nyuuzyou/wikis huggingface dataset:
Condition for Cosets of Subgroup of Monoid to be Partition

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$.
Let the identity element of $\struct {H, \circ}$ be $e$.
Then:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$.
== Proof ==
Because of the fact that it is not necessarily the case that $x \in S$ has an inverse, we cannot invoke the result Left Congruence Modulo Subgroup is Equivalence Relation.
Instead we prove partitionhood directly.
First we recall the definition of left coset and right coset:
:The left coset of $H$ by $x$, is:
::$x \circ H = \set {y \in S: \exists h \in H: y = x \circ h}$
:The right coset of $H$ by $x$, is:
::$H \circ x = \set {y \in S: \exists h \in H: y = h \circ x}$
Next we recall the definition of partition:
:A partition of $S$ is a set of non-empty subsets $\Bbb S$ of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
Testing each of the criteria for a partition as follows:
=== No coset is empty, and each element in at least one coset ===
Thus:
:no left coset of $\struct {H, \circ}$ in $\struct {S, \circ}$ is empty.
:every element of $S$ is in at least one left coset of $\struct {H, \circ}$.
Also:
Thus:
:no right coset of $\struct {H, \circ}$ in $\struct {S, \circ}$ is empty.
:every element of $S$ is in at least one right coset of $\struct {H, \circ}$.
=== Each element in no more than one coset ===
Let $r, s \in S$ such that:
:$r \in s \circ H$
for some $s \in S$.
It will be shown that $r \circ H$ and $s \circ H$ are the same left coset.
We have that:
:$r \in r \circ H$
from above.
By definition of left coset:
:$r \in \set {y \in S: \exists h \in H: y = s \circ h}$
That is:
:$r = s \circ h$
and so:
:$r \circ h^{-1} = s \circ \paren {h \circ h^{-1} }$
leading to:
:$r \circ h^{-1} = s$
That is:
:$s \in \set {y \in S: \exists h^{-1} \in H: y = r \circ h^{-1} }$
and so:
:$s \in r \circ H$
That is:
:$r \circ H = s \circ H$
Similarly, if $r \in H \circ s$, then $H \circ r$ and $H \circ s$ are the same right coset.
Thus:
:each element of $S$ is in no more than one left coset of $\struct {H, \circ}$ in $\struct {S, \circ}$
and
:each element of $S$ is in no more than one right coset of $\struct {H, \circ}$ in $\struct {S, \circ}$.
Thus:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
as we were to prove.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.15$
Category:Cosets
Category:Monoids
Condition for Cross-Ratio to be Harmonic

== Theorem ==
Let $A$, $B$, $C$ and $D$ be points on a straight line.
Let their cross-ratio $\set {A, B; C, D}$ be a harmonic ratio.
Then:
:$\set {A, B; C, D} = \set {B, A; C, D}$
== Proof ==
Let $\set {A, B; C, D}$ form a harmonic ratio.
We have:
== Sources ==
* 
* 
Category:Definitions/Cross-Ratios
Category:Definitions/Harmonic Ratios
Condition for Darboux Integrability/Historical Note

== Historical Note on Condition for Darboux Integrability ==
The necessary and sufficient conditions for the existence of a Riemann integral were determined by in his paper of $1854$, on the subject of Fourier series.
It is noted that the proof given on this page refers specifically to the Darboux integral, which is a simpler but equivalent concept to that introduced by .
That they are equivalent is given in Equivalence of Definitions of Riemann and Darboux Integrals
== Sources ==
* : Chapter $\text {A}.32$: Riemann ()
Category:Historical Notes
Condition for Denesting of Square Root

