hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\left({S, \circ}\right)$ be a group. $\left({S, \circ}\right)$ is a semigroup by the definition of a group. By Group has Latin Square Property, the Latin square property holds in $S$. {{qed|lemma}} === Sufficient Condition ===  Let $\left({S, \circ}\right)$ be a semigroup on which the Latin square property holds. Taking the group axioms in turn: ==== G0: Closure ==== $\left({S, \circ}\right)$ is semigroup. Hence $\left({S, \circ}\right)$ is closed by definition. {{qed|lemma}} ==== G1: Associativity ==== $\left({S, \circ}\right)$ is semigroup. Hence $\circ$ is associative by definition. {{qed|lemma}} ==== G2: Identity ==== Let $a \in S$. Then there is an $x \in S$ such that: {{begin-eqn}} {{eqn | n = 1       | l = a \circ x       | r = a }} {{end-eqn}} Let $b \in S$. Then there is a $y \in S$ such that: {{begin-eqn}} {{eqn | n = 2       | l = y \circ a       | r = b }} {{end-eqn}} Proving that $x$ is a right identity: {{begin-eqn}} {{eqn | l = b \circ x       | r = \left({y \circ a}\right) \circ x       | c = by $(2)$ }} {{eqn | r = y \circ \left({a \circ x}\right)       | c = Definition of Associativity }} {{eqn | r = y \circ a       | c = by $(1)$ }} {{eqn | r = b       | c = by $(2)$ }} {{end-eqn}} Let $c \in S$. Then there is a $z \in S$ such that: {{begin-eqn}} {{eqn | n = 3       | l = x \circ z       | r = c }} {{end-eqn}} Proving that $x$ is a left identity and thus an identity: {{begin-eqn}} {{eqn | l = x \circ c       | r = x \circ \left({x \circ z}\right)       | c = by $(3)$ }} {{eqn | r = \left({x \circ x}\right) \circ z       | c = Definition of Associativity }} {{eqn | r = x \circ z       | c = $x$ is a right identity }} {{eqn | r = c       | c = by $(3)$ }} {{end-eqn}} Relabel $x$ as $e$. {{qed|lemma}} ==== G3: Inverses ==== Let $x \in S$. Then $x$ has a right and left inverse by the property as there exists a $y$ and $z$ in $S$ such that: {{begin-eqn}} {{eqn | n = 4       | l = x \circ y       | r = e }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 5       | l = z \circ x       | r = e }} {{end-eqn}} We shall prove that $y$ is also the left inverse of $x$: {{begin-eqn}} {{eqn | l = y \circ x       | r = \left({z \circ x}\right) \circ \left({y \circ x}\right)       | c = Definition of Identity }} {{eqn | r = z \circ \left({\left({x \circ y}\right) \circ x}\right)       | c = Definition of Associativity }} {{eqn | r = z \circ \left({e \circ x}\right)       | c = by $(4)$ }} {{eqn | r = z \circ x       | c = Definition of Identity }} {{eqn | r = e       | c = by $(5)$ }} {{end-eqn}} Hence every element of $S$ has a inverse. {{qed}}
=== Necessary Condition === Let $\left({S, \circ}\right)$ be a group. $\left({S, \circ}\right)$ is a semigroup by the definition of a group. By Group has Latin Square Property, the Latin square property holds in $S$. {{qed|lemma}} === Sufficient Condition ===  Let $a \in G$. By hypothesis: : $\exists e \in S: a \circ e = a$ and such an $e$ is unique. Let $b \in G$. By hypothesis: : $\exists x \in G: x \circ a = b$ and such an $x$ is unique. Hence: {{begin-eqn}} {{eqn | l = b       | r = x \circ a       | c =  }} {{eqn | r = x \circ a \circ e       | c =  }} {{eqn | r = b \circ e       | c =  }} {{end-eqn}} Also by hypothesis: : $\exists a' \in G: a' \circ a = e$ Thus $\left({S, \circ}\right)$ satisfies the right-hand Group Axioms Thus $\left({S, \circ}\right)$ is a group. {{qed}}
