hcju/proofwikiqa huggingface dataset:
Let $\triangle ABC$ be a triangle. === Case 1: $AC$ greater than $AB$ === Using $AC$ as the radius, we construct a circle whose center is $A$. Now we extend: : $CB$ to $D$ : $AB$ to $F$ : $BA$ to $G$ : $CA$ to $E$. $D$ is joined with $E$, thus: :300px Using the Intersecting Chord Theorem we have: : $GB \cdot BF = CB \cdot BD$ $AF$ is a radius, so $AF = AC = b = GA$ and thus: : $GB = GA + AB = b + c$ : $BF = AF - AB = b - c$ Thus: {{begin-eqn}} {{eqn | l = \left({b + c}\right) \left({b - c}\right)       | r = a \cdot BD }} {{eqn | ll= \implies       | l = \frac {b^2 - c^2} a        | r = BD }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = CD       | r = CB + BD }} {{eqn | r = a + \frac {b^2 - c^2} a }} {{eqn | r = \frac {a^2 + b^2 - c^2} a }} {{end-eqn}} As $CA$ is a radius, $CE$ is a diameter. By Thales' Theorem, it follows that $\angle CDE$ is a right angle. Then using the definition of cosine, we have {{begin-eqn}} {{eqn | l = \cos C        | r = \frac {CD} {CE} }} {{eqn | r = \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2 b} }} {{eqn | r = \frac {a^2 + b^2 - c^2} {2 a b} }} {{eqn | ll= \implies       | l = c^2       | r = a^2 + b^2 - 2 a b \cos C }} {{end-eqn}} {{qed|lemma}} === Case 2: $AC$ less than $AB$ === When $AC$ is less than $AB$, the point $B$ lies outside the circle and so the diagram needs to be modified accordingly: :300px {{expand|Draw a diagram for the case where $\angle ACB$ is a right angle and where it is a convex angle to show that the formula will be the same.}} Now we extend: : $BA$ to $G$ : $CA$ to $E$. Then we construct: :$D$ as the point at which $CB$ intersects the circle :$F$ as the point at which $AB$ intersects the circle. Finally $D$ is joined to $E$. Using the Secant Secant Theorem we have: : $GB \cdot BF = CB \cdot BD$ $AF$ is a radius, so $AF = AC = b = GA$ and thus: : $GB = GA + AB = b + c$ : $BF = AB - AF = b - c$ Thus: {{begin-eqn}} {{eqn | l = \left({b + c}\right) \left({b - c}\right)       | r = CB \cdot BD       | c = Secant Secant Theorem }} {{eqn | l = \left({b + c}\right) \left({b - c}\right)       | r = a \cdot BD }} {{eqn | ll= \implies       | l = \frac {b^2 - c^2} a        | r = BD }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = CD       | r = CB - BD }} {{eqn | r = a - \frac {b^2 - c^2} a }} {{eqn | r = \frac {a^2 - b^2 + c^2} a }} {{end-eqn}} As $CA$ is a radius, $CE$ is a diameter. By Thales' Theorem, it follows that $\angle CDE$ is a right angle. Then using the definition of cosine, we have {{begin-eqn}} {{eqn | l = \cos C        | r = \frac {CD} {CE} }} {{eqn | r = \frac {\left({\dfrac {a^2 - b^2 + c^2} a}\right)} {2 b} }} {{eqn | r = \frac {a^2 - b^2 + c^2} {2 a b}       | c =  }} {{eqn | ll= \implies       | l = c^2       | r = a^2 + b^2 - 2 a b \cos C }} {{end-eqn}} {{qed|lemma}} === Case 3: $AC = AB$ === When $AC = AB$ the points $B$, $D$ and $F$ coincide on the circumference of the circle: :300px We extend: : $BA$ to $G$ : $CA$ to $E$ and immediately: : $GB = CB$ {{finish}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin c \sin a \cos B       | r = \cos b - \cos c \cos a       | c = Spherical Law of Cosines }} {{eqn | r = \cos b - \cos c \paren {\cos b \cos c + \sin b \sin c \cos A}       | c = Spherical Law of Cosines }} {{eqn | r = \cos b \paren {1 - \cos^2 c} - \sin b \sin c \cos c \cos A       | c = rearranging }} {{eqn | r = \sin^2 c \cos b - \sin b \sin c \cos c \cos A       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \sin a \cos B       | r = \sin c \cos b - \sin b \cos c \cos A       | c = simplifying }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sin a \sin b \cos C       | r = \cos c - \cos a \cos b       | c = Spherical Law of Cosines }} {{eqn | r = \cos c - \cos b \paren {\cos b \cos c + \sin b \sin c \cos A}       | c = Spherical Law of Cosines }} {{eqn | r = \cos c \paren {1 - \cos^2 b} - \sin b \sin c \cos b \cos A       | c = rearranging }} {{eqn | r = \sin^2 b \cos c - \sin b \sin c \cos b \cos A       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \sin a \cos C       | r = \cos b \sin c - \sin c \cos b \cos A       | c = simplifying }} {{end-eqn}} {{qed}}
