nyuuzyou/wikis huggingface dataset:
Equivalent Conditions for Entropic Structure/Pointwise Operation is Homomorphism

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
Let $\struct {T, \circledast}$ be an arbitrary algebraic structure.
Let $f$ and $g$ be mappings from $\struct {T, \circledast}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^T$ induced by $\odot$.
Then:
:If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism
:
:$\struct {S, \odot}$ is an entropic structure.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: T \to S$ and $g: T \to S$ be arbitrary homomorphisms.
Let $a, b \in T$ be arbitrary.
Because $T$ is arbitrary, and $f$ and $g$ are arbitrary, it follows that:
Thus we have:
As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.
=== Necessary Condition ===
Let $\struct {S, \odot}$ be an entropic structure.
Let $f: T \to S$ and $g: T \to S$ be arbitrary homomorphisms.
Then:
Hence by definition $f \odot g$ is a homomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.13$
Category:Equivalent Conditions for Entropic Structure
Category:Pointwise Operations
Category:Homomorphisms (Abstract Algebra)
Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^{S \times S}$ induced by $\odot$.
Then:
:If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism
:
:$\struct {S, \odot}$ is an entropic structure.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.
Let $w$, $x$, $y$ and $d$ in $S$ be arbitrary.
Then as $f$ and $g$ are arbitrary:
Then we have:
As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.
=== Necessary Condition ===
Let $\struct {S, \odot}$ be an entropic structure.
Let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.
Then:
As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.13$
Category:Equivalent Conditions for Entropic Structure
Category:Pointwise Operations
Category:Homomorphisms (Abstract Algebra)
Category:External Direct Products
Equivalent Expressions for Scalar Triple Product

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
Then this identity applies to the scalar triple product:
while:
== Proof ==
The remaining identities follow from Dot Product Operator is Commutative.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors: $(2.21)$
* : Chapter $\text I$: Vector Algebra: $\S 4$
Category:Scalar Triple Product
Equivalent Matrices have Equal Rank

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$.
Let $\map \phi {\mathbf A}$ denote the rank of $\mathbf A$.
Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are matrix equivalent.
Then:
:$\mathbf A \equiv \mathbf B$
:
:$\map \phi {\mathbf A} = \map \phi {\mathbf B}$
 
== Proof ==
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$ such that $\mathbf A \equiv \mathbf B$.
Let $S$ and $T$ be vector spaces of dimensions $n$ and $m$ over $K$.
Let $\mathbf A$ be the matrix of a linear transformation $u: S \to T$ relative to the ordered bases $\sequence {a_n}$ of $S$ and $\sequence {b_m}$ of $T$.
Let $\psi: K^m \to T$ be the isomorphism defined as:
:$\ds \map \psi {\sequence {\lambda_m} } = \sum_{k \mathop = 1}^m \lambda_k b_k$
Then $\psi$ takes the $j$th column of $\mathbf A$ into $\map u {a_j}$.
Hence it takes the subspace of $K^m$ generated by the columns of $\mathbf A$ onto the codomain of $u$.
Thus $\map \rho {\mathbf A} = \map \rho u$, and equivalent matrices over a field have the same rank.
Now let $\map \LL {K^n, K^m}$ be the set of all linear transformations from $K^n$ to $K^m$.
Let $u, v \in \map \LL {K^n, K^m}$ such that $\mathbf A$ and $\mathbf B$ are respectively the matrices of $u$ and $v$ relative to the standard ordered bases of $K^n$ and $K^m$.
Let $r = \map \phi {\mathbf A}$.
By Linear Transformation from Ordered Basis less Kernel, there exist ordered bases $\sequence {a_n}, \sequence {a'_n}$ of $K^n$ such that:
:$\sequence {\map u {a_r} }$ and $\sequence {\map v {a'_r} }$ are ordered bases of $\map u {K^n}$ and $\map v {K^n}$ respectively
and such that:
:$\set {a_k: k \in \closedint {r + 1} n}$ and $\set {a'_k: k \in \closedint {r + 1} n}$ are respectively bases of the kernels of $u$ and $v$.
Thus, by Results concerning Generators and Bases of Vector Spaces there exist ordered bases $\sequence {b_m}$ and $\sequence {b'_m}$ of $K^m$ such that $\forall k \in \closedint 1 r$:
Let $z$ be the automorphism of $K^n$ which satisfies $\forall k \in \closedint 1 n: \map z {a'_k} = a_k$.
Let $w$ be the automorphism of $K^m$ which satisfies $\forall k \in \closedint 1 m: \map w {b'_k} = b_k$.
Then:
:$\map {\paren {w^{-1} \circ u \circ z} } {a'_k} = \begin {cases}
\map {w^{-1} } {b_k} = b'_k = \map v {a'_k} & : k \in \closedint 1 r \\
0 = \map v {a'_k} & : k \in \closedint {r + 1} n
\end {cases}$
So:
:$w^{-1} \circ u \circ z = v$
Now let $\mathbf P$ be the |matrix of $z$ relative to the standard ordered bases of $K^n$, and let $\mathbf Q$ be the matrix of $w$ relative to the standard ordered bases of $K^m$.
Then $\mathbf P$ and $\mathbf Q$ are nonsingular and:
:$\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$
and thus:
:$\mathbf A \equiv \mathbf B$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.6$
Category:Matrix Equivalence
Category:Rank of Matrix
Equivalent Matrices may not be Similar