=== Closure === Let $p, q \in n \Z$. Then for some $p', q' \in \Z$: :$p = n p'$ :$q = n q'$ Hence: :$p q = \paren {n p'} \paren {n q'}$ By the commutativity and associativity of integer multiplication: :$p q = n \paren {n \paren {p' q'} }$ Hence $p q \in n \Z$ and $n \Z$ is closed under $\times$. {{qed|lemma}} === Assocativity === By definition $n \Z \subseteq \Z$. Hence $\times$ is associative on $n \Z$ as a direct results of Restriction of Associative Operation is Associative. {{qed|lemma}} === Identity Element === If $\size n = 1$ then $\paren {n \Z, \times}$ is a monoid from Integers under Multiplication form Countably Infinite Commutative Monoid. If $n \ne 1$ then $\left({n \Z, \times}\right)$ is a semigroup but not a monoid as it is without an identity as $1 \notin n \Z$. {{qed}}
By definition the order of $\struct {S \times T, \circ}$ is $\card S \times \card T$ is the cardinality of the underlying set $S \times T$. The result follows directly from Cardinality of Cartesian Product. {{qed}}
Suppose $a \equiv b \bmod m$ and $c \equiv d \bmod m$. Then by the definition of congruence there exists $k, k' \in \Z$ such that: :$\left({a - b}\right) = km$ :$\left({c - d}\right) = k'm$ Hence: :$\left({a - b}\right) + \left({c - d}\right) = km + k'm$ Using the properties of the integers: :$\left({a + c}\right) - \left({b + d}\right) = m \left({k + k'}\right)$ Hence $\left({a + c}\right) \equiv \left({b + d}\right) \bmod m$ and congruence modulo $m$ is a congruence relation. {{qed}}
{{improve|For a neater proof, suggest the One-Step Subgroup Test or Two-Step Subgroup Test be used.}} Taking each of the group axioms in turn: === $\text G 0$: Closure === {{wtd|It is very likely that there are a small set of pages already on here that would make this into a one or two line subproof.}} {{qed|lemma}} === $\text G 1$: Associativity === As $\Q_p \subseteq \Q$ the result follows directly from Rational Addition is Associative and Restriction of Associative Operation is Associative. {{qed|lemma}} === $\text G 2$: Identity === By Integer is Coprime to 1: : $\dfrac 0 1 \in \Q_p$ regardless of our choice of $p$. By the definition of addition on $\Q$: :$\dfrac a b + \dfrac 0 1 = \dfrac a b$ and :$\dfrac 0 1 + \dfrac a b = \dfrac a b$ for all $\dfrac a b \in \Q$. Hence $\dfrac 0 1$ is the identity. {{qed|lemma}} === $\text G 3$: Inverses === For $\dfrac a b$ we have that $\dfrac {-a} b$ is the inverse of $\dfrac a b$. As it has the same denominator as $\dfrac a b$ we have that $\dfrac {-a} b \in \Q_p$ as well. {{qed}}
=== Definition from Triangle implies Definition from Circle === Let $\cot \theta$ be defined as $\dfrac {\text{Adjacent}} {\text{Opposite}}$ in a right triangle. Consider the triangle $\triangle OAB$. By construction, $\angle OAB$ is a right angle. From Parallelism implies Equal Alternate Angles: :$\angle OBA = \theta$ Thus: {{begin-eqn}} {{eqn | l = \cot \theta       | r = \frac {AB} {OA}       | c =  }} {{eqn | r = \frac {AB} 1       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = AB       | c =  }} {{end-eqn}} That is: :$\cot \theta = AB$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\cot \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$. Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above. From Parallelism implies Equal Alternate Angles: :$\angle OBA = \theta$ We have that: :$\angle CAB = \angle OBA = \theta$ :$\angle ABC = \angle OAB$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text{Adjacent} } {\text{Opposite} }       | r = \frac {AB} {BC}       | c = by definition }} {{eqn | r = \frac {AB} {OA}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = AB       | c = $OA$ is Radius of the Unit Circle }} {{eqn | r = \cot \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text{Adjacent} } {\text{Opposite} } = \cot \theta$ {{qed}} Category:Cotangent Function q0f9zl83ur4pv8c76gkdkonq97atwhl
