avewright/proofwiki-math huggingface dataset:
Theorem: Proportion is Equivalence Relation

Proportion is an equivalence relation.

Proof:
:Proportion is Reflexive: $\forall x \in \R: x \propto x$
:Proportion is Symmetric: $\forall x, y \in \R: x \propto y \implies y \propto x$
:Proportion is Transitive: $\forall x, y, z \in \R: x \propto y \land y \propto z \implies x \propto z$
The result follows from the definition of an equivalence relation.
{{qed}}
Category:Proportion
Category:Examples of Equivalence Relations
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Theorem: Parallelism implies Supplementary Interior Angles

Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the interior angles on the same side of the transversal are supplementary.

Proof:
:200px
Let $AB$ and $CD$ be parallel infinite straight lines.
Let $EF$ be a transversal that cuts them.
From Parallelism implies Equal Corresponding Angles and Euclid's second common notion:
:$\angle EGB + \angle BGH = \angle DHG + \angle BGH$
From Two Angles on Straight Line make Two Right Angles, $\angle EGB + \angle BGH$ equals two right angles.
So by definition, $\angle BGH$ and $\angle DHG$ are supplementary.
{{qed}}
{{Euclid Note|29|I|{{EuclidNoteConverse|prop = 28|title = Supplementary Interior Angles implies Parallel Lines|part = second}}|part = third}}
Theorem: Proportion of Power

Let $x$ and $y$ be proportional.
{{explain|Establish what types of object $x$ and $y$ are. As it stands here, they could be anything.}}
Let $n \in \Z$. 
Then $x^n \propto y^n$.

Proof:
Let $x \propto y$.
Then $\exists k \ne 0: x = k \times y$ by the definition of proportion.
Raising both sides of this equation to the $n$th power:
{{begin-eqn}}
{{eqn | l = x^n
      | r = \paren {k \times y}^n
}}
{{eqn | r = k^n \times y^n
}}
{{end-eqn}}
so $k^n$ is the desired constant of proportion.
The result follows from the definition of proportion.
{{qed}}
Category:Proportion
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Theorem: Sine of 18 Degrees

:$\sin 18 \degrees = \sin \dfrac \pi {10} = \dfrac 1 {2 \phi} = 2^{-1} \phi^{-1} = \dfrac {\sqrt 5 - 1} 4$
where $\sin$ denotes the sine function.
and $\phi$ denotes the golden mean.

Proof:
From Sine of $90 \degrees$:
:$\map \sin {5 \times 18 \degrees} = \sin 90 \degrees = 1$.
Consider the equation:
:$\sin 5x = 1$
where $x = 18 \degrees$ is one of the solutions.
From Quintuple Angle Formula for Sine:
:$16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$
Let $s = \sin \theta$:
:$16 s^5 - 20 s^3 + 5s - 1 = 0$
That is:
:$\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$
Therefore:
{{begin-eqn}}
{{eqn | l = \sin 18 \degrees
      | r = \frac {-2 \pm \sqrt {\paren {-2}^2 + 4 \times 4} } {2 \times 4}
      | c = Quadratic Formula
}}
{{eqn | r = \frac {-2 \pm \sqrt {20} } 8
}}
{{eqn | r = \frac {\sqrt 5 - 1} 4
      | c = negative root is rejected as $0 < \sin 18 \degrees < 1$
}}
{{eqn | r = \frac {\sqrt 5 - 1} 4 \times \frac {\sqrt 5 + 1} {\sqrt 5 + 1}
      | c = multiplying top and bottom by $\sqrt 5 + 1$ 
}}
{{eqn | r = \frac 1 {\sqrt 5 + 1}
      | c = 
}}
{{eqn | r = \frac 1 {2 \phi}
      | c = {{Defof|Golden Mean}}
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Golden Mean
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Theorem: Cosine of 72 Degrees

:$\cos 72 \degrees = \cos \dfrac {2 \pi} 5 = 2^{-1} \phi ^{-1} = 2^{-1} \paren {\phi - 1} = \dfrac {\sqrt 5 - 1} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 72 \degrees
      | r = \map \cos {90 \degrees - 18 \degrees}
}}
{{eqn | r = \sin 18 \degrees
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \dfrac {\sqrt 5 - 1} 4
      | c = {{sin|18}}
}}
{{eqn | r = \frac 1 2 \times \paren {\dfrac {\sqrt 5 + 1} 2 - 1}
      | c = 
}}
{{eqn | r = \frac 1 2 \times \paren {\phi - 1}
      | c = 
}}
{{eqn | r = \frac 1 2 \times \frac 1 \phi
      | c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = 2^{-1} \phi^{-1}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 72 Degrees