== Theorem ==
Let $a, b \in \Q_{\ge 0}$.
Suppose $\sqrt b \notin \Q$.
Then:
:$\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
:
:$\exists n \in \Q: a^2 - b = n^2$.
== Proof ==
=== Lemma ===
=== Necessary condition ===
This shows that, in order for $p \in \Q$, we must have $\sqrt {a^2 - b} \in \Q$ as well.
The result follows from $\paren {\sqrt {a^2 - b} }^2 = a^2 - b$.
=== Sufficient condition ===
Let $n^2 = a^2 - b$.
As $a^2 = b + n^2$ it follows that:
:$a \ge \sqrt b$
Then:
By Rational Addition is Closed and Rational Subtraction is Closed:
:$\dfrac a 2 + \dfrac n 2$ and $\dfrac a 2 - \dfrac n 2$
are rational.
Category:Square Roots
Category:Condition for Denesting of Square Root
Condition for Denesting of Square Root/Lemma

== Theorem ==
Let $a, b, c, d \in \Q_{\ge 0}$.
Suppose $\sqrt b \notin \Q$.
Then:
:$\sqrt {a + \sqrt b} = \sqrt {c + \sqrt d} \implies a = c, b = d$
== Proof ==
 $b \ne d$.
Then:
But this contradicts our assertion that $\sqrt b \notin \Q$.
Hence our supposition that $b \ne d$ must be false.
Therefore we must have $b = d$. 
Consequently:
:$a - c = \sqrt d - \sqrt b = 0$
This implies $a = c$.
Category:Condition for Denesting of Square Root
Condition for Difference of Field Elements to be Zero

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a, b \in F$
Then:
:$a - b = 0_F$
:$a = b$
where $a - b$ denotes subtraction.
== Proof ==
=== Necessary Condition ===
Let $a = b$.
Then:
=== Sufficient Condition ===
Let $a - b = 0_F$.
Then:
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(ii)}$
Category:Field Theory
Condition for Differentiable Functional to have Extremum

== Theorem ==
Let $S$ be a set of mappings.
Let $y, h \in S: \R \to \R$ be real functions. 
Let $J \sqbrk y: S \to \R$ be a differentiable functional.
Then a necessary condition for the differentiable functional $J \sqbrk {y; h}$ to have an extremum for $y = \hat y$ is:
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$ 
== Proof ==
Suppose $J \sqbrk {y; h}$ attains a minimum for $y = \hat y$.
Then:
:$\Delta J \sqbrk {\hat y; h} \ge 0$
By definition of the differentiable functional:
:$\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \epsilon \size h$
where:
 
:$\ds \lim_{\size h \mathop \to 0} \epsilon = 0$
Hence, there exists $\size h$ small enough, that signs of $\Delta J \sqbrk {y; h}$ and $\delta J \sqbrk {y; h}$ match.
Therefore, for $\size h$ small enough it holds that $\delta J \sqbrk {\hat y; h}\ge 0$.
 $\delta J \sqbrk {y; h_0} \ne 0$ for some $h_0 \in S$.
Then: 
:$\forall \alpha > 0: \delta J \sqbrk {y; -\alpha h_0} = -\delta J \sqbrk {y; \alpha h_0}$
Therefore, for $\size h$ however small, $\Delta J \sqbrk {y; h}$ can be made to have any sign.
However, by assumption that $J \sqbrk {y; h}$ has a minimum, for sufficiently small $\size h$ it holds that:
:$\Delta J \sqbrk {\hat y; h} = J \sqbrk {\hat y + h} - J \sqbrk {\hat y} \ge 0$
This automatically fixes the sign of $\delta J \sqbrk {y; h}$.
This is in contradiction with the previous statement.
Hence for all $h$ it holds that:
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$ 
== Sources ==
* : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Category:Calculus of Variations
Condition for Divisibility of Powers of Prime