=== Definition from Triangle implies Definition from Circle === Let $\sec \theta$ be defined as $\dfrac {\text{Hypotenuse}} {\text{Adjacent}}$ in a right triangle. Consider the triangle $\triangle OAB$. By construction, $\angle OAB$ is a right angle. Thus: {{begin-eqn}} {{eqn | l = \sec \theta       | r = \frac {OB} {OA}       | c =  }} {{eqn | r = \frac {OB} 1       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = OB       | c =  }} {{end-eqn}} That is: : $\sec \theta = OB$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\sec \theta$ be defined as the length of $OB$ in the triangle $\triangle OAB$. Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above. We have that: :$\angle CAB = \angle BOA = \theta$ :$\angle ABC = \angle OAB$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text{Hypotenuse} } {\text{Adjacent} }       | r = \frac {AC} {AB}       | c = by definition }} {{eqn | r = \frac {OB} {OA}       | c = by definition of similarity }} {{eqn | r = OB       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = \sec \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text{Hypotenuse} } {\text{Adjacent} } = \sec \theta$ {{qed}} Category:Secant Function eqyy632t30dy4add1ddu3fqsgk4at16
=== Definition from Triangle implies Definition from Circle === Let $\csc \theta$ be defined as $\dfrac {\text{Hypotenuse}} {\text{Opposite}}$ in a right triangle. Consider the triangle $\triangle OAB$. By construction, $\angle OAB$ is a right angle. From Parallelism implies Equal Alternate Angles: :$\angle OBA = \theta$ Thus: {{begin-eqn}} {{eqn | l = \csc \theta       | r = \frac {OB} {OA}       | c =  }} {{eqn | r = \frac {OB} 1       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = OB       | c =  }} {{end-eqn}} That is: :$\csc \theta = OB$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\csc \theta$ be defined as the length of $OB$ in the triangle $\triangle OAB$. Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above. From Parallelism implies Equal Alternate Angles: :$\angle OBA = \theta$ We have that: :$\angle CAB = \angle OBA = \theta$ :$\angle ABC = \angle OAB$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text{Hypotenuse} } {\text{Opposite} }       | r = \frac {AC} {BC}       | c = by definition }} {{eqn | r = \frac {OB} {OA}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = OB       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = \csc \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text{Hypotenuse} } {\text{Opposite} } = \csc \theta$ {{qed}} Category:Cosecant Function hp6pd5dh8ygsuwwfoys0c6htkxdlnfm
:500px Let a unit circle $C$ be drawn with its center at the origin $O$. Let $Q$ be the point on $C$ which intersects $OP$. From Parallelism implies Equal Alternate Angles, $\angle OQR = \theta$. Thus: :$(1): \quad \cos \theta = RQ$ by definition of cosine $\angle OSP = \angle ORQ$, as both are right angles. Both $\triangle OSP$ and $\triangle ORQ$ share angle $\angle SOP$. By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar. By definition of similarity: Then: {{begin-eqn}} {{eqn | l = \frac x r       | r = \frac {SP} {OP}       | c =  }} {{eqn | r = \frac {RQ} {OQ}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = RQ       | c = as $OP$ is the radius of the unit circle }} {{eqn | r = \cos \theta       | c = from $(1)$ above }} {{end-eqn}} {{qed}} :500px When $\theta$ is obtuse, the same argument holds, except that this time both $x$ and $\cos \theta$ are negative. When $\theta = \dfrac \pi 2$ we have that $x = 0$ and $\cos \theta = 0 = \dfrac x r$. Thus $y = r$ and $\sin \theta = 1 \dfrac y r$. Thus the relation holds for $\theta = \dfrac \pi 2$. When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis. Thus the relation continues to hold. When $\theta = 0$ we have that $y = 0$ and $\cos \theta = 1 = \dfrac x r$. When $\theta = \pi$ we have that $y = 0$ and $x = -r$, while $\cos \theta = -1 = \dfrac x r$. Hence the result. {{qed}}