== Theorem ==
If two square matrices of order $n > 1$ over a ring with unity $R$ are equivalent, they are not necessarily similar.
== Proof ==
Proof by Counterexample:
Let $\mathbf A = \mathbf I_n$ be the unit matrix of order $n > 1$.
Let $\mathbf B$ be an arbitrary nonsingular matrix over $R$ of order $n$ that is different from the unit matrix.
Then:
:$\mathbf I_n^{-1} \mathbf A \mathbf B = \mathbf I_n^{-1} \mathbf I_n \mathbf B = \mathbf B$
showing that $\mathbf A$ and $\mathbf B$ are equivalent.
Let $\mathbf P$ be an nonsingular square matrix of order $n$.
Then:
:$\mathbf P^{-1} \mathbf A \mathbf P = \mathbf P^{-1} \mathbf P = \mathbf I_n \ne \mathbf B$
Hence, $\mathbf A$ is not similar to $\mathbf B$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Equivalence
Category:Matrix Similarity
Equivalent Norms are both Non-Archimedean or both Archimedean

== Theorem ==
Let $R$ be a division ring with unity $1_R$.
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.
Then $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are either both non-Archimedean or both Archimedean.
== Proof ==
By Characterisation of Non-Archimedean Division Ring Norms then:
:$\norm {\,\cdot\,}_1$ is non-Archimedean $\iff \forall n \in \N_{>0}: \norm{n \cdot 1_R}_1 \le 1$.
By the definition of norm equivalence then:
:$\forall n \in \N: \norm {n \cdot 1_R}_1 \le 1 \iff \norm {n \cdot 1_R}_2 \le 1$
Similarly, by Characterisation of Non-Archimedean Division Ring Norms then:
:$\forall n \in \N_{>0}: \norm {n \cdot 1_R}_2 \le 1 \iff \norm {\,\cdot\,}_2$ is non-Archimedean.
The result follows.
== Sources ==
* : $\S 1.2$ Normed fields: Proposition $1.18$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Equivalent Norms on Direct Product of Normed Vector Spaces

