nyuuzyou/wikis huggingface dataset:
Condition for Points in Complex Plane to form Isosceles Triangle

== Theorem ==
Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the vertices of a triangle.
Then $\triangle ABC$ is isosceles, where $A$ is the apex, :
:${x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2} = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}$
== Proof ==
By definition of isosceles triangle:
:$\triangle ABC$ is isosceles, where $A$ is the apex, $AB = AC$.
Hence:
== Examples ==
Category:Isosceles Triangles
Category:Geometry of Complex Plane
Condition for Points in Complex Plane to form Isosceles Triangle/Examples/1+2i, 4-2i, 1-6i

== Examples of Use of Condition for Points in Complex Plane to form Isosceles Triangle ==
Let $A = z_1 = 1 + 2 i$, $B = z_2 = 4 - 2 i$ and $C = z_3 = 1 - 6 i$ represent on the complex plane the vertices of a triangle.
Then $\triangle ABC$ is isosceles, where $B$ is the apex.
== Proof ==
By Condition for Points in Complex Plane to form Isosceles Triangle:
:$\triangle ABC$ is isosceles, where $A$ is the apex, $AB = AC$.
Hence:
Similarly:
So $AB = BC$ and so $\triangle ABC$ is isosceles.
Finally note that:
demonstrating that $\triangle ABC$ is not equilateral.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $64$
Category:Examples of Use of Condition for Points in Complex Plane to form Isosceles Triangle
Condition for Points in Complex Plane to form Parallelogram

== Theorem ==
Let $A = z_1$, $B = z_2$, $C = z_3$ and $D = z_4$ represent on the complex plane the vertices of a quadrilateral.
Then $ABCD$ is a parallelogram :
:$z_1 - z_2 + z_3 - z_4 = 0$
== Proof ==
:
$ABCD$ is a parallelogram :
:$\vec {AB} = \vec {DC}$
Then:
== Examples ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $65$
:(although see Condition for Points in Complex Plane to form Parallelogram: Mistake for analysis of an error in that work)
Category:Parallelograms
Category:Geometry of Complex Plane
Category:Condition for Points in Complex Plane to form Parallelogram
Condition for Points in Complex Plane to form Parallelogram/Examples/2+i, 3+2i, 2+3i, 1+2i

== Theorem ==
The points in the complex plane represented by the complex numbers:
:$2 + i, 3 + 2 i, 2 + 3 i, 1 + 2 i$
are the vertices of a square.
== Proof ==
Let us label the points:
From Geometrical Interpretation of Complex Subtraction, we have that the difference $p - q$ between two complex numbers $p, q$ represents the vector $\vec {q p}$.
Let us take the differences of the complex numbers given:
So:
:$\vec {AB} = \vec {DC} = 1 + i$
:$\vec {DA} = \vec {CB} = 1 - i$
Thus by definition $ABCD$ forms a parallelogram.
Next it is noted that:
:$\paren {1 + i} i = i + i^2 = 1 - i$
and so $AB$ and $DC$ are perpendicular to $DA$ and $CB$.
Thus by definition $ABCD$ is a rectangle.
Finally note that:
:$\cmod {1 + i} = \cmod {1 - i} = \sqrt {1^2 + 1^2} = \sqrt 2$
and so:
:$\cmod {AB} = \cmod {DC} = \cmod {DC} = \cmod {DC}$
That is, all four sides of $ABCD$ are the same length.
Thus by definition $ABCD$ is a square.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $2$.
Category:Condition for Points in Complex Plane to form Parallelogram
Condition for Points in Complex Plane to form Parallelogram/Mistake