:500px Let a unit circle $C$ be drawn with its center at the origin $O$. Let a tangent line be drawn to $C$ parallel to $PS$ meeting $C$ at $R$. Let $Q$ be the point on $OP$ which intersects this tangent line. $\angle OSP = \angle ORQ$, as both are right angles. Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$. By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar. Thus: Then: {{begin-eqn}} {{eqn | l = \frac y x       | r = \frac {SP} {OS}       | c =  }} {{eqn | r = \frac {RQ} {OR}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = RQ       | c = $OP$ is Radius of the Unit Circle }} {{eqn | r = \tan \theta       | c = {{Defof|Tangent Function|subdef = Definition from Circle}} }} {{end-eqn}} When $\theta$ is obtuse, the same argument holds, but both $x$ and $\tan \theta$ are negative. When $\theta = \dfrac \pi 2$ we have that $x = 0$. Then $OP$ is parallel to the tangent line at $R$ which it therefore does not meet. Thus when $\theta = \dfrac \pi 2$, it follows that $\tan \theta$ is not defined. Likewise $\dfrac y x$ is not defined when $x = 0$. Thus the relation holds for $\theta = \dfrac \pi 2$. When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis. In this case, $y$ is negative. Thus the relation continues to hold. When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\tan \theta = 0 = \dfrac y x$. Hence the result. {{qed}}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis. Then: {{begin-eqn}} {{eqn | l = \frac {\sin \theta} {\cos \theta}       | r = \frac {y / r} {x / r}       | c = Sine of Angle in Cartesian Plane and Cosine of Angle in Cartesian Plane }} {{eqn | r = \frac y r \frac r x       | c =  }} {{eqn | r = \frac y x       | c =  }} {{eqn | r = \tan \theta       | c = Tangent of Angle in Cartesian Plane }} {{end-eqn}} When $\cos \theta = 0$ the expression $\dfrac {\sin \theta} {\cos \theta}$ is not defined. {{qed}}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis. Then: {{begin-eqn}} {{eqn | l = \frac {\cos \theta} {\sin \theta}       | r = \frac {x / r} {y / r}       | c = Cosine of Angle in Cartesian Plane and Sine of Angle in Cartesian Plane }} {{eqn | r = \frac x r \frac r y       | c =  }} {{eqn | r = \frac x y       | c =  }} {{eqn | r = \cot \theta       | c = Cotangent of Angle in Cartesian Plane }} {{end-eqn}} When $\sin \theta = 0$ the expression $\dfrac {\cos \theta} {\sin \theta}$ is not defined. {{qed}}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis. Then: {{begin-eqn}} {{eqn | l = \cot \theta       | r = \frac x y       | c = Cotangent of Angle in Cartesian Plane }} {{eqn | r = \frac 1 {y / x}       | c =  }} {{eqn | r = \frac 1 {\tan \theta}       | c = Tangent of Angle in Cartesian Plane }} {{end-eqn}} $\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$. {{qed}}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis. Then: {{begin-eqn}} {{eqn | l = \sec \theta       | r = \frac r x       | c = Secant of Angle in Cartesian Plane }} {{eqn | r = \frac 1 {x / r}       | c =  }} {{eqn | r = \frac 1 {\cos \theta}       | c = Cosine of Angle in Cartesian Plane }} {{end-eqn}} When $\cos \theta = 0$, $\dfrac 1 {\cos \theta}$ is not defined. {{qed}}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis. Then: {{begin-eqn}} {{eqn | l = \csc \theta       | r = \frac r y       | c = Cosecant of Angle in Cartesian Plane }} {{eqn | r = \frac 1 {y / r}       | c =  }} {{eqn | r = \frac 1 {\sin \theta}       | c = Sine of Angle in Cartesian Plane }} {{end-eqn}} When $\sin \theta = 0$, $\dfrac 1 {\sin \theta}$ is not defined. {{qed}}