:$\cos 72 \degrees = \cos \dfrac {2 \pi} 5 = 2^{-1} \phi ^{-1} = 2^{-1} \paren {\phi - 1} = \dfrac {\sqrt 5 - 1} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 72 \degrees
      | r = 2 \paren {\cos 36 \degrees}^2 - 1
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | r = 2 \paren {\dfrac \phi 2}^2 - 1
      | c = {{cos|36}}
}}
{{eqn | r = \dfrac {\phi^2} 2 - 1
      | c = 
}}
{{eqn | r = \dfrac {\phi + 1} 2 - 1
      | c = Square of Golden Mean equals One plus Golden Mean
}}
{{eqn | r = \dfrac {\phi - 1} 2
      | c = 
}}
{{eqn | r = -\dfrac {1 - \phi} 2
      | c = 
}}
{{eqn | r = \dfrac {\phi^{-1} } 2
      | c = Reciprocal Form of One Minus Golden Mean
}}
{{eqn | r = \dfrac {\sqrt 5 - 1} 4
      | c = {{Defof|Golden Mean|index = 2}}, and algebra
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 72 Degrees

:$\sin 72 \degrees = \sin \dfrac {2 \pi} 5 = 2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} = \dfrac {\sqrt{10 + 2 \sqrt 5} } 4$
where $\sin$ denotes the sine function.

Proof:
{{begin-eqn}}
{{eqn | l = \sin 72 \degrees
      | r = \sqrt {1 - \cos^2 72 \degrees}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sqrt {1 - \paren {\frac{\sqrt 5 - 1} 4}^2}
      | c = Cosine of $72 \degrees$
}}
{{eqn | r = \sqrt {1 - \frac {6 - 2 \sqrt 5} {16} }
}}
{{eqn | r = \frac {\sqrt {10 + 2 \sqrt 5} } 4
      | c = 
}}
{{eqn | r = \frac {\sqrt {\paren {2 \sqrt 5 } \paren {\sqrt 5 + 1} \times \frac 2 2 } } 4
      | c = 
}}
{{eqn | r = \frac {\sqrt {\paren {4 \sqrt 5 } \phi } } 4
      | c = 
}}
{{eqn | r = 2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 18 Degrees

:$\cos 18 \degrees = \cos \dfrac \pi {10} = \dfrac {\sqrt {10 + 2 \sqrt 5} } 4$
where $\cos$ denotes the cosine function.

Proof:
{{begin-eqn}}
{{eqn | l = \cos 18 \degrees
      | r = \map \cos {90 \degrees - 72 \degrees}
}}
{{eqn | r = \sin 72 \degrees
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \frac {\sqrt {10 + 2 \sqrt 5} } 4
      | c = Sine of $72 \degrees$
}}
{{end-eqn}}
{{qed}}
Category:Cosine Function
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Theorem: Sine of 3 Degrees

:$\sin 3^\circ = \sin \dfrac \pi {60} = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}$
where $\sin$ denotes the sine function.

Proof:
{{begin-eqn}}
{{eqn | l = \sin 3^\circ
      | r = \sin \left({75^\circ - 72^\circ}\right)
}}
{{eqn | r = \sin 75^\circ \cos 72^\circ - \cos 75^\circ \sin 72^\circ
      | c = Sine of Difference
}}
{{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \cos 75^\circ \sin 72^\circ
      | c = Sine of $75^\circ$, Cosine of $72^\circ$ ...
}}
{{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt 5} } 4
      | c = ... Cosine of $75^\circ$, Sine of $72^\circ$
}}
{{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt 6 + \sqrt 2}\right) \times \left({\sqrt 5 - 1}\right) - \left({\sqrt 6 - \sqrt 2}\right) \times \left({\sqrt {10 + 2 \sqrt 5} }\right) }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2}\right) - \left({\sqrt {60 + 12 \sqrt 5} - \sqrt {20 + 4 \sqrt 5} }\right) }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - \sqrt {60 + 12 \sqrt5 } + \sqrt {20 + 4 \sqrt 5} }\right)
}}
{{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2 \sqrt {5 + \sqrt 5} }\right)
}}
{{eqn | r = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2\sqrt {5 + \sqrt 5} } {16}
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
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Theorem: Cosine of 3 Degrees

:$\cos 3 \degrees = \cos \dfrac {\pi} {60} = \dfrac {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2 + 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}$
where $\cos$ denotes the cosine function.