:500px Suppose $c$ is less than $\dfrac \pi 2$. Let $BA$ be produced to $D$ so that $BD = \dfrac \pi 2$. Then: :$AD = \dfrac \pi 2 - c$ and: :$\angle CAD = pi - A$ Let $C$ and $D$ be joined by an arc of a great circle, denoted $x$. From the triangle $\sphericalangle DAC$, using the Spherical Law of Cosines: {{begin-eqn}} {{eqn | l = \cos x       | r = \map \cos {\dfrac \pi 2 - c} \cos b + \map \sin {\dfrac \pi 2 - c} \sin b \, \map \cos {\pi - A}       | c =  }} {{eqn | r = \sin c \cos b - \cos c \sin b \cos A       | c =  }} {{end-eqn}} From the triangle $\sphericalangle DBC$, using the Spherical Law of Cosines: {{begin-eqn}} {{eqn | l = \cos x       | r = \cos \dfrac \pi 2 \cos a + \sin \pi 2 \sin a \cos B       | c =  }} {{eqn | r = \sin a \cos B       | c =  }} {{end-eqn}} Hence: :$\sin a \cos B = \sin c \cos b - \cos c \sin b \cos A$ The case where $c > \dfrac \pi 2$ is worked similarly, but by making $D$ the point between $A$ and $B$ such that $BD$ is $\dfrac \pi 2$. {{qed}}
:500px Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$. By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$. By definition of a great circle, the center of each of these great circles is $O$. Let $AD$ be the tangent to the great circle $AB$. Let $AE$ be the tangent to the great circle $AC$. Thus the radius $OA$ of $S$ is perpendicular to $AD$ and $AE$. By construction, $AD$ lies in the same plane as $AB$. Thus when $OB$ is produced, it will intersect $AD$ at $D$, say. Similarly, $OC$ can be produced to intersect $AE$ at $E$, say. The spherical angle $\sphericalangle BAC$ is defined as the angle between the tangents $AD$ and $AE$. Thus: :$\sphericalangle BAC = \angle DAE$ or, denoting that spherical angle $\sphericalangle BAC$ as $A$: :$A = \angle DAE$ In the (plane) triangle $OAD$, we have that $\angle OAD$ is a right angle. We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the length of a side of a spherical triangle. Thus: {{begin-eqn}} {{eqn | l = AD       | r = OA \tan c       | c =  }} {{eqn | l = OD       | r = OA \sec c       | c =  }} {{end-eqn}} and by similar analysis of $\triangle OAE$, we have: {{begin-eqn}} {{eqn | l = AE       | r = OA \tan b       | c =  }} {{eqn | l = OE       | r = OA \sec b       | c =  }} {{end-eqn}} From consideration of $\triangle DAE$: {{begin-eqn}} {{eqn | l = DE^2       | r = AD^2 + AE^2 - 2 AD \cdot AE \cos \angle DAE       | c = Law of Cosines }} {{eqn | n = 1       | r = OA^2 \paren {\tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A}       | c =  }} {{end-eqn}} From consideration of $\triangle DOE$: {{begin-eqn}} {{eqn | l = DE^2       | r = OD^2 + OE^2 - 2 OD \cdot OE \cos \angle DOE       | c = Law of Cosines }} {{eqn | n = 2       | r = OA^2 \paren {\sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a}       | c = as $\angle DOE = \angle BOC$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a       | r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A       | c = from $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = \paren {1 + \tan^2 c} + \paren {1 + \tan^2 b} - 2 \sec b \sec c \cos a       | r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = 1 - \sec b \sec c \cos a       | r = \tan b \tan c \cos A       | c = simplifying }} {{eqn | ll= \leadsto       | l = \cos b \cos c - \cos a       | r = \sin b \sin c \cos A       | c = multiplying both sides by $\cos b \cos c$ }} {{end-eqn}} and the result follows. {{qed}}