== Source Work ==
:Chapter $1$: Complex Numbers
::Supplementary Problems: $65$
== Mistake ==
:Let $z_1, z_2, z_3, z_4$ be the position vectors of the vertices for quadrilateral $ABCD$. Prove that $ABCD$ is a parallelogram $z_1 - z_2 - z_3 + z_4 = 0$.
== Correction ==
The unspoken assumption here is that $z_1 = A, z_2 = B, z_3 = C$ and $z_4 = D$. Without such an assumption, the question is ambiguous.
When defining a polygon in this manner, its vertices are cited in order around the perimeter of the polygon.
Thus the object defined should look like this:
:
By Condition for Points in Complex Plane to form Parallelogram it is seen that the correct condition is:
:$z_1 - z_2 + z_3 - z_4 = 0$
Were the quadrilateral specified as $ABDC$, then:
== Also see ==
* Condition for Points in Complex Plane to form Parallelogram
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $65$
Category:Mistakes/Theory and Problems of Complex Variables
Condition for Power Set to be Totally Ordered

== Theorem ==
Let $\powerset S$ be the power set of a set $S$.
Let $\struct {\powerset S, \subseteq}$ be the set $\powerset S$ ordered by $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is totally ordered $S$ is either the empty set or a singleton.
== Proof ==
From Subset Relation on Power Set is Partial Ordering we have that $\struct {\powerset S, \subseteq}$ is an ordered set.
We now need to show that $\struct {\powerset S, \subseteq}$ is a totally ordered set exactly when $S = \O$ or $S$ has exactly one element.
When $S = \O$ then $\powerset S = \set \O$ and it follows trivially that $\struct {\powerset S, \subseteq}$ is totally ordered.
Let $S = \set x$.
Then $\powerset S = \set {\O, \set x}$.
From Empty Set is Subset of All Sets we have $\O \subseteq \set x$.
Again it follows by definition that $\struct {\powerset S, \subseteq}$ is totally ordered.
Now suppose $S$ is neither the empty set nor a singleton.
Then:
:$\exists x, y \in S: x \ne y$
and so:
:$\exists \set x, \set y \in \powerset S: \set x \ne \set y$
But further, $\set x \nsubseteq \set y$ and $\set y \nsubseteq \set x$.
That is, $\set x$ and $\set y$ are non-comparable by $\subseteq$.
Thus, by definition, $\struct {\powerset S, \subseteq}$ is not totally ordered.
Hence the result.
== Sources ==
* : $\S 7$: Exercise $5$
Category:Power Set
Category:Subset Relation
Category:Total Orderings
Condition for Power of Element of Quotient Group to be Identity

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $N$ be a normal subgroup of $G$.
Let $a \in G$.
Then:
:$\paren {a N}^n$ is the identity of the quotient group $G / N$
:
:$a^n \in N$
== Proof ==
Let $\paren {a N}^n$ be the identity of $G / N$.
Then:
== Sources ==
* : Chapter $6$: Cosets: Exercise $13$
Category:Coset Product
Condition for Quadratic Equation to describe Perpendicular Straight Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Let $\LL_1$ and $\LL_2$ be perpendicular.
Then:
:$a + b = 0$
That is, $E$ is of the form:
:$a \paren {x^2 - y^2} + 2 h x y + 2 g x + 2 f y + c = 0$
== Proof ==
From Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
As $\LL_1$ and $\LL_2$ are parallel respectively to $\LL'_1$ and $\LL'_2$, it follows that $\LL'_1$ and $\LL'_2$ are themselves perpendicular.
Hence from Condition for Homogeneous Quadratic Equation to describe Perpendicular Straight Lines, $\LL'_1$ and $\LL'_2$ are represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y - a y^2$
That is:
:$a + b = 0$
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $17$.
Category:Straight Lines
Category:Perpendiculars
Condition for Quartic with Real Coefficients to have Wholly Imaginary Root