Proof:
{{begin-eqn}}
{{eqn | l = \cos 3 \degrees
      | r = \map \cos {75 \degrees - 72 \degrees}
}}
{{eqn | r = \cos 75 \degrees \cos 72 \degrees + \sin 75 \degrees \sin 72 \degrees
      | c = Cosine of Difference
}}
{{eqn | r = \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 + \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt5 } } 4
      | c = Cosine of $75 \degrees$, Cosine of $72 \degrees$, Sine of $75 \degrees$, Sine of $72 \degrees$
}}
{{eqn | r = \dfrac 1 {16} \times \paren {\paren {\sqrt 6 - \sqrt 2} \times \paren {\sqrt 5 - 1} + \paren {\sqrt 6 + \sqrt 2} \times \paren {\sqrt {10 + 2 \sqrt 5} } }
}}
{{eqn | r = \dfrac 1 {16} \times \paren {\paren {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2} + \paren {\sqrt {60 + 12 \sqrt 5} + \sqrt {20 + 4 \sqrt 5} } }
}}
{{eqn | r = \dfrac 1 {16} \times \paren {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2 + \sqrt {60 + 12 \sqrt 5} + \sqrt {20 + 4 \sqrt 5} }
}}
{{eqn | r = \dfrac 1 {16} \times \paren {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2 + 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} }
}}
{{eqn | r = \dfrac {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2 + 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}
}}
{{end-eqn}}
{{qed}}
Category:Cosine Function
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Theorem: Square Root of Sum as Sum of Square Roots

Let $a, b \in \R, a \ge b$.
Then:
:$\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$

Proof:
Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$.
From Square of Sum:
: $a + b = c + d + 2 \sqrt {c d}$
We now need to solve the simultaneous equations:
: $a = c + d$
: $b = 2 \sqrt {c d}$
First:
{{begin-eqn}}
{{eqn | l = a
      | r = c + d
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = d
      | r = a - c
      | c = subtracting $c$ from both sides
}}
{{end-eqn}}
Solving for $c$:
{{begin-eqn}}
{{eqn | l = b
      | r = 2 \sqrt {c d}
}}
{{eqn | ll= \leadsto
      | l = b^2
      | r = 4 c d
      | c = squaring both sides
}}
{{eqn | r = 4 c \paren {a - c}
      | c = substituting $d = a - c$ from $(1)$
}}
{{eqn | r = 4 a c - 4 c^2
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | ll= \leadsto
      | l = 4 c^2 - 4 a c + b^2
      | r = 0
      | c = adding $4 c^2 - 4 a c$ to both sides
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = c
      | r = \frac a 2 \pm \frac {\sqrt {a^2 - b^2} } 2
      | c = Quadratic Formula
}}
{{end-eqn}}
Solving for $d$:
{{begin-eqn}}
{{eqn | l = d
      | r = a - c
}}
{{eqn | r = a - \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2
      | c = substituting $c = \dfrac a 2 \pm \dfrac {\sqrt {a^2 - b^2} } 2$ from $(2)$
}}
{{eqn | r = \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2
}}
{{end-eqn}}
From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$.
{{qed}}
Theorem: Square Root of Sum as Sum of Square Roots

Let $a, b \in \R, a \ge b$.
Then:
:$\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$

Proof:
From Sum of Square Roots as Square Root of Sum:
:$\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$
Let
:$p = \dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2$,
:$q = \dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2$.
Then
{{begin-eqn}}
{{eqn | l = p + q
      | r = \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2
}}
{{eqn | r = a
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sqrt {4pq}
      | r = \sqrt {4 \paren {\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2} \paren {\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2} }
}}
{{eqn | r = \sqrt {\paren {a + \sqrt {a^2 - b^2} } \paren {a - \sqrt {a^2 - b^2} } }
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = \sqrt {a^2 - \paren {a^2 - b^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \sqrt {b^2}
}}
{{eqn | r = b
}}
{{end-eqn}}
Therefore:
:$\sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2} = \sqrt {a + b}$
{{qed}}
Theorem: Rectangle is Parallelogram

Let $ABCD$ be a rectangle.
Then $ABCD$ is a parallelogram.