:500px Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$. By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$. By definition of a great circle, the center of each of these great circles is $O$. Let $O$ be joined to each of $A$, $B$ and $C$. Let $P$ be an arbitrary point on $OC$. Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$. Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$. In the plane $OAB$: :construct $QS$ perpendicular to $OA$ :construct $RS$ perpendicular to $OB$ where $S$ is the point where $QS$ and $RS$ intersect. Let $OS$ and $PS$ be joined. Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$. These tangents contain the spherical angle $A$. But by construction, $QS$ and $QP$ are parallel to these tangents Hence $\angle PQS = \sphericalangle A$. Similarly, $\angle PRS = \sphericalangle B$. Also we have: {{begin-eqn}} {{eqn | l = \angle COB       | r = a }} {{eqn | l = \angle COA       | r = b }} {{eqn | l = \angle AOB       | r = c }} {{end-eqn}} It is to be proved that $PS$ is perpendicular to the plane $AOB$. By construction, $OQ$ is perpendicular to both $PQ$ and $QS$. Thus $OQ$ is perpendicular to the plane $PQS$. Similarly, $OR$ is perpendicular to the plane $PRS$. Thus $PS$ is perpendicular to both $OQ$ and $OR$. Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$. That is, $PS$ is perpendicular to the plane $OAB$. In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$ It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles. From the right triangles $\triangle OQP$ and $\triangle ORP$, we have: {{begin-eqn}} {{eqn | n = 1       | l = PQ       | r = OP \sin b }} {{eqn | n = 2       | l = PR       | r = OP \sin a }} {{eqn | n = 3       | l = OQ       | r = OP \cos b }} {{eqn | n = 4       | l = OR       | r = OP \cos a }} {{end-eqn}} Let us denote the angle $\angle SOQ$ by $x$. Then: :$\angle ROS = c - x$ We have that: {{begin-eqn}} {{eqn | l = OS       | r = OQ \sec x }} {{eqn | l = OS       | r = OR \, \map \sec {c - x} }} {{eqn | ll= \leadsto       | l = OR \cos x       | r = OQ \, \map \cos {c - x}       | c =  }} {{eqn | ll= \leadsto       | l = OP \cos a \cos x       | r = OP \cos b \, \map \cos {c - x}       | c = from $(3)$ and $(4)$ }} {{eqn | ll= \leadsto       | l = \cos a \cos x       | r = \cos b \paren {\cos c \cos x - \sin c \sin x}       | c = Cosine of Difference }} {{eqn | n = 5       | ll= \leadsto       | l = \cos a       | r = \cos b \cos c + \cos b \sin c \tan x       | c = dividing both sides by $\cos x$ and multiplying out }} {{end-eqn}} But we also have: {{begin-eqn}} {{eqn | l = \tan x       | r = \dfrac {QS} {OQ}       | c =  }} {{eqn | r = \dfrac {PQ \cos A} {OQ}       | c =  }} {{eqn | r = \tan b \cos A       | c =  }} {{eqn | ll= \leadsto       | l = \cos a       | r = \cos b \cos c + \cos b \sin c \tan b \cos A       | c = substituting for $\tan x$ from $(5)$ }} {{eqn | r = \cos b \cos c + \sin b \sin c \cos A       | c =  }} {{end-eqn}} Hence the result. {{qed}}
As Sine Function is Absolutely Convergent and Cosine Function is Absolutely Convergent, we have: {{begin-eqn}} {{eqn | l = \cos z + i \sin z       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n + 1} } {\paren {2 n + 1}!}       | c = {{Defof|Complex Cosine Function}} and {{Defof|Complex Sine Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {z^{2 n + 1} } {\paren {2 n + 1}!} }       | c = Sum of Absolutely Convergent Series }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i z}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = {{Defof|Imaginary Unit}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {\paren {i z}^n} {n!} }} {{eqn | r = e^{i z}       | c = {{Defof|Exponential Function/Complex|subdef = Sum of Series|Complex Exponential Function}} }} {{end-eqn}} {{qed}}