== Theorem ==
Let $Q$ be the quartic equation:
:$(1): \quad z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$
such that all of $a_1, a_2, a_3, a_4$ are real numbers.
Then $Q$ has a root which is wholly imaginary :
:$\text {(a)}: \quad a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$
:$\text {(b)}: \quad a_1 a_3 > 0$
== Proof ==
=== Necessary Condition ===
We have:
This leads to the factorisation of $(1)$:
Two solutions to this are found from:
The other two solutions are found by solving:
As $a_1 \ne 0$ it follows that $(3)$ always has a real part.
So only the $(2)$ can produce roots which are wholly imaginary, and that can happen only when:
:$a_3 / a_1 > 0$
that is:
:$a_1 a_3 > 0$
Thus, if $a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$ and $a_1 a_3 > 0$, $Q$ has wholly imaginary roots.
=== Sufficient Condition ===
Let $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ have wholly imaginary roots.
Then it can be written:
:$\paren {z^2 + b^2} \paren {z^2 + m z + n} = 0$
By long division of $Q$ by $z^2 + b^2$:
:$\paren {z^2 + b^2} \paren {z^2 + a_1 z + \paren {a_2 + b^2} } + \paren {a_3 - a_1 b^2} z + a_4 - \paren {a_2 - b^2} b^2 = 0$
For this to be in the above format, it is necessary that:
:$\paren {a_3 - a_1 b^2} z + a_4 - \paren {a_2 - b^2} b^2 \equiv 0$
That is:
Hence:
Hence if $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ has wholly imaginary roots, then:
:$a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$
We also see that the roots themselves are $\pm b$, where:
:$b = \sqrt {-\dfrac {a_3} {a_1}}$
Thus we have also shown that $a_3 a_1 > 0$ in order for those roots to indeed be wholly imaginary as required.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $129$
::(although see Condition for Quartic with Real Coefficients to have Wholly Imaginary Root/Mistake for analysis of an error in that work)
Category:Quartic Equations
Condition for Quartic with Real Coefficients to have Wholly Imaginary Root/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: $129$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:$\text{(a)} \quad$ Show that the equation $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ where $a_1, a_2, a_3, a_4$ are real constants different from zero, has a
:::pure imaginary root if ${a_3}^2 + {a_1}^2 a_4 = a_1 a_2 a_3$.
As demonstrated in Condition for Quartic with Real Coefficients to have Wholly Imaginary Root it is also necessary for $a_1 a_3 > 0$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $129$
Category:Mistakes/Theory and Problems of Complex Variables
Condition for Rational Cut to be Less than Given Cut

== Theorem ==
Let $\alpha$ be a cut.
Let $p^*$ be the rational cut associated with a rational number $p$.
Then:
:$p \in \alpha$
:
:$p^* < \alpha$
where $<$ denotes the strict ordering on cuts.
== Proof ==
Let $p$ be a rational number such that $p \in \alpha$.
Then by definition of rational cut:
:$p \notin p^*$
Thus:
:$p \in \alpha \implies p^* < \alpha$
Let $p^* < \alpha$.
Then there exists a rational number $q$ such that $q \in \alpha$ and $q \notin p$.
Thus $q \ge p$.
But as $q \in \alpha$ it follows that $p \in \alpha$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.30$. Theorem
Category:Cuts
Condition for Rational Number to be Square of Rational Number

== Theorem ==
Let $m$ and $n$ be (strictly) positive integers which are coprime.
Then:
:$\dfrac m n$ is the square of a rational number
:
:both $m$ and $n$ are square numbers.
== Proof ==
Let $m$ and $n$ be (strictly) positive integers which are coprime.
=== Sufficient Condition ===
Suppose that $\dfrac m n$ is the square of a rational number.
Then there exists a rational number $r$ such that:
:$\dfrac m n = r^2$
By Existence of Canonical Form of Rational Number, we can find $p \in \mathbb Z, q \in \mathbb Z_{>0}, p \perp q$ such that:
:$r = \dfrac p q$
Now we have:
:$\dfrac m n = \paren {\dfrac p q}^2 = \dfrac {p^2} {q^2}$
By Powers of Coprime Numbers are Coprime, $p^2$ and $q^2$ are coprime.
Hence both $\dfrac m n$ and $\dfrac {p^2} {q^2}$ are in canonical form.
By Canonical Form of Rational Number is Unique, we must have:
:$m = p^2, n = q^2$
which shows that both $m, n$ are square numbers.
=== Necessary Condition ===
Suppose that both $m$ and $n$ are square numbers.
Then there exists integers $x, y$ such that:
:$m = x^2, n = y^2$
Since $n > 0$, we must have $y \ne 0$.
Therefore we can write:
:$\dfrac m n = \dfrac {x^2} {y^2} = \paren {\dfrac x y}^2$
By definition, $\dfrac x y$ is a rational number.
Hence we have shown that $\dfrac m n$ is the square of a rational number.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 5$
Category:Square Roots
Category:Rational Numbers
Category:Square Numbers
Condition for Rational to be Convergent