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $\struct {X \times Y, \norm {\, \cdot \,}_{X \times Y} }$ be the direct product of $X$ and $Y$ with the direct product norm. 
Let $p \ge 1$. 
Let $\norm {\, \cdot \,}_p$ be the $p$-norm on $\R^2$.
Define:
:$\norm {\tuple {x, y} }' = \norm {\tuple {\norm x_X, \norm y_Y} }_p = \begin{cases}\paren {\norm x_X^p + \norm y_Y^p}^{1/p} & p \ne \infty \\ \max \set {\norm x_X, \norm y_Y} & p = \infty\end{cases}$
for each $x \in X$ and $y \in Y$. 
Then $\norm {\, \cdot \,}'$ is a norm on $X \times Y$ that is equivalent to $\norm {\, \cdot \,}_{X \times Y}$. 
== Proof ==
We first show that $\norm {\, \cdot \,}'$ is a norm.
Towards proving , suppose that $\norm {\tuple {x, y} }' = 0$. 
Then we have $\norm {\tuple {\norm x_X, \norm y_Y} }_p = 0$.
Hence $\norm x_X = 0$ and $\norm y_Y = 0$ by .
From , we have $x = {\mathbf 0}_X$ and $y = {\mathbf 0}_Y$. 
Then $\tuple {x, y} = \tuple { {\mathbf 0}_X, {\mathbf 0}_Y} = {\mathbf 0}_{X \times Y}$.
Hence we have proven .
Now we prove , let $\tuple {x, y} \in X \times Y$ and $\lambda \in \GF$.
We have:
:$\norm {\lambda \tuple {x, y} }' = \norm {\tuple {\norm {\lambda x}_X, \norm {\lambda y}_Y} }_p$
From , we have:
:$\norm {\tuple {\norm {\lambda x}_X, \norm {\lambda y}_Y} }_p = \norm {\tuple {\cmod \lambda \norm x_X, \cmod \lambda \norm y_Y} }_p = \cmod \lambda \norm {\tuple {\norm x_X, \norm y_Y} }_p$
Hence:
:$\norm {\lambda \tuple {x, y} }' = \cmod \lambda \norm {\tuple {x, y} }'$
Finally, we look to prove . 
Let $\tuple {x_1, y_1} \in X \times Y$ and $\tuple {x_2, y_2} \in X \times Y$. 
We have:
:$\norm {\tuple {x_1, y_1} + \tuple {x_2, y_2} }' = \paren {\norm {x_1 + x_2}_X^p + \norm {y_1 + y_2}_Y^p}^{1/p}$
From , we have:
:$\paren {\norm {x_1 + x_2}_X^p + \norm {y_1 + y_2}_Y^p}^{1/p} \le \paren {\paren {\norm {x_1}_X + \norm {x_2}_X}^p + \paren {\norm {y_1}_Y + \norm {y_2}_Y}^p}^{1/p} = \norm {\tuple {\norm {x_1}_X + \norm {x_2}_X, \norm {y_1}_Y + \norm {y_2}_Y} }_p$
Applying for $\norm {\, \cdot \,}_p$, we have:
:$\norm {\tuple {\norm {x_1}_X + \norm {x_2}_X, \norm {y_1}_Y + \norm {y_2}_Y} }_p \le \norm {\tuple {\norm {x_1}_X, \norm {y_1}_Y} }_p + \norm {\tuple {\norm {x_2}_X, \norm {y_2}_Y} }_p = \norm {\tuple {x_1, y_1} }' + \norm {\tuple {x_2, y_2} }'$
Hence $\norm {\, \cdot \,}'$ is a norm on $X \times Y$. 
From Norms on Finite-Dimensional Real Vector Space are Equivalent, for each $p \ge 1$ there exists $M_1, M_2 > 0$ such that:
:$M_1 \norm {\tuple {u, v} }_p \le \norm {\tuple {u, v} }_\infty \le M_2 \norm {\tuple {u, v} }_p$
for each $\tuple {u, v} \in \R^2$. 
Setting $u = \norm x_X$ and $v = \norm y_Y$, we obtain:
:$M_1 \norm {\tuple {x, y} }' \le \norm {\tuple {x, y} }_{X \times Y} \le M_2 \norm {\tuple {x, y} }'$
Hence $\norm {\, \cdot \,}'$ is equivalent to $\norm {\, \cdot \,}_{X \times Y}$.
Category:Direct Product of Vector Spaces
Category:Normed Vector Spaces
Equivalent Norms on Lipschitz Space/Shift of Finite Type