Proof:
Let $ABCD$ be a rectangle.
{{AimForCont}} $ABCD$ is ''not'' a parallelogram.
{{WLOG}}, let line segments $AB$ and $CD$ ''not'' be parallel.
By {{EuclidPostulateLink|Second}}, let us produce $AB$ and $CD$ into two infinite straight lines.
By the Parallel Postulate, the $AD$ and $BC$ will eventually meet at one side or the other.
Let their point of intersection be $E$.
There are two possibilities:
:$ADE$ is a triangle
:$CBE$ is a triangle.
{{WLOG}}, let $\triangle ADE$ be a triangle:
:600px
By Sum of Angles of Triangle equals Two Right Angles, $\angle ADE + \angle DEA + \angle EAD$ add to two right angles.
But by the definition of a rectangle, $\angle ADE + \angle EAD$ add to two right angles.
As $\angle DEA$ is not a zero angle (else $\triangle ADE$ would not be a triangle), this cannot be so.
Hence $ABCD$ is not a rectangle.
But {{hypothesis}}, $ABCD$ is a rectangle.
This is a contradiction.
Hence by Proof by Contradiction, $AB$ and $CD$ are parallel.
The same argument can be used to demonstrate that $AD$ and $BC$ are also parallel.
Hence the result by definition of parallelogram.
{{qed}}
Category:Rectangles
Category:Parallelograms
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Theorem: Condition for Denesting of Square Root

Let $a, b \in \Q_{\ge 0}$.
Suppose $\sqrt b \notin \Q$.
Then:
:$\exists p, q \in \Q: \sqrt {a + \sqrt b} = \sqrt p + \sqrt q$
{{iff}}:
:$\exists n \in \Q: a^2 - b = n^2$.

Proof:
=== Lemma ===
{{:Condition for Denesting of Square Root/Lemma}}{{qed|lemma}}
Theorem: Characteristic Function of Union/Variant 1

:$\chi_{A \mathop \cup B} = \min \set {\chi_A + \chi_B, 1}$

Proof:
By Characteristic Function Determined by 1-Fiber, it suffices to show:
:$\min \set {\map {\chi_A} s + \map {\chi_B} s, 1} = 1 \iff s \in A \cup B$
By the nature of the min operation, this amounts to showing that:
:$\map {\chi_A} s + \map {\chi_B} s \ge 1 \iff s \in A \cup B$
As $\chi_A, \chi_B$ are characteristic functions, the {{LHS}} amounts to:
:$s \in A \lor s \in B$
which is precisely the definition of $s \in A \cup B$.
{{qed}}
Theorem: Characteristic Function of Union/Variant 2

:$\chi_{A \mathop \cup B} = \chi_A + \chi_B - \chi_{A \mathop \cap B}$

Proof:
From Subset of Union:
:$A, B \subseteq A \cup B$
From Intersection is Subset of Union:
:$A \cap B \subseteq A \cup B$
Thus from Characteristic Function of Subset:
:$\map {\chi_{A \mathop \cup B} } s = 0 \implies \map {\chi_A} s = \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s = 0$
Now suppose that $\map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$.
That is:
:$\map {\chi_A} s + \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s$
Suppose the latter equals $1$, that is $s \in A \cap B$.
Then both $s \in A$ and $s \in B$, so by definition of characteristic function:
:$\map {\chi_A} s + \map {\chi_B} s = 1 + 1 = 2$
Since $2 \ne 1$, it follows that $\map {\chi_{A \mathop \cap B} } s \ne 1$, that is it equals $0$.
Thence, it follows that:
:$\map {\chi_A} s + \map {\chi_B} s = 0$.
This only happens when $\map {\chi_A} s = \map {\chi_B} s = 0$.
Thus, $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.
It finally follows that $\map {\chi_{A \mathop \cup B} } s = 0$.
It is now established that:
:$\map {\chi_{A \mathop \cup B} } s = 0 \iff \map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$
and from Characteristic Function Determined by 0-Fiber, the result follows.
{{qed}}
Theorem: Characteristic Function of Union/Variant 3