== Theorem ==
Let $x$ be an irrational number.
Let the rational number $\dfrac a b$ satisfy the inequality:
:$\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$
Then $\dfrac a b$ is a convergent of $x$.
== Proof ==
 $\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, but that $\dfrac a b$ is not one of the convergents $\dfrac {p_n} {q_n}$ of $x$.
Let $r$ be the unique integer for which $q_r \le b \le q_{r + 1}$.
Then:
Therefore:
:$q_r \size {x - \dfrac {p_r} {q_r} }< \dfrac 1 {2 b}$
and so:
:$\size {x - \dfrac {p_r} {q_r} } < \dfrac 1 {2 q_r b}$
Hence:
Now note that $q_r a - p_r b$ is a integer.
However, $\dfrac a b$ is not one of the convergents $\dfrac {p_n} {q_n}$ of $x$.
Thus $\dfrac a b \ne \dfrac {p_r} {q_r}$.
But we have:
So, combining results $(1)$ and $(2)$, we get:
:$\dfrac 1 {q_r b} < \dfrac 1 {2 q_r b} + \dfrac 1 {2 b^2}$
This simplifies to $q_r > b$.
But $r$ be the unique integer for which $q_r \le b \le q_{r + 1}$.
From this contradiction it follows that $\dfrac a b$ is one of the convergents of $x$
Category:Continued Fractions
Condition for Relation to be Transitive and Antitransitive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is both transitive and antitransitive
:
:$\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$
== Proof ==
=== Necessary Condition ===
Suppose $\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$.
Then $\RR$ is both transitive and antitransitive vacuously.
=== Sufficient Condition ===
Suppose $\RR$ is both transitive and antitransitive.
 it is not the case that $\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$.
Then $\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z$.
By transitivity:
:$x \mathrel \RR z$
By antitransitivity:
:$\neg \paren {x \mathrel \RR z}$
This is a contradiction.
Hence we must have $\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $7 \ \text{(a)}$
Category:Transitive Relations
Category:Antitransitive Relations
Condition for Repunits to be Coprime

== Theorem ==
Let $R_p$ and $R_q$ be repunit numbers with $p$ and $q$ digits respectively.
Then $R_p$ and $R_q$ are coprime $p$ and $q$ are coprime.
== Proof ==
=== Necessary Condition ===
We show the contrapositive.
Suppose $p$ and $q$ are not coprime.
Then there exists some $d > 1$ such that $p = d m$ and $q = d n$.
Then by Divisors of Repunit with Composite Index:
:$R_d \divides R_p$
and:
:$R_d \divides R_q$
where $\divides$ denotes divisibility.
That is, $R_d$ is a common divisor of $R_p$ and $R_q$.
Therefore $R_p$ and $R_q$ are not coprime.
Thus we have shown the contrapositive of the necessary condition.
=== Sufficient Condition ===
Let $p$ and $q$ be coprime.
Let $R_p$ and $R_q$ be repunits in base $b$.
By Integer Combination of Coprime Integers, there exists integers $m, n$ such that $m p + n q = 1$.
One of $m, n$ is strictly positive while the other is negative.
, suppose $m > 0$ and $n \le 0$.
Then from $m p = 1 - n q$:
We have $R_{m p} - b R_{- n q} = 1$.
By Divisors of Repunit with Composite Index, $R_p$ is a divisor of $R_{m p}$, while $R_q$ is a divisor of $R_{- n q}$.
By definition of divisibility, there exists integers $c, d$ such that $R_{m p} = c R_p, R_{- n q} = d R_q$.
Therefore $c R_p - b d R_q = 1$.
Note that both $c$ and $- b d$ are integers.
By Integer Combination of Coprime Integers, $R_p$ and $R_q$ are coprime.
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Repunits
Condition for Resonance in Forced Vibration of Underdamped System