{{begin-eqn}} {{eqn | l = \cos \theta       | r = 1 - 2 \ \sin^2 \frac \theta 2       | c = Double Angle Formula for Cosine: Corollary 2 }} {{eqn | ll= \leadsto       | l = 2 \ \sin^2 \frac \theta 2       | r = 1 - \cos \theta }} {{eqn | ll= \leadsto       | l = \sin \frac \theta 2       | r = \pm \sqrt {\frac {1 - \cos \theta} 2} }} {{end-eqn}} We also have that: :In quadrant I and quadrant II, $\sin \theta > 0$ :In quadrant III and quadrant IV, $\sin \theta < 0$. {{qed}}
Define: :$u = \dfrac \theta 2$ Then: {{begin-eqn}} {{eqn | l = \sin^2 u       | r = \frac {1 - \cos 2 u} 2       | c = Power Reduction Formulas }} {{eqn | ll= \leadsto       | l = \sin \frac \theta 2       | r = \pm \sqrt {\frac {1 - \cos \theta} 2} }} {{end-eqn}} We also have that: :In quadrant I, and quadrant II, $\sin \theta > 0$ :In quadrant III, and quadrant IV, $\sin \theta < 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \cos \theta       | r = 2 \cos^2 \frac \theta 2 - 1       | c = Double Angle Formula for Cosine: Corollary 1 }} {{eqn | ll= \leadsto       | l = 2 \cos^2 \frac \theta 2       | r = 1 + \cos \theta }} {{eqn | ll= \leadsto       | l = \cos \frac \theta 2       | r = \pm \sqrt {\frac {1 + \cos \theta} 2} }} {{end-eqn}} We also have that: :In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$ :In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$. {{qed}}
Define: :$u = \dfrac \theta 2$ Then: {{begin-eqn}} {{eqn | l = \cos^2 u       | r =  \frac {1 + \cos 2 u} 2       | c = Power Reduction Formulas }} {{eqn | ll= \leadsto       | l = \cos \frac \theta 2       | r = \pm \sqrt {\frac {1 + \cos \theta} 2} }} {{end-eqn}} We also have that: :In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$ :In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | r = \frac {\sin \frac \theta 2} {\cos \frac \theta 2}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\pm \sqrt {\frac {1 - \cos \theta} 2} } {\pm \sqrt {\frac {1 + \cos \theta} 2} }       | c = Half Angle Formula for Sine and Half Angle Formula for Cosine }} {{eqn | r = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} } }} {{end-eqn}} Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$. When $\cos \theta = -1$ it follows that $\cos \theta + 1 = 0$ and then $\tan \dfrac \theta 2$ is undefined. This happens when $\theta = \paren {2 k + 1} \pi$. We have that: :$\tan \dfrac \theta 2 = \dfrac {\sin \dfrac \theta 2} {\cos \dfrac \theta 2}$ === Quadrant $\text I$ === In quadrant $\text I$, we have: :Sine in First Quadrant: $\sin \dfrac \theta 2 > 0$ :Cosine in first Quadrant: $\cos \dfrac \theta 2 > 0$ and so in quadrant $\text I$: {{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \tan \frac \theta 2       | r = +\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} } }} {{end-eqn}} === Quadrant $\text {II}$ === In quadrant $\text {II}$, we have: :Sine in Second Quadrant: $\sin \dfrac \theta 2 > 0$ :Cosine in Second Quadrant: $\cos \dfrac \theta 2 < 0$ and so in quadrant $\text {II}$: {{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = \tan \frac \theta 2       | r = -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} } }} {{end-eqn}} === Quadrant $\text {III}$ === In quadrant $\text {III}$, we have: :Sine in Third Quadrant: $\sin \dfrac \theta 2 < 0$ :Cosine in Third Quadrant: $\cos \dfrac \theta 2 < 0$ and so in quadrant $\text {III}$: {{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \tan \frac \theta 2       | r = + \sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} } }} {{end-eqn}} === Quadrant $\text {IV}$ === In quadrant $\text {IV}$, we have: :Sine in Fourth Quadrant: $\sin \dfrac \theta 2 < 0$ :Cosine in Fourth Quadrant: $\cos \dfrac \theta 2 > 0$ and so in quadrant $\text {IV}$: {{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = \tan \frac \theta 2       | r = -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} } }} {{end-eqn}} {{qed}}