== Theorem ==
Let $p$ be a prime.
Let $k, l \in \Z_{>0}$.
Then:
:$p^k \divides p^l \iff k \le l$
== Proof ==
=== Necessary Condition ===
Let $k \le l$.
Then:
:$l - k \ge 0$
Thus $p^k, p^{l - k} \in \Z$ such that $p^l = p^k p^{l - k}$.
Thus:
:$p^k \divides p^l$
=== Sufficient Condition ===
Let $p^k \divides p^l$.
Then:
:$\exists b \in \Z_{>0}: p^l = p^k b$
By the Fundamental Theorem of Arithmetic, $b$ has a unique prime decomposition.
Either $b = 1$ (in which case $k = l$) or $b$ has a prime decomposition consisting entirely of $p$'s.
In this case:
:$\exists m \in \Z: b = p^m$
Hence:
:$p^{l - k} = p^m$
Thus from the Fundamental Theorem of Arithmetic:
:$l - k = m > 0$
Thus:
:$l > k$
The result follows from combining the two results.
Category:Prime Numbers
Condition for Division by Field Elements to be Unity

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b \in F$.
Then:
:$\dfrac a b = 1_F$
:
:$a = b$
where $\dfrac a b$ denotes division.
== Proof ==
=== Necessary Condition ===
Let $a = b$.
Then:
=== Sufficient Condition ===
Let $\dfrac a b = 1_F$.
Then:
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(ii)}$
Category:Field Theory
Condition for Edge to be Bridge

== Theorem ==
Let $G = \struct {V, E}$ be a connected graph.
Let $e \in E$ be an edge of $G$.
Then $e$ is a bridge $e$ does not lie on any circuit of $G$.
== Proof ==
Let $G - e$ denote the edge deletion of $e$ from $G$.
=== Necessary Condition ===
We use a Proof by Contraposition.
To that end, suppose $e = u v$ such that $e$ lies on a circuit, say:
:$C = \tuple {u, v, w, \ldots, x, u}$
Thus $G - e$ contains a $u - v$ path, that is, $\tuple {v, x, \ldots, w, u}$.
So $u$ is connected to $v$ in $G - e$.
Now we need to show that $G - e$ is connected.
Let $u_1, v_1 \in V$.
Since $G$ is connected, there is a $u_1 - v_1$ path $P$ in $G$.
If $e \notin P$, then $P$ is also a path in $G - e$ and $u_1$ is connected to $v_1$ in $G - e$.
On the other hand, suppose $e \notin P$.
Then $P$ can be expressed as $\tuple {u_1, \ldots, u, v, \ldots, v_1}$ or $\tuple {u_1, \ldots, v, u, \ldots, v_1}$.
In the first case, $u_1$ is connected to $u$ and $v$ is connected to $v_1$ in $G - e$.
In the second case, $u_1$ is connected to $v$ and $u$ is connected to $v_1$ in $G - e$.
But we have already established that $u$ is connected to $v$ in $G - e$.
From Graph Connectedness is Equivalence Relation it follows that $u_1$ is connected to $v_1$.
So, if $e$ lies on a circuit, then $G - e$ is connected.
Therefore $e$ is not a bridge.
From Rule of Transposition it follows that if $e$ is a bridge, it does not lie on a circuit.
=== Sufficient Condition ===
We use a Proof by Contraposition.
To that end, suppose $e = u v$ is not a bridge.
Then $G - e$ is connected, and thus there is a $u-v$ path $P$ in $G - e$.
But $P$ together with $e = u v$ produces a circuit containing $e$.
Thus $e$ lies on a circuit of $G$.
From Rule of Transposition it follows that if $e$ does not lie on a circuit, it is a bridge.
== Also presented as ==
This result is usually given that:
:$e$ is a bridge $e$ does not lie on any cycle of $G$.
As a cycle is by definition a circuit, the result as given in this entry is slightly stronger.
== Sources ==
* : $\S 2.4$: Cut-Vertices and Bridges: Theorem $2.5$
Category:Graph Theory
Category:Proofs by Contraposition
Condition for Element of Quotient Group of Additive Group of Reals by Integers to be of Finite Order