:$\chi_{A \mathop \cup B} = \max \set {\chi_A, \chi_B}$

Proof:
Suppose $\map {\chi_{A \mathop \cup B} } s = 0$.
Then $s \notin A \cup B$, so $s \notin A$ and $s \notin B$.
Hence:
:$\map {\chi_A} s = \map {\chi_B} s = 0$
and by definition of max operation:
:$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
Conversely, suppose:
:$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
Then it follows that:
:$\map {\chi_A} s = \map {\chi_B} s = 0$
because characteristic functions are $0$ or $1$.
Hence $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.
That is:
:$\map {\chi_{A \mathop \cup B} } s = 0$
Above considerations give:
:$\map {\chi_{A \mathop \cup B} } s = 0 \iff \max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
and applying Characteristic Function Determined by 0-Fiber, the result follows.
{{qed}}
Theorem: Characteristic Function of Intersection/Variant 1

:$\chi_{A \mathop \cap B} = \chi_A \chi_B$

Proof:
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
:$\map {\chi_A} s \map {\chi_B} s = 1 \iff s \in A \cap B$
Now, both $\chi_A$ and $\chi_B$ are characteristic functions.
It follows that, for any $s \in S$:
:$\map {\chi_A} s \map {\chi_B} s = 1 \iff \map {\chi_A} s = \map {\chi_B} s = 1$
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is, $s \in A \cap B$, by definition of set intersection.
{{qed}}
Theorem: Characteristic Function of Intersection/Variant 2

:$\chi_{A \mathop \cap B} = \min \set {\chi_A, \chi_B}$

Proof:
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
:$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1 \iff s \in A \cap B$
By definition of characteristic function, we have:
:$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1$
{{iff}}:
:$\map {\chi_A} s = \map {\chi_B} s = 1$
because $\chi_A, \chi_B$ take values in $\set {0, 1}$.
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is:
:$s \in A \cap B$
by definition of set intersection.
{{qed}}
Theorem: Multiplication of Real Numbers is Right Distributive over Subtraction

{{:Euclid:Proposition/V/6}}
That is, for any number $a$ and for any integers $m, n$:
:$m a - n a = \paren {m - n} a$

Proof:
Let two magnitudes $AB, CD$ be equimultiples of two magnitudes $E, F$.
Let $AG, CH$ subtracted from them be equimultiples of the same two $E, F$.
We need to show that the remainders $GB, HD$ are either equal to $E, F$ or are equimultiples of them.
:350px
First let $GB = E$.
Let $CK$ be made equal to $F$.
We have that $AG$ is the same multiple of $E$ that $CH$ is of $F$, while $GB = E$ and $KC = F$.
Therefore from the Distributive Laws of Arithmetic, $AB$ is the same multiple of $E$ that $KH$ is of $F$.
But {{hypothesis}}, $AB$ is the same multiple of $E$ that $CD$ is of $F$.
Since then, each of the magnitudes $KH, CD$ is the same multiple of $F$.
Therefore $KH = CD$.
Let $CH$ be subtracted from each.
Therefore the remainder $KC$ equals the remainder $HD$.
But $F = KC$, so $HD = F$.
Hence, if $GB = E$ then $HD = F$.
Similarly we can prove that, even if $GB$ is a multiple of $E$, then $HD$ is also the same multiple of $F$.
{{qed}}
Theorem: Sum of Square Roots as Square Root of Sum

:$\sqrt a + \sqrt b = \sqrt {a + b + \sqrt {4 a b} }$

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt a + \sqrt b
      | r = \sqrt {\paren {\sqrt a + \sqrt b}^2}
}}
{{eqn | r = \sqrt {\sqrt a^2 + \sqrt b^2 + 2 \sqrt a \sqrt b}
      | c = Square of Sum
}}
{{eqn | r = \sqrt {a + b + \sqrt {4 a b} }
      | c = Power of Product
}}
{{end-eqn}}
{{qed}}
Category:Algebra
grdxa15a4j79pyvvs7dfyrqst3wc5fw
Theorem: Divisor of One of Coprime Numbers is Coprime to Other

Let $a, b \in \N$ be numbers such that $a$ and $b$ are coprime:
:$a \perp b$
Let $c > 1$ be a divisor of $a$:
:$c \divides a$
Then $c$ and $b$ are coprime:
:$c \perp b$
{{:Euclid:Proposition/VII/23}}

