hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = y       | r = \cosh^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = \cosh y       | r = x       | c = {{Defof|Inverse Hyperbolic Cosine}} }} {{eqn | ll= \leadsto       | l = \map \cosh {\pm \, y}       | r = x       | c = Hyperbolic Cosine Function is Even }} {{eqn | ll= \leadsto       | l = \map \cos {\pm \, i y}       | r = x       | c = Hyperbolic Cosine in terms of Cosine }} {{eqn | ll= \leadsto       | l = \pm \, i y       | r = \cos^{-1} x       | c = {{Defof|Inverse Cosine}} }} {{eqn | ll= \leadsto       | l = y       | r = \pm \, i \cos^{-1} x       | c = multiplying both sides by $\pm \, i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \tan^{-1} \paren {i x}       | c =  }} {{eqn | ll= \implies       | l = \tan y       | r = i x       | c = {{Defof|Inverse Tangent}} }} {{eqn | ll= \implies       | l = i \tan y       | r = - x       | c = $i^2 = -1$ }} {{eqn | ll= \implies       | l = \tanh \paren {i y}       | r = -x       | c = Tangent in terms of Hyperbolic Tangent }} {{eqn | ll= \implies       | l = i y       | r = \tanh^{-1} \paren {-x}       | c = {{Defof|Inverse Hyperbolic Tangent}} }} {{eqn | ll= \implies       | l = i y       | r = -\tanh^{-1} x       | c = Inverse Hyperbolic Tangent is Odd Function }} {{eqn | ll= \implies       | l = y       | r = i \tanh^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \tan^{-1} \left({i x}\right) 	| r = \frac i 2 \ln \left({\frac {1 - i \left({i x}\right)} {1 + i \left({ i x}\right)} }\right) 	| c = Arctangent Logarithmic Formulation }} {{eqn	| r = \frac i 2 \ln \left({\frac{1 + x} {1 - x} }\right) }} {{eqn	| r = i \tanh^{-1} x 	| c = {{Defof|Inverse Hyperbolic Tangent}}  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \map {\tanh^{-1} } {i x}       | c =  }} {{eqn | ll= \leadsto       | l = \tanh y       | r = i x       | c = {{Defof|Inverse Hyperbolic Tangent}} }} {{eqn | ll= \leadsto       | l = i \tanh y       | r = -x       | c = $i^2 = -1$ }} {{eqn | ll= \leadsto       | l = \map \tan {i y}       | r = -x       | c = Hyperbolic Tangent in terms of Tangent }} {{eqn | ll= \leadsto       | l = i y       | r = \map {\tan^{-1} } {-x}       | c = {{Defof|Inverse Tangent}} }} {{eqn | ll= \leadsto       | l = i y       | r = -\tan^{-1} x       | c = Inverse Tangent is Odd Function }} {{eqn | ll= \leadsto       | l = y       | r = i \tan^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \cot^{-1} \paren {i x}       | c =  }} {{eqn | ll= \implies       | l = \cot y       | r = i x       | c = {{Defof|Inverse Cotangent}} }} {{eqn | ll= \implies       | l = i \cot y       | r = - x       | c = $i^2 = -1$ }} {{eqn | ll= \implies       | l = \coth \paren {i y}       | r = x       | c = Cotangent in terms of Hyperbolic Cotangent }} {{eqn | ll= \implies       | l = i y       | r = \coth^{-1} x       | c = {{Defof|Inverse Hyperbolic Cotangent}} }} {{eqn | ll= \implies       | l = y       | r = -i \coth^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \map {\coth^{-1} } {i x}       | c =  }} {{eqn | ll= \leadsto       | l = \coth y       | r = i x       | c = {{Defof|Inverse Hyperbolic Cotangent}} }} {{eqn | ll= \leadsto       | l = i \coth y       | r = - x       | c = $i^2 = -1$ }} {{eqn | ll= \leadsto       | l = \map \cot {i y}       | r = x       | c = Hyperbolic Cotangent in terms of Cotangent }} {{eqn | ll= \leadsto       | l = i y       | r = \cot^{-1} x       | c = {{Defof|Inverse Cotangent}} }} {{eqn | ll= \leadsto       | l = y       | r = -i \cot^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \sec^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = \sec y       | r = x       | c = {{Defof|Inverse Secant}} }} {{eqn | ll= \leadsto       | l = \map \sec {\pm \, y}       | r = x       | c = Secant Function is Even }} {{eqn | ll= \leadsto       | l = \map \sech {\pm \, i y}       | r = x       | c = Secant in terms of Hyperbolic Secant }} {{eqn | ll= \leadsto       | l = \pm \, i y       | r = \sech^{-1} x       | c = {{Defof|Inverse Hyperbolic Secant}} }} {{eqn | ll= \leadsto       | l = y       | r = \pm \, i \sech^{-1} x       | c = multiplying both sides by $\pm \, i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \sech^{-1} x       | c =  }} {{eqn | ll= \leadsto       | l = \sech y       | r = x       | c = {{Defof|Inverse Hyperbolic Secant}} }} {{eqn | ll= \leadsto       | l = \map \sech {\pm \, y}       | r = x       | c = Hyperbolic Secant Function is Even }} {{eqn | ll= \leadsto       | l = \map \sec {\pm \, i y}       | r = x       | c = Hyperbolic Secant in terms of Secant }} {{eqn | ll= \leadsto       | l = \pm \, i y       | r = \sec^{-1} x       | c = {{Defof|Inverse Secant}} }} {{eqn | ll= \leadsto       | l = y       | r = \pm \, i \sec^{-1} x       | c = multiplying both sides by $\pm \, i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \map {\csc^{-1} } {i x}       | c =  }} {{eqn | ll= \leadsto       | l = \csc y       | r = i x       | c = {{Defof|Inverse Cosecant}} }} {{eqn | ll= \leadsto       | l = i \csc y       | r = -x       | c = $i^2 = -1$ }} {{eqn | ll= \leadsto       | l = \map \csch {i y}       | r = -x       | c = Cosecant in terms of Hyperbolic Cosecant }} {{eqn | ll= \leadsto       | l = i y       | r = \map {\csch^{-1} } {-x}       | c = {{Defof|Inverse Hyperbolic Cosecant}} }} {{eqn | ll= \leadsto       | l = i y       | r = -\csch^{-1} x       | c = Inverse Hyperbolic Cosecant is Odd Function }} {{eqn | ll= \leadsto       | l = y       | r = i \sinh^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \map {\csch^{-1} } {i x}       | c =  }} {{eqn | ll= \leadsto       | l = \csch y       | r = i x       | c = {{Defof|Inverse Hyperbolic Cosecant}} }} {{eqn | ll= \leadsto       | l = i \csch y       | r = - x       | c = $i^2 = -1$ }} {{eqn | ll= \leadsto       | l = \map \csc {i y}       | r = x       | c = Hyperbolic Cosecant in terms of Cosecant }} {{eqn | ll= \leadsto       | l = i y       | r = \csc^{-1} x       | c = {{Defof|Inverse Cosecant}} }} {{eqn | ll= \leadsto       | l = y       | r = -i \csc^{-1} x       | c = multiplying both sides by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arccsc {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \csc y       | c = {{Defof|Arccosecant}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \sin y       | c = Cosecant is Reciprocal of Sine }} {{eqn | ll= \leadstoandfrom       | l = \arcsin x       | r = y       | c = {{Defof|Arcsine}} }} {{end-eqn}} {{qed}} Category:Arcsine Function Category:Arccosecant Function Category:Reciprocals lz7oto4eo0iirm8inhbwwmpbsdcjvy9
{{begin-eqn}} {{eqn | l = \map \arcsec {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \sec y       | c = {{Defof|Arcsecant}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \cos y       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadstoandfrom       | l = \arccos x       | r = y       | c = {{Defof|Arcsecant}} }} {{end-eqn}} {{qed}} Category:Arccosine Function Category:Arcsecant Function Category:Reciprocals hh86ajd1zpklrkor5kvciuco5plyud3
{{begin-eqn}} {{eqn | l = \map \arccot {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \cot y       | c = {{Defof|Arccotangent}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \tan y       | c = Cotangent is Reciprocal of Tangent }} {{eqn | ll= \leadstoandfrom       | l = \arctan x       | r = y       | c = {{Defof|Arctangent}} }} {{end-eqn}} {{qed}} Category:Arctangent Function Category:Arccotangent Function Category:Reciprocals n10gxqlrgjdluw6byu4ca16nbxrtpop
From Solution to Quadratic Equation: :$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$ If the discriminant $b^2 - 4 a c > 0$ then $\sqrt {b^2 - 4 a c}$ has two values and the result follows. If the discriminant $b^2 - 4 a c = 0$ then $\sqrt {b^2 - 4 a c} = 0$ and $x = \dfrac {-b} {2 a}$. If the discriminant $b^2 - 4 a c < 0$, then it can be written as: :$b^2 - 4 a c = \paren {-1} \size {b^2 - 4 a c}$ Thus: :$\sqrt {b^2 - 4 a c} = \pm i \sqrt {\size {b^2 - 4 a c} }$ and the two solutions are: :$x = \dfrac {-b} {2 a} + i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}, x = \dfrac {-b} {2 a} - i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}$ and once again the result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \alpha       | r = \frac {-b + \sqrt {b^2 - 4 a c} } {2 a}       | c = Solution to Quadratic Equation }} {{eqn | l = \beta       | r = \frac {-b - \sqrt {b^2 - 4 a c} } {2 a}       | c = {{WLOG}}, selecting $\alpha$ and $\beta$ as such }} {{eqn | ll= \leadsto       | l = \alpha + \beta       | r = \frac {-b - \sqrt {b^2 - 4 a c} - b + \sqrt {b^2 - 4 a c} } {2 a}       | c =  }} {{eqn | r = \frac {-2 b} {2 a}       | c =  }} {{eqn | r = -\frac b a       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \alpha       | r = \frac {-b + \sqrt {b^2 - 4 a c} } {2 a}       | c = Solution to Quadratic Equation }} {{eqn | l = \beta       | r = \frac {-b - \sqrt {b^2 - 4 a c} } {2 a}       | c = {{WLOG}}, selecting $\alpha$ and $\beta$ as such }} {{eqn | ll= \leadsto       | l = \alpha \beta       | r = \frac {\paren {-b - \sqrt {b^2 - 4 a c} } \paren {- b + \sqrt {b^2 - 4 a c} } } {4 a^2}       | c =  }} {{eqn | r = \frac {b^2 - \paren {b^2 - 4 a c} } {4 a^2}       | c = Difference of Two Squares }} {{eqn | r = \frac {4 a c