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\R / \Z$ denote the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$.
Let $x + \Z$ denote the coset of $\Z$ by $x \in \R$.
Then $x + \Z$ is of finite order $x$ is rational.
== Proof ==
From Additive Group of Integers is Normal Subgroup of Reals, we have that $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.
Hence $\R / \Z$ is indeed a quotient group.
By definition of rational number, what is to be proved is:
:$x + \Z$ is of finite order :
:$x = \dfrac m n$
for some $m \in \Z, n \in \Z_{> 0}$.
Let $x + \Z$ be of finite order in $\R / \Z$.
Then:
== Sources ==
* : Chapter $6$: Cosets: Exercise $14$
Category:Integers
Category:Real Numbers
Category:Examples of Quotient Groups
Condition for Elements of Group to be in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$ be such that $2$ elements of $\set {x, y, x y}$ are elements of $h$.
Then all the elements of $\set {x, y, x y}$ are in $H$.
== Proof ==
As $H$ is a subgroup of $G$, it is a group in its own right.
Thus the group axioms all apply to $H$.
Let $x, y \in H$.
Then by , $x y \in H$.
Let $x, x y \in H$.
As $x \in H$, it follows that $x^{-1} \in H$ by .
Thus by , $x^{-1} \paren {x y} = y \in H$.
Let $y, x y \in H$.
As $y \in H$, it follows that $y^{-1} \in H$ by .
Thus by , $\paren {x y} y^{-1} = x \in H$.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $6$
Category:Subgroups
Condition for Equal Angles contained by Elevated Straight Lines from Plane Angles

== Theorem ==
=== Porism ===
== Proof ==
:
Let the angles $\angle BAC$ and $\angle EDF$ be equal.
From the points $A$ and $D$ let the elevated straight lines $AG$ and $DM$ be set up so they contain equal angles with the original straight lines:
:$\angle MDE = \angle GAB$
:$\angle MDF = \angle GAC$
Let $G$ and $M$ be arbitrary points on $AG$ and $DM$.
Let $GL$ and $MN$ be drawn from $G$ and $M$ perpendicular to the planes holding $\angle BAC$ and $\angle EDF$ respectively, meeting them at $L$ and $N$.
Let $LA$ and $ND$ be joined.
It is to be demonstrated that $\angle GAL = \angle MDN$.
Let $AH = DM$.
Let $HK$ be drawn through $H$ parallel to $GL$.
But $GL$ is perpendicular to the plane holding $\angle BAC$.
Therefore from :
:$HK$ is perpendicular to the plane holding $\angle BAC$.
From the points $K, N$ let $KN, NF, KB, NE$ be drawn perpendicular to the straight lines $AC, DF, AB, DE$.
Let $HC, CB, MF, FE$ be joined.
From :
:$HA^2 = HK^2 + KA^2$
and:
:$KA^2 = KC^2 + CA^2$
Therefore:
:$HA^2 = HK^2 + KC^2 + CA^2$
But from :
:$HC^2 = HK^2 + KC^2$
Therefore:
:$HA^2 = HC^2 + CA^2$
Therefore from :
:$\angle HCA$ is a right angle.
For the same reason:
:$\angle DFM$ is a right angle.
Therefore:
:$\angle HCA = \angle DFM$
But:
:$\angle HAC = \angle MDF$
Therefore from :
:$\triangle MDF = \triangle HAC$
Therefore:
:$AC = DF$
Similarly it can be proved that $AB = DE$.
So we have that:
:$AC = DF$
:$AB = DE$
We also have that $\angle CAB = \angle FDE$
So from :
:$\triangle CAB = \triangle FDE$
and so:
:$\angle ACB = \angle DFE$
But the right angle $\angle ACK$ equals the right angle $\angle DFN$.
Therefore the remaining angles are equal:
:$\angle BCK = \angle EFN$
For the same reason:
:$\angle CBK = \angle FEN$
Therefore from :
:$\triangle BCK = \triangle EFN$
Therefore:
:$CK = FN$
But we have that:
:$AC = DF$
Therefore from :
:$\triangle ACK = \triangle DFN$
and so:
:$AK = DN$
We have that:
:$AH = DM$
and so:
:$AH^2 = DM^2$
But $\angle AKH$ is a right angle.
So from :
:$AK^2 + KH^2 = AH^2$
Also $\angle DNM$ is a right angle.
So from :
:$DM^2 + DN^2 = NM^2$
But as:
:$AK = DN$
it follows that:
:$AK^2 = DN^2$
Therefore:
:$KH^2 = NM^2$
and so:
:$KH = NM$
So we have that:
:$AH = MD$
:$AK = DN$
:$HK = MN$
So from :
:$\angle HAK = \angle MDN$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Geometry
Condition for Equality of Adjacent Binomial Coefficients