== Theorem ==
Let $\struct {X, \sigma}$ be a shift of finite type.
Let $F_\theta$ be the Lipschitz space on $X$.
Let $x_0 \in F_\theta$.
Then the following norms on $F_\theta$ are equivalent:
:$(1): \quad$ Lipschitz norm $\norm\cdot_\theta$
:$(2): \quad \forall f \in F_\theta : \norm f '_\theta := \cmod {\map f {x_0} } + \size f_\theta$
== Proof ==
Category:Functional Analysis
Equivalent Norms on Rational Numbers

== Theorem ==
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent :
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$: Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem: Theorem $1.50$
Category:Normed Division Rings
Equivalent Norms on Rational Numbers/Necessary Condition

== Theorem ==
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent norms.
Then:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
== Proof ==
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent.
By Norm is Power of Other Norm then:
:$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$
In particular:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski's Theorem, Theorem $1.50$
Category:Normed Division Rings
Equivalent Norms on Rational Numbers/Sufficient Condition

== Theorem ==
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent
== Proof ==
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
By Norm of Negative then:
:$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$
Hence:
:$\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$
By Norm of Quotient then:
:$\forall \dfrac a b \in \Q: \norm {\dfrac a b}_1 = \dfrac {\norm a_1} {\norm b_1} = \dfrac {\norm a_2^\alpha} {\norm b_2^\alpha} = \norm {\dfrac a b}_2^\alpha$
By Norm is Power of Other Norm then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Normed Division Rings
Equivalent Properties of Nondegenerate Subspace of Scalar Product Space

== Theorem ==
Let $\struct {V, q}$ be a scalar product space.
Let $S \subseteq V$ be a linear subspace.
Let $S^\perp$ be the vector subspace perpendicular to $S$ with respect to $q$.
Then the following are equivalent:
:$S$ is nondegenerate;
:$S^\perp$ is nondegenerate;
:$S \cap S^\perp = \set 0$
:$V = S \oplus S^\perp$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Scalar Product Spaces
Equivalent Properties of Nondegenerate Symmetric Covariant 2-Tensor

== Theorem ==
Let $V$ and $V^*$ be a finite dimensional vector space and its dual.
Let $q$ be a symmetric covariant 2-tensor on $V$.
Let $\tuple {\epsilon^i}$ be any basis of $V^*$
Then the following properties are equivalent:
:$q$ is nondegenerate.
:$\forall v \in V : v \ne 0 : \exists w \in V : \map q {v, w} \ne 0$
:If $q = q_{ij} \epsilon^i \epsilon^j$, then the matrix $\paren {q_{ij}}$ is nonsingular.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Riemannian Geometry
Equivalent Statements for Congruence Modulo Subgroup/Left

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x \equiv^l y \pmod H$ denote that $x$ is left congruent modulo $H$ to $y$.
Then the following statements are equivalent:
== Proof ==
== Sources ==
* : $\S 42.2$ Another approach to cosets
Category:Congruence Modulo Subgroup
Equivalent Statements for Congruence Modulo Subgroup/Right

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x \equiv^r y \pmod H$ denote that $x$ is right congruent modulo $H$ to $y$.
Then the following statements are equivalent:
== Proof ==
== Sources ==
* : $\S 42.2$ Another approach to cosets
Category:Congruence Modulo Subgroup
Equivalent Statements for Transitive Set

== Theorem ==
Let $a$ be a set.
:$(1): \quad \bigcup a \subseteq a$
:$(2): \quad a \subseteq \powerset a$
:$(3): \quad a$ is a transitive set
== Proof ==
=== $(1) \iff (3)$ ===
This is shown in Class is Transitive iff Union is Subclass.
=== $(2) \iff (3)$ ===
This is shown in Set is Transitive iff Subset of Power Set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Remark
Category:Transitive Classes
Equivalent Statements for Vector Subspace Dimension One Less