== Theorem ==
Consider a physical system $S$ whose behaviour is defined by the second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$
where:
:$K \in \R: k > 0$
:$a, b \in \R_{>0}: b < a$
Then $S$ is in resonance when:
:$\omega = \sqrt {a^2 - 2 b^2}$
and thus the resonance frequency is:
:$\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$
This resonance frequency exists only when $a^2 - 2 b^2 > 0$.
== Proof ==
From Linear Second Order ODE: $y' ' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ the general solution of $(1)$ is:
:$(2): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
By definition, $S$ is in resonance when the amplitude of the steady-state component of $(1)$ is at a maximum.
We are to determine the value of $\omega$ when this is the case.
Consider the steady-state component of $(2)$:
:$y_s = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
By definition, the amplitude of $y_s$ is:
:$A = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} }$
We have that both $a$ and $b$ are greater than $0$.
It is also assumed that $\omega$ is also greater than $0$.
Thus:
:$4 b^2 \omega^2 > 0$
:$a^2 > 0$
:$\omega^2 > 0$
Then $y_s$ is at a maximum when:
:$\map f \omega = 4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2$
is at a minimum.
Differentiating $\map f \omega$ $\omega$:
Setting $\map {f'} \omega = 0$:
As $\omega > 0$ the first of these cannot apply.
If $a^2 < 2 b^2$ then $a^2 - 2 b^2 < 0$ and so $\omega^2$ has no real roots.
Hence the resonance frequency is given by:
:$\omega = \sqrt {a^2 - 2 b^2}$
 $a^2 > 2 b^2$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems
Category:Resonance
Category:Mathematical Physics
Condition for Set Equivalent to Cardinal Number

== Theorem ==
Let $S$ be a set.
Let $\card S$ denote the cardinality of $S$.
That is, let $\card S$ be the intersection of all ordinals equivalent to $S$.
Note that in the absence of the Axiom of Choice, $\card S$ may be the class of all sets.
Then the following are equivalent:
:$(1): \quad S \sim \card S$
:$(2): \quad \exists x \in \On: S \sim x$
:$(3): \quad \exists x \in \On: \exists y: \paren {y \subseteq x \land S \sim y}$
where $A \sim B$ means that there is a bijection from $A$ onto $B$, and the quantification over $y$ is unbounded (so $y$ may be any set).
== Proof ==
=== $2 \implies 1$ ===
If $\exists x \in \On: S \sim x$, then by Class of All Ordinals is Well-Ordered by Subset Relation there is a smallest ordinal $x$ such that $S \sim x$.
This smallest ordinal $x$ is the cardinal number of $S$, by definition.
=== $3 \implies 2$ ===
Suppose that $y \subseteq x$ and $S \sim y$ for some ordinal $x$.
Since $y \subseteq x$, it follows that $y \sim z$ for some $z \in \On$ by Unique Isomorphism between Ordinal Subset and Unique Ordinal.
Therefore, by Set Equivalence behaves like Equivalence Relation, $S \sim z$.
=== $1 \implies 3$ ===
Suppose that $3$ is not true.
It follows that $S \not \sim x$ for any ordinal $x$.
But $S \not \sim \mathbb U$, so $S \not \sim \card S$ by the definition of cardinal number.
The result follows by contraposition.
== Sources ==
* : $\S 10.9$
Category:Cardinals
Condition for Set Union Equivalent to Associated Cardinal Number

