definition that $O_n = 0 = \map P {8, n}$.
Then:
Thus $\map P {8, n}$ and $O_n$ are generated by the same recurrence relation.
Category:Octagonal Numbers
Equivalence of Definitions of Odd-Times Odd Integer

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $n$ be an odd-times odd integer by definition 1.
Then by definition:
:$n > 1$
:$n$ is not prime.
Thus:
:$n = r s$
where $1 < r < p$ and $1 < s < p$.
As $n$ is odd:
:$2 \nmid n$
and so:
:$2 \nmid r$
and:
:$2 \nmid s$
So both $r$ and $s$ are odd integers greater than $1$ such that $n = r s$.
Thus $n$ is an odd-times odd integer by definition 2.
=== $(2)$ implies $(1)$ ===
Let $n$ be an odd-times odd integer by definition 2.
Then by definition:
:$n = r s$
where $r > 1, s > 1$ are odd integers.
Thus $n$ is not prime by definition.
As $r$ and $s$ are odd:
:$2 \nmid r$
and
:$2 \nmid s$
and so:
:$2 \nmid r s$
So $n = r s$ is odd.
As $r > 1$ and $s > 1$ it follows from Divisor Relation on Positive Integers is Partial Ordering that $n > 1$.
Thus $n$ is an odd-times odd integer by definition 1.
Category:Odd-Times Odd Integers
Equivalence of Definitions of Odd Integer

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ $(2)$ ===
From the Division Theorem, setting $b = 2$:
:$\forall a \in \Z: \exists! q, r \in \Z: a = 2 q + r, 0 \le r < 2$
Thus, either:
:$\exists q \in \Z: n = 2 q$
or:
:$\exists q \in \Z: n = 2 q + 1$
When $n = 2 q$, $n$ is even by definition.
When $n$ is not even, $n = 2 q + 1$
Likewise, when $n = 2 q + 1$ it follows that $n$ is not even.
Hence both definitions of odd integer are equivalent.
=== $(2)$ $(3)$ ===
By definition of congruence modulo $2$:
:$x \equiv y \pmod 2 \iff \exists r \in \Z: x - y = 2 r$
Setting $y = 1$:
:$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x - y = 2 r$
from which:
:$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x = 2 r + 1$
Thus definition 2 is logically equivalent to definition 3.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm
Category:Odd Integers
Equivalence of Definitions of Odd Permutation

== Theorem ==
Let $n \in \N$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho \in S_n$ be a permutation in $S_n$.
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
The sign of $\rho$ is defined as:
:$\map \sgn \rho = \begin {cases} 1 & : \text {$k$ even} \\ -1 & : \text {$k$ odd} \\ \end {cases}$
The result follows.
Category:Odd Permutations
Equivalence of Definitions of Open Set (Complex Analysis)

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $S \subseteq \C$.
=== Definition 1 implies Definition 2 ===
Let $S$ be an open set in $\C$ by definition 1.
That is:
: $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$
Thus every point in $S$ has an $\epsilon$-neighborhood $N_{\epsilon} \left({z_0}\right)$ such that $N_{\epsilon} \left({z_0}\right) \subseteq S$.
That is, by definition, every point of $S$ is an interior point.
Thus $S$ is an open set in $\C$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $S$ be an open set in $\C$ by definition 2.
That is, every point of $S$ is an interior point.
This means:
: $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$
Thus $S$ is an open set in $\C$ by definition 1.
Category:Complex Analysis