== Theorem ==
Let $K$ be a field.
Let $M$ be a subspace of the $n$-dimensional vector space $K^n$.
The following statements are equivalent:
:$(1): \quad \map \dim M = n - 1$
:$(2): \quad M$ is the kernel of a nonzero linear form
:$(3): \quad$ There exists a sequence $\sequence {\alpha_n} $ of scalars, not all of which are zero, such that:
:::$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$
=== Existence of Scalar for Vector Subspace Dimension One Less ===
Suppose the above hold.
== Proof ==
Let $M^\circ$ be the annihilator of $M$.
Let $N = M^{\circ}$.
By Results Concerning Annihilator of Vector Subspace, $N$ is one-dimensional and $M = \map {J^{-1} } {N^\circ}$.
Let $\phi \in N: \phi \ne 0$.
Then $N$ is the set of all scalar multiples of $\phi$.
Because:
:$\map {J^{-1} } {N^\circ} = \set {x \in K^n: \forall \psi \in N: \map \psi x = 0}$
it follows that $\map {J^{-1} } {N^\circ}$ is simply the kernel of $\phi$.
Hence $(1)$ implies $(2)$.
By Rank Plus Nullity Theorem, $(2)$ also implies $(1)$.
Suppose $\sequence {\alpha_n}$ is any sequence of scalars.
Let $\sequence { {e_n}'}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$.
Let $\ds \phi = \sum_{k \mathop = 1}^n \alpha_k e'_k$.
Then, by simple calculation:
:$\map \ker \phi = \set {\tuple {\lambda_1, \ldots, \lambda_n}: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$
It follows that:
:$\phi \ne 0 \iff \exists k \in \closedint 1 n: \alpha_k \ne 0$
Thus $(2)$ and $(3)$ are equivalent.
== Sources ==
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations: Theorem $28.11$
Category:Linear Algebra
Equivalent Subobjects have Isomorphic Domains

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ be the category of subobjects of $C$.
Let $m, m'$ be equivalent subobjects of $C$.
Then there exists an isomorphism $f: m \to m'$.
== Proof ==
For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$.
Since $m$ and $m'$ are equivalent subobjects of $C$, we have that:
:$m \subseteq m'$ and $m' \subseteq m$
where $\subseteq$ denotes the inclusion relation on subobjects.
That is, there exist morphisms $f: m \to m'$ and $f': m' \to m$ of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$.
Hence, we have morphisms:
:$f \circ f': m' \to m'$
:$f' \circ f: m \to m$
By Category of Subobjects is Preorder Category, these are the unique such morphisms.
It follows that both are identity morphisms:
:$f \circ f' = \operatorname{id}_{m'}$
:$f' \circ f = \operatorname{id}_m$
That is, $f: m \to m'$ is an isomorphism.
== Sources ==
* : $\S 5.1$
Category:Categories of Subobjects
Category:Subobjects
Equivocation of Nothing

== Fallacy ==
This is a fallacy in the following form.
Consider a universe of discourse $\mathbb U$ whose elements are strictly ordered by a relation, here denoted better.
Let $M \in \mathbb U$ be a maximal element with respect to better.
Let $m \in \mathbb U$ be any other element.
The fallacy of equivocation of nothing goes as follows:
:$m$ is better than nothing.
:Nothing is better than $M$.
:Therefore, $m$ is better than $M$.
As $M$ is maximal, this is a contradiction.
== Examples ==
== Resolution ==
This is an example of a falsidical paradox based on the equivocation of the meaning of the word nothing.
In the first statement:
:$m$ is better than nothing
the word nothing means the element (possibly hypothetical) against which every element of $\mathbb U$ is better.
In the second statement:
:Nothing is better than $M$
the word nothing means there exist no elements.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(9)$
Category:Falsidical Paradoxes
Equivocation of Nothing/Examples