Let $x \in H$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \ker A       | c =  }} {{eqn | ll= \leadstoandfrom       | l = A x       | r = \mathbf 0_H       | c = {{Defof|Kernel of Linear Transformation}} }} {{eqn | ll= \leadstoandfrom       | l = \gen {A x, A x}       | r = 0       | c = {{Defof|Inner Product}} }} {{eqn | ll= \leadstoandfrom       | l = \gen {A^*Ax, x}       | r = 0       | c = {{Defof|Adjoint Linear Transformation}} }} {{eqn | ll= \leadstoandfrom       | l = \gen {A A^* x, x}       | r = 0       | c = {{Defof|Normal Operator}} }} {{eqn | ll= \leadstoandfrom       | l = \gen {x, AA^*x}       | r = 0       | c = {{Defof|Inner Product}} }} {{eqn | ll= \leadstoandfrom       | l = \gen {A^*x, A^* x}       | r = 0       | c = {{Defof|Adjoint Linear Transformation}} }} {{eqn | ll= \leadstoandfrom       | l = A^*x       | r = \mathbf 0_H       | c = {{Defof|Inner Product}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \ker A^*       | c = {{Defof|Kernel of Linear Transformation}} }} {{end-eqn}} Hence, by definition of set equality: :$\ker A = \ker A^*$ {{qed}} Category:Adjoints Category:Linear Transformations on Hilbert Spaces 8nnktytpzh3yrk7r6howhqwdvwzeglx
{{begin-eqn}} {{eqn | l = \tan \frac \theta 2       | r = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }       | c = Half Angle Formula for Tangent }} {{eqn | r = \pm \sqrt {\frac {\paren {1 - \cos \theta} \paren {1 + \cos \theta} } {\paren {1 + \cos \theta}^2} }       | c = multiplying top and bottom by $\sqrt {1 + \cos \theta}$ }} {{eqn | r = \pm \sqrt {\frac {1 - \cos^2 \theta} {\paren {1 + \cos \theta}^2} }       | c = Difference of Two Squares }} {{eqn | r = \pm \sqrt {\frac {\sin^2 \theta} {\paren {1 + \cos \theta}^2} }       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \pm \frac {\sin \theta} {1 + \cos \theta} }} {{end-eqn}} Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$. When $\cos \theta = -1$ it follows that $1 + \cos \theta = 0$. This happens when $\theta = \paren {2 k + 1} \pi$, for $k \in \Z$. At these points, $\sin \theta = 0$ as well. Then: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\sin \theta} {1 + \cos \theta}       | o = =       | r = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\dfrac \d {\d \theta} \sin \theta} {\map {\dfrac \d {\d \theta} } {1 + \cos \theta} }       | c = L'Hôpital's Rule }} {{eqn | r = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\cos \theta} {\sin \theta}       | c =  }} {{eqn | r = \frac 1 0       | c =  }} {{end-eqn}} So it follows that at $\theta = \paren {2 k + 1} \pi$, $\dfrac {\sin \theta} {1 + \cos \theta}$ is undefined. {{qed|lemma}} At all other values of $\theta$, $\cos \theta + 1 > 0$. Therefore the sign of $\dfrac {\sin \theta} {1 + \cos \theta}$ is equal to the sign of $\sin \theta$. We recall: :In quadrant $\text I$ and quadrant $\text {II}$: $\sin \theta > 0$ :In quadrant $\text {III}$ and quadrant $\text {IV}$: $\sin \theta < 0$ Thus it follows that the same applies to $\dfrac {\sin \theta} {1 + \cos \theta}$. Let $\dfrac \theta 2$ be in quadrant I or quadrant $\text {III}$. Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text I$ or quadrant $\text {II}$. Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is positive. Let $\dfrac \theta 2$ be in quadrant $\text {II}$ or quadrant $\text {IV}$. Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text {III}$ or quadrant $\text {IV}$. Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is negative. {{qed}}