Proof:
Let $a \perp b$ and $c > 1: c \divides a$.
{{AimForCont}} $c \not \perp b$.
So by definition of not coprime:
:$\exists d > 1: d \divides c, d \divides b$
But from Divisor Relation is Transitive:
:$d \divides c, c \divides a \implies d \divides a$
So $d$ is a common divisor of both $a$ and $b$.
So, by definition, $a$ and $b$ are not coprime.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Divisor of One of Coprime Numbers is Coprime to Other

Let $a, b \in \N$ be numbers such that $a$ and $b$ are coprime:
:$a \perp b$
Let $c > 1$ be a divisor of $a$:
:$c \divides a$
Then $c$ and $b$ are coprime:
:$c \perp b$
{{:Euclid:Proposition/VII/23}}

Proof:
Let $A, B$ be two numbers which are prime to one another.
Let $C$ be any number greater than $1$ which measures $A$.
:250px
Suppose $C$ and $B$ are not prime to one another.
Then some number $D$ will measure them both.
We have that $D$ measures $C$ and $C$ measures $A$.
So $D$ measures $A$.
But $D$ also measures $B$.
So $D$ measures $A$ and $B$ which are prime to one another.
By {{EuclidDefLink|VII|12|Relatively Prime}}, this is a contradiction.
Therefore there can be no such $D$ that measures both $B$ and $C$.
That is, $B$ and $C$ are prime to one another.
{{qed}}
Theorem: Integer Coprime to Factors is Coprime to Whole

Let $a, b, c \in \Z$ be integers.
Let:
:$a \perp b$
:$a \perp c$
where $\perp$ denotes coprimality.
Then:
:$a \perp b c$

Proof:
''This proof follows the structure of {{AuthorRef|Euclid}}'s proof, if not its detail.''
Let $a, b, c \in \Z$ such that $c$ is coprime to each of $a$ and $b$.
Let $d = a b$.
{{AimForCont}} $c$ and $d$ are not coprime.
Then:
:$\exists e \in \Z_{>1}: e \divides c, e \divides d$
We have that $c \perp a$ and $e \divides c$
From {{EuclidPropLink|book = VII|prop = 23|title = Divisor of One of Coprime Numbers is Coprime to Other}}:
:$e \perp a$
As $e \divides d$ it follows by definition of divisibility that:
:$\exists f \in \Z: e f = d$
We also have that:
:$a b = d$
So:
:$a b = e f$
But from {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}:
:$e : a = b : f$
or in more contemporary language:
:$\dfrac a e = \dfrac b f$
But $a \perp e$.
From {{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}, $\dfrac a e$ is in canonical form.
From {{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}}:
:$e \divides b$
But we already have that:
:$e \divides c$
So $e$ is a divisor of both $b$ and $c$.
But this is a contradiction of the assertion that $c$ and $b$ are coprime.
Hence the result.
{{qed}}
Theorem: Integer Coprime to Factors is Coprime to Whole

Let $a, b, c \in \Z$ be integers.
Let:
:$a \perp b$
:$a \perp c$
where $\perp$ denotes coprimality.
Then:
:$a \perp b c$

Proof:
Let $a, b, c \in \Z$ such that $a$ is coprime to each of $b$ and $c$.
We have:
{{begin-eqn}}
{{eqn | l = a
      | o = \perp
      | r = b
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | q = \exists x, y \in \Z
      | l = 1
      | r = a x + b y
      | c = Integer Combination of Coprime Integers
}}
{{eqn | l = a
      | o = \perp
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists u, v \in \Z
      | l = 1
      | r = a u + c v
      | c = Integer Combination of Coprime Integers
}}
{{eqn | r = \paren {a x + b y} \paren {a u + c v}
      | c = from $(1)$
}}
{{eqn | r = a^2 u x + a c v x + a b y u + b c y v
      | c = multiplying out
}}
{{eqn | r = a \paren {a u x + c v x + b y u} + b c \paren {y v}
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \perp
      | r = b c
      | c = Integer Combination of Coprime Integers
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Coprime Number is Coprime

Let $a$ and $b$ be coprime integers:
:$a, b \in \Z: a \perp b$
Then:
:$a^2 \perp b$
{{:Euclid:Proposition/VII/25}}

Proof:
Let $a \perp b$.
Let $a^2 = c$.
Let $d = a$.
As $a \perp b$ it follows that $d \perp b$.
From {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to Factors is Coprime to Whole}}:
:$a d \perp b$
But $a d = c = a^2$.
Hence the result.
{{qed}}
{{Euclid Note|25|VII}}