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {k + 1}$
:
:$n$ is an odd integer
:$k = \dfrac {n - 1} 2$
== Proof ==
=== Sufficient Condition ===
Let $n$ be odd and $k = \dfrac {n - 1} 2$.
Let $n = 2 m + 1$ for some $m \in \Z_{\ge 0}$.
We have:
Hence:
=== Necessary Condition ===
Let $n$ and $k$ be such that:
:$\dbinom n k = \dbinom n {k + 1}$
From Condition for Increasing Binomial Coefficients, it is not the case that:
:$0 \le k < \dfrac {n - 1} 2$
as under such a condition we would have:
:$\dbinom n k = \dbinom n {k + 1}$
So:
:$k \ge \dfrac {n - 1} 2$
By the Symmetry Rule for Binomial Coefficients, we also have that:
:$\dfrac {n - 1} 2 < k \le n \iff \dbinom n k > \dbinom n {k + 1}$
and so for $\dbinom n k = \dbinom n {k + 1}$ it is not the case that $\dfrac {n - 1} 2 < k \le n$.
So:
:$k \le \dfrac {n - 1} 2$
Hence that means:
:$k = \dfrac {n - 1} 2$
which means:
:$n = 2 k + 1$
and so $n$ is odd.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $1 \ \text{(b)}$
Category:Binomial Coefficients
Condition for Equilibrium using Force Polygon

== Theorem ==
Let $\SS$ be a system of forces $\SS$ act at a point.
Let $\SS$ give rise to a force polygon $A_0 A_1 \ldots A_n A_0$.
Then $\SS$ is in equilibrium $A_n$ and $A_0$ coincide.
== Proof ==
== Sources ==
* 
Category:Force Polygons
Condition for Equivalence Relation for Max Operation on Natural Numbers to be Congruence