== Theorem ==
Let $S$ and $T$ be sets.
Let $\left|{S}\right|$ denote the cardinal number of $S$.
Let $\sim$ denote set equivalence.
Then:
:$S \cup T \sim \left|{S \cup T}\right| \iff S \sim \left|{S}\right| \land T \sim \left|{T}\right|$
== Proof ==
=== Necessary Condition ===
Let $S \cup T \sim \left|{S \cup T}\right|$.
By definition of set equivalence, there exists a bijection:
: $f: S \cup T \to \left|{S \cup T}\right|$
Since $f$ is a bijection, it follows that:
:$S$ is equivalent to the image of $S$ under $f$.
This, in turn, is a subset of the ordinal $\left|{S \cup T}\right|$.
$\left|{S \cup T}\right|$ is an ordinal by Cardinal Number is Ordinal.
By Condition for Set Equivalent to Cardinal Number, it follows that:
: $S \sim \left|{S}\right|$
Similarly:
:$T \sim \left|{T}\right|$
=== Sufficient Condition ===
Suppose $S \sim \left|{S}\right|$ and $T \sim \left|{T}\right|$.
Let $f: S \to \left|{S}\right|$ and $g: T \to \left|{T}\right|$ be bijections.
By definition of set equivalence, these bijections are known to exist.
Define the mapping $F : S \cup T \to \left|{S}\right| + \left|{T}\right|$ to be:
:$F \left({x}\right) = \begin{cases}
f \left({x}\right) & : x \in S \\
\left|{S}\right| + g \left({x}\right) & : x \notin S
\end{cases}$
Suppose $F \left({x}\right) = F \left({y}\right)$.
Let $x, y \in S$.
Then:
: $f \left({x}\right) = f\left({y}\right)$
Since $f$ is a bijection, it follows that:
: $x = y$
Let $x \in S, y \in T$.
Then:
: $f \left({x}\right) = \vert S \vert + g\left({x}\right)$
But this is a contradiction, since $f\left({x}\right)$ has to be an element of $\left\vert{S}\right\vert$.
Finally, let $x, y \in T$.
Then:
: $\left\vert{S}\right\vert + g \left({x}\right) = \left\vert{S}\right\vert + g\left({y}\right)$
From Ordinal Addition is Left Cancellable:
:$g \left({x}\right) = g\left({y}\right)$
By the definition of a bijection:
:$x = y$
It follows that $F: S \cup T \to \left|{S}\right| + \left|{T}\right|$ is an injection, where $\left|{S}\right| + \left|{T}\right|$ denotes ordinal addition.
Therefore, $S \cup T$ is equivalent to some subset of the ordinal $\left|{S}\right| + \left|{T}\right|$.
By Condition for Set Equivalent to Cardinal Number, it follows that $S \cup T \sim \left|{S \cup T}\right|$.
== Also see ==
* Set Equivalent to Cardinal, which requires the axiom of choice.
== Sources ==
* : $\S 10.15 \ (1)$
Category:Cardinals
Condition for Solubility of Galois Group of Polynomial

== Theorem ==
Let $P$ be a polynomial.
Let $G$ be the Galois group of $P$.
Then $G$ is soluble the roots of $P$ can be obtained from the coefficients of $P$ using just the arithmetic operations and raising to powers of the form $\dfrac 1 n$ for a natural number $n$.
== Proof ==
== Sources ==
* 
* 
Category:Galois Groups of Polynomials
Category:Solvable Groups
Condition for Straight Lines in Plane to be Parallel/Examples/Arbitrary Example 1

== Examples of Use of Condition for Straight Lines in Plane to be Parallel ==
Let $\LL_1$ be the straight line whose equation in general form is given as:
:$3 x - 4 y = 7$
Let $\LL_2$ be the straight line parallel to $\LL_1$ which passes through the point $\tuple {1, 2}$.
The equation for $\LL_2$ is:
:$3 x - 4 y = -5$
== Proof ==
From Condition for Straight Lines in Plane to be Parallel, $\LL_2$ has an equation of the form:
:$(1): \quad 3 x - 4 y = C$
We have that $\tuple {1, 2}$ is on $\LL_2$.
Hence substituting $x = 1$ and $y = 2$ into $(1)$:
Hence the result.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines: Example $\text{(i)}$
Category:Condition for Straight Lines in Plane to be Parallel