== Examples of Equivocation of Nothing ==
These are examples of the fallacy of Equivocation of Nothing.
The classic often-cited argument is:
:Candlelight is brighter than nothing.
:Nothing is brighter than the light from the sun.
:Therefore, candlelight is brighter than the light from the sun.
This version is given in :
:Any soap is better than no soap.
:No soap is better than Wonder-Bubble.
:Therefore, any soap is better than Wonder-Bubble.
This version is given in :
:Time waits for no man.
:No man is an island.
:Therefore, time waits for an island.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(9)$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1.1$: 'You talking to me?'
Category:Examples of Equivocation
Eratosthenes' Measurement of Earth

== Theorem ==
The circumference of Earth was measured by to be of the order of $45 \, 000 \, \mathrm{km}$.
== Proof ==
:
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.16$: The Sequence of Primes: Footnote $1$
* 
* 
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Geodesy
Eratosthenes' Measurement of Earth/Historical Note

== Historical Note on Eratosthenes' Measurement of Earth ==
It is not certain what the length of the unit used by when he made his measurement of the circumference of the Earth.
The Olympic stade of $176.4 \, \mathrm m$ would give a circumference of $44 \, 100 \, \mathrm {km}$, an error of $10 \%$.
The Italian stade of $184.8 \, \mathrm m$ would give a circumference of $46 \, 100 \, \mathrm {km}$, an error of $15 \%$.
== Sources ==
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Historical Notes
Erdős-Moser Equation

== Theorem ==
The Erdős-Moser Equation is the Diophantine equation:
where $k, p \in \N$.
Its only solution is $k = 2$ and $p = 1$:
:$1^1 + 2^1 = 3^1$
== Proof ==
Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial:
:$\ds \map {B_n} x = \sum_{k \mathop = 0}^n \binom n k b_{n - k} x^k$
where $b_n$ denotes the $n$th Bernoulli number.
=== Lemma 1 ===
Erdös-Moser equation is equivalent:
:$(1): \quad \ds \sum_{k \mathop = 0}^x k^p \equiv \dfrac {\map {B_{p + 1} } {x + 1} } {p + 1} = \paren {x + 1}^p$
==== Proof ====
By Faulhaber's Formula:
:$\ds \sum_{k \mathop = 0}^x k^p = \dfrac {\map {B_{p + 1} } {x + 1} - \map {B_{p + 1} } 0} {p + 1}$
It is proved that for another solution of the Erdös-Moser equation $p + 1$ must be odd and $\map {B_{p + 1} } 0$ with odd subscripts is equal to zero.
=== Lemma 2 ===
:$(2): \quad \map {B_{p + 1} } {x + 1} - \map {B_p} x = \paren {p + 1} x^p$
:$(3): \quad \map {B_{p + 1} } {x + 2} - \map {B_p} {x + 1} = \paren {p + 1} \paren {x + 1}^p$
==== Proof ====
General identity involving Bernoulli polynomials:
:$\map {B_n} {x + 1} - \map {B_n} x = n x^{n - 1}$
=== Lemma 3 ===
:$(4): \quad \dfrac {\map {B_{p + 1} } {x + 1} } {\map {B_{p + 1} } x} = \dfrac {\paren {x + 1}^p} {\paren {x + 1}^p - x^p}$
==== Proof ====
=== Theorem 1 ===
The Erdös-Moser equation has other solution than $ 1 + 2 = 3$ :
:$(6): \quad \dfrac {\map {B_{p + 1} } {x + 2} } {\map {B_{p + 1} } {x + 1} } = 2$
where $x, p \in \N \land x > 2 \land p > 1$.
==== Proof ====
=== Theorem 2 ===
The Erdös-Moser equation has no solution for $p > 1$.
==== Proof ====
The sum of $p$th powers for $p = 1$:
To prove this theorem we use $(9)$:
 
 
== Sources ==
* : $127$
*
Erdős-Straus Conjecture

== Unproven Hypothesis ==
Let $n$ be an integer.
It is conjectured that $\dfrac 4 n$ can always be expressed as the sum of $3$ unit fractions.
=== Sierpiński Variant ===
 
== Sources ==
* 
* : $2 / 3$
* 
* : $2 / 3$
Category:Unit Fractions
Category:Unproven Hypotheses