== Theorem ==
Let $\RR$ be an equivalence relation on the set of natural numbers $\N$.
Let $\vee$ denote the max operation on $\N$:
:$\forall a, b \in \N: a \vee b := \max \set {a, b}$
Then:
:$\RR$ is a congruence relation for $\vee$ on $\N$
:
:every equivalence class under $\RR$ is a convex subset of $\N$.
== Proof ==
=== Necessary Condition ===
Suppose every equivalence class under $\RR$ is a convex subset of $\N$.
Let $x_1, x_2, y_1, y_2 \in \N$ such that $x_1 \mathrel \RR x_2$ and $y_1 \mathrel \RR y_2$.
, suppose:
:$x_1 \le x_2$
:$y_1 \le y_2$
:$x_1 \le y_1$
Then we have two cases:
==== Case $1$: $y_1 \le x_2$ ====
In this case we have $x_1 \le y_1 \le x_2$.
Since $\eqclass {x_1} \RR$ is a convex subset of $\N$ and $x_1, x_2 \in \eqclass {x_1} \RR$, we must have:
:$y_1 \in \eqclass {x_1} \RR$
Thus, by Equivalence Class Equivalent Statements $(2)$ and $(5)$:
:$\eqclass {x_1} \RR = \eqclass {y_1} \RR$.
So $x_1, x_2, y_1, y_2$ all belongs to the same equivalence class, and so does $x_1 \vee y_1$ and $x_2 \vee y_2$.
Therefore $\paren {x_1 \vee y_1} \mathrel \RR \paren {x_2 \vee y_2}$.
==== Case $2$: $x_2 \le y_1$ ====
In this case we have $x_1 \le x_2 \le y_1 \le y_2$.
Thus:
:$x_1 \vee y_1 = y_1$
:$x_2 \vee y_2 = y_2$
Since $y_1 \mathrel \RR y_2$:
:$\paren {x_1 \vee y_1} \mathrel \RR \paren {x_2 \vee y_2}$
In both cases we have $\paren {x_1 \vee y_1} \mathrel \RR \paren {x_2 \vee y_2}$.
Therefore $\RR$ is a congruence relation for $\vee$ on $\N$.
=== Sufficient Condition ===
Suppose that $\RR$ is a congruence relation for $\vee$ on $\N$.
Let $x, y, z \in \N$ and suppose that $x, z \in \eqclass z \RR$ and $x \le y \le z$.
Then:
Therefore $\eqclass z \RR$ is a convex subset of $\N$.
As the choice of $x, y, z$ is arbitrary, every equivalence class under $\RR$ is a convex subset of $\N$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.20 \ \text {(a)}$
Category:Congruence Relations
Category:Max Operation
Category:Natural Numbers
Condition for Equivalence of Norms that induce Banach Spaces

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_1$ and $\norm \cdot_2$ be norms on $X$ such that $\struct {X, \norm \cdot_1}$ and $\struct {X, \norm \cdot_2}$ are Banach spaces.
Suppose that, for some real number $C > 0$: 
:$\norm x_2 \le C \norm x_1$ for all $x \in X$. 
Then $\norm \cdot_1$ and $\norm \cdot_2$ are equivalent.
== Proof ==
Consider $I : \struct {X, \norm \cdot_1} \to \struct {X, \norm \cdot_2}$ defined by: 
:$I x = x$ for each $x \in X$. 
Then for $x, y \in X$ and $\lambda \in \Bbb F$. 
Then we have: 
:$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$
so $I$ is linear. 
Similarly, we have:
:$\norm {I x}_2 = \norm x_2 \le C \norm x_1$
so $I$ is a bounded linear transformation.
Clearly $I$ is a bijection, so by the Banach Isomorphism Theorem, the inverse $I^{-1} : \struct {X, \norm \cdot_2} \to \struct {X, \norm \cdot_1}$ is bounded.
So, there exists a real number $C' > 0$ such that: 
:$\norm {I x}_1 = \norm x_1 \le C' \norm x_2$
Then we have: 
:$\norm x_1 \le C' \norm x_2 \le C C' \norm x_1$
so that: 
:$\ds \frac 1 {C'} \norm x_1 \le \norm x_2 \le C \norm x_1$
for $C, C' > 0$. 
So $\norm \cdot_1$ and $\norm \cdot_2$ are equivalent.
Category:Banach Spaces
Condition for Existence of Cardinal Sum

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
: $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$
where $\mathbf c$ is also a cardinal.
== Proof ==
Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some sets $A$ and $B$.
From Equivalence of Definitions of Dominate (Set Theory), there exists a bijection from $A$ onto a subset $E$ of $B$.
Thus:
: $\mathop a = \card E$
Let $F = \relcomp B E$
From Set with Relative Complement forms Partition:
: $B = E \cup F$
: $E \cap F = \O$
By definition of sum of cardinals, it follows that:
By defining $\mathbf c := \card F$ it follows that:
:$\mathbf a + \mathbf c = \mathbf b$
for the $\mathbf c$ that has been demonstrated to exist.
Now suppose there exists a cardinal $\mathbf c$ such that $\mathbf a + \mathbf c = \mathbf b$.
Then by definition of sum of cardinals:
:$\card B = \card {A \cup C}$
for some sets $A$, $B$ and $C$ such that $A \cap C = \O$.
Let $f: A \to B$ be an injection, proved to exist by Equivalence of Definitions of Dominate (Set Theory).
Then it follows that $\mathbf a \le \mathbf b$.
== Sources ==
* : $\S 8$: Theorem $8.8$
Category:Cardinals
Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image