Let $G$ be a group. Let $H$ be a subgroup of $G$. From Cosets are Equivalent, a left coset $y H$ has the same number of elements as $H$, namely $\order H$. Since left cosets are identical or disjoint, each element of $G$ belongs to exactly one left coset. From the definition of index of subgroup, there are $\index G H$ left cosets, and therefore: :$\order G = \index G H \order H$ Let $G$ be of finite order. All three numbers are finite, and the result follows. Now let $G$ be of infinite order. If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order. If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite. {{qed}}
Follows directly from the Orbit-Stabilizer Theorem applied to Group Action on Coset Space. {{Qed}}
Let $G$ be finite. Consider the mapping $\phi: G \to G / H^l$, defined as: :$\phi: G \to G / H^l: \map \phi x = x H^l$ where $G / H^l$ is the left coset space of $G$ modulo $H$. For every $y H \in G / H^l$, there exists a corresponding $y \in G$, so $\phi$ is a surjection. From Cardinality of Surjection it follows that $G / H^l$ is finite. From Cosets are Equivalent, $G / H^l$ has the same number of elements as $H$. We have that the $G / H^l$ is a partition of $G$. It follows from Number of Elements in Partition that $\index G H = \dfrac {\order G} {\order H}$ {{Qed}}
=== Necessary Condition === Suppose $f$ is differentiable at $c$. Then by definition $f' \left({c}\right)$ exists. So we can define $\varphi$ by: :$\varphi \left({x}\right) = \begin{cases} \dfrac {f \left({x}\right) - f \left({c}\right)} {x - c} & : x \ne c, x \in I \\ f' \left({c}\right) & : x = c \end{cases}$ Condition $(2)$, that $\varphi$ is continuous at $c$, is satisfied, since: {{begin-eqn}} {{eqn | l = \varphi \left({c}\right)       | r = f' \left({c}\right)       | c = Definition of $\varphi$ }} {{eqn | r = \lim_{x \to c} \frac {f \left({x}\right) - f \left({c}\right)} {x - c}       | c = Definition of derivative }} {{eqn | r = \lim_{x \to c} \varphi \left({x}\right)       | c = Definition of $\varphi$ }} {{end-eqn}} Finally, condition $(1)$ is vacuous for $x = c$. For $x \ne c$, it follows from the definition of $\varphi$ by dividing both sides of $(1)$ by $x - c$. {{qed|lemma}} === Sufficient Condition === Suppose a $\varphi$ as in the theorem statement exists. Then for $x \ne c$, we have that: :$\varphi \left({x}\right) = \dfrac {f \left({x}\right) - f \left({c}\right)} {x - c}$ Since $\varphi$ is continuous at $c$: :$\displaystyle \varphi \left({c}\right) = \lim_{x \to c} \varphi \left({x}\right) = \lim_{x \to c} \frac {f \left({x}\right) - f \left({c}\right)} {x - c}$ That is, $f$ is differentiable at $c$, and $f' \left({c}\right) = \varphi \left({c}\right)$. {{qed}} {{Namedfor|Constantin Carathéodory|cat = Carathéodory}} Category:Direct Proofs Category:Real Analysis Category:Differential Calculus kuvgsqeze4423l7w2tclrn6y4dqellz
=== Necessary Condition === Let $H$ be a subset of $G$ that fulfils the conditions given. It is noted that the fact that $H$ is non-empty is one of the conditions. It is also noted that the group operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$. So it remains to show that $\struct {H, \circ}$ is a group. We check the four group axioms: '''$\text G 1$: Associativity''': From Subset Product within Semigroup is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$. '''$\text G 2$: Identity''' Let $e$ be the identity of $\struct {G, \circ}$. Since $H$ is non-empty, $\exists x \in H$. If we take $a = x$ and $b = x$, then $a \circ b^{-1} = x \circ x^{-1} = e \in H$, where $e$ is the identity element. '''$\text G 3$: Inverses''' If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$. Thus every element of $H$ has an inverse also in $H$. '''$\text G 0$: Closure''' Let $x, y \in H$. Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$. So: :$a \circ b^{-1} = x \circ \paren {y^{-1} }^{-1} = x \circ y \in H$ Thus, $H$ is closed. Therefore, $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group. Therefore $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$. {{qed|lemma}} === Sufficient Condition === Now suppose $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$. : $(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a group and therefore can not be empty. : $(2): \quad$ As $\struct {H, \circ}$ is a group, it is closed and every element has an inverse. So it follows that $\forall a, b \in H: a \circ b^{-1} \in H$. {{qed}}
=== Necessary Condition === Let $H$ be a subset of $G$ that fulfils the conditions given. It is noted that the fact that $H$ is nonempty is one of the conditions. It is also noted that the group operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$. So it remains to show that $\struct {H, \circ}$ is a group. We check the four group axioms: ==== $\text G 0$: Closure ==== The closure condition is given by condition $(2)$. {{qed|lemma}} ==== $\text G 1$: Associativity ==== From Closed Substructure of Semigroup is Semigroup, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$. {{qed|lemma}} ==== $\text G 2$: Identity ==== Let $e$ be the identity of $\struct {G, \circ}$. From condition $(1)$, $H$ is non-empty. Therefore $\exists x \in H$. From condition $(3)$, $\struct {H, \circ}$ is closed under inversion. Therefore $x^{-1} \in H$. Since $\struct {H, \circ}$ is closed under $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$. {{qed|lemma}} ==== $\text G 3$: Inverses ==== From condition $(3)$, every element of $H$ has an inverse. {{qed|lemma}} So $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group. So by definition $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$. {{qed|lemma}} === Sufficient Condition === Now suppose $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$. :$(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a group and therefore can not be empty. :$(2): \quad a, b \in H \implies a \circ b \in H$ follows from {{GroupAxiom|0}} as applied to the group $\struct {H, \circ}$. :$(3): \quad a \in H \implies a^{-1} \in H$ follows from {{GroupAxiom|3}} as applied to the group $\struct {H, \circ}$. {{qed}}
In Integer is Expressible as Product of Primes it is proved that every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes. In Prime Decomposition of Integer is Unique, it is proved that this prime decomposition is unique up to the order of the factors. {{qed}}
We have that the integers form a Euclidean domain. Then from Irreducible Elements of Ring of Integers we have that the irreducible elements of $\Z$ are the primes and their negatives. The result then follows directly from Euclid's Lemma for Irreducible Elements. {{qed}}
Let $p \divides a b$. Suppose $p \nmid a$. Then from the definition of prime: :$p \perp a$ where $\perp$ indicates that $p$ and $a$ are coprime. Thus from Euclid's Lemma it follows that: :$p \divides b$ Similarly, if $p \nmid b$ it follows that $p \divides a$. So: :$p \divides a b \implies p \divides a$ or $p \divides b$ as we needed to show. {{qed}}
Let $a, b, c \in \Z$. We have that $a \perp b$. That is: :$\gcd \set {a, b} = 1$ where $\gcd$ denotes greatest common divisor. From Bézout's Lemma, we may write: :$a x + b y = 1$ for some $x, y \in \Z$. Upon multiplication by $c$, we see that: :$c = c \paren {a x + b y} = c a x + c b y$ Now note that $c a x + c b y$ is an integer combination of $a c$ and $b c$. So, since: :$a \divides a c$ and: :$a \divides b c$ it follows from Common Divisor Divides Integer Combination that: :$a \divides \paren {c a x + c b y}$ However: : $c a x + c b y = c \paren {a x + b y} = c \cdot 1 = c$ Therefore: : $a \divides c$ {{qed}}