== Theorem ==
Let $\struct {A, \odot}$ and $\struct {B, \otimes}$ be algebraic structures.
Let $\RR$ be a congruence relation on $\struct {A, \odot}$.
Let $f: \struct {A, \odot} \to \struct {B, \otimes}$ be an epimorphism.
Let $\struct {A / \RR, \odot_\RR}$ denote the quotient structure defined by $\RR$.
Let $q_\RR: A \to A / \RR$ denote the quotient mapping induced by $\RR$:
:$\forall x \in A: \map {q_\RR} x = \eqclass x \RR$
where $\eqclass x \RR$ denotes the equivalence class of $x$ under $\RR$.
Then:
:there exists an epimorphism $g$ from $\struct {A / \RR, \odot_\RR}$ to $\struct {B, \otimes}$ which satisfies $g \circ q_\RR = f$
:
:$\RR \subseteq \RR_f$
where $\RR_f$ denotes the equivalence relation induced by $f$.
== Proof ==
=== Necessary Condition ===
Let $\RR \subseteq \RR_f$.
Recall the definition of $\RR_f$:
:$\forall x, y \in A: x \mathrel {\RR_f} y \iff \map f x = \map f y$ 
Let us define $g: \struct {A / \RR, \odot_\RR} \to \struct {B, \otimes}$ as:
:$\forall \eqclass x \RR \in A / \RR: \map g {\eqclass x \RR} = \map f x$
We show that $g$ is well-defined.
Let $x \mathrel \RR y$.
Then as $\RR \subseteq \RR_f$, it follows that:
:$x \mathrel {\RR_f} y$
By definition of the equivalence class of $x$:
:$\eqclass x \RR = \eqclass y \RR$
Thus:
Thus we have that:
:$\forall x, y \in A: x \mathrel \RR y \implies \map g {\eqclass x \RR} = \map g {\eqclass y \RR}$
and so $g$ is well-defined.
Then:
and so we have that:
:$g \circ q_\RR = f$
Let $x, y \in A$.
We have:
Thus we see that $g$ is a homomorphism.
We have that $g \circ q_\RR = f$ is an epimorphism.
Hence $g \circ q_\RR$ is a surjection.
So from Surjection if Composite is Surjection we have that $g$ is likewise a surjection.
Thus we have that $g$ is a surjective homomorphism.
Hence by definition $g$ is an epimorphism.
We note that if it is not the case that $\RR \subseteq \RR_f$, then there exist $\tuple {x, y} \in \RR$ such that $\map f x \ne \map f y$.
In that case we would see that while $x \mathrel y$ we would have that:
:$\map g {\eqclass x \RR} \ne \map g {\eqclass y \RR}$
meaning that $g$ is not a well-defined mapping.
So, in summary, given that $\RR \subseteq \RR_f$, we have demonstrated the existence of an epimorphism $g$ which satisfies $g \circ q_\RR = f$.
=== Sufficient Condition ===
Let there exist an epimorphism $g$ from $\struct {A / \RR, \odot_\RR}$ to $\struct {B, \otimes}$ which satisfies $g \circ q_\RR = f$.
Let $x, y \in A$ be arbitrary such that $x \mathrel \RR y$.
Then we have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.18 \ \text {(a)}$
Category:Quotient Structures
Category:Congruence Relations
Category:Epimorphisms (Abstract Algebra)