Let $p \divides a b$. Suppose $p \nmid a$. Then by {{EuclidPropLink|book = VII|prop = 29|title = Prime not Divisor implies Coprime}}: :$p \perp a$ As $p \divides a b$, it follows by definition of divisor: :$\exists e \in \Z: e p = a b$ So by {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}: :$p : a = b : e$ But as $p \perp a$, by: :{{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}} and:  :{{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}} it follows that: :$p \divides b$ Similarly, if $p \perp b$ then $p \divides a$. Hence the result. {{qed}}
Consider the sequence of integers $n, 2n, 3n, \dots, (p-1)n$. Note that none of these integers are congruent modulo $p$ to the others. If this were the case, we would have $a n \equiv b n \pmod p$ for some $1 \le a < b \le p-1$. Then as $\gcd \left\{{n, p}\right\} = 1$, and we can cancel the $n$, we get $a \equiv b \pmod p$ and so $a = b$. Also, since $p \nmid n$ and $p \nmid c$, for any $1 \le c \le p-1$, then by Euclid's Lemma $p \nmid c n$ for any such $c n$, which means $c n \not \equiv 0 \pmod p$. Thus, each integer in the sequence can be reduced modulo $p$ to exactly one of $1, 2, 3, \ldots, p-1$. So $\left\{{1, 2, 3, \ldots, p-1}\right\}$ is the set of Reduced Residue System modulo $p$. So, upon taking the product of these congruences, we see that $n \times 2n \times 3n \times \dots \times (p-1)n \equiv 1 \times 2 \times 3 \times \cdots \times (p-1) \pmod p$. This simplifies to $n^{p-1} \times (p-1)! \equiv (p-1)! \pmod p$. Since $p \nmid (p-1)!$, we can cancel $(p-1)!$ from both sides, leaving us with $n^{p-1}\equiv 1 \pmod p$. {{qed}}
By Prime not Divisor implies Coprime: : $p \nmid n \implies p \perp n$ and Euler's Theorem can be applied. Thus: : $n^{\map \phi p} \equiv 1 \pmod p$ But from Euler Phi Function of Prime Power: : $\map \phi p = p \paren {1 - \dfrac 1 p} = p - 1$ and the result follows. {{qed}}
Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$. From the corollary to Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_p, \times}$ forms a group of order $p - 1$ under modulo multiplication. By Element to Power of Group Order is Identity, we have: : $n^{p - 1} \equiv 1 \pmod p$ {{qed}}
Proof by induction over $n$. Induction base: :$1^p \equiv 1 \pmod p$ Induction step: Assume $n^p \equiv n \pmod p$ {{begin-eqn}} {{eqn | l = \paren {n + 1}^p       | r = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} \cdot 1^k       | c = Binomial Theorem }} {{eqn | l = \forall k: 0 < k < p: {p \choose k}       | o = \equiv       | r = 0 \pmod p       | c = Binomial Coefficient of Prime }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k}       | o = \equiv       | r = n^p + n^0 \pmod p }} {{eqn | o = \equiv       | r = n^p + 1 \pmod p }} {{eqn | o = \equiv       | r = n + 1 \pmod p       | c = by induction step }} {{end-eqn}} Dividing by $n$: :$\forall n: n^p \equiv n \pmod p \implies n^{p - 1} \equiv 1 \pmod p$ {{qed}}
{{begin-eqn}} {{eqn | l = \paren {r \paren {\cos x + i \sin x} }^\omega       | r = \paren {r e^{i x} }^\omega       | c = Euler's Formula }} {{eqn | r = r^\omega e^{i \omega x}       | c = Power of Power }} {{eqn | r = r^\omega \paren {\map \cos {\omega x} + i \, \map \sin {\omega x} }       | c = Euler's Formula }} {{end-eqn}} {{qed}}
This can be done in sections. === Proof for Natural Number Index === {{:Power Rule for Derivatives/Natural Number Index/Proof by Binomial Theorem}} === Proof for Integer Index === {{:Power Rule for Derivatives/Integer Index}} === Proof for Fractional Index === {{:Power Rule for Derivatives/Fractional Index/Proof 1}} === Proof for Rational Index === {{:Power Rule for Derivatives/Rational Index}} === Proof for Real Number Index === {{:Power Rule for Derivatives/Real Number Index/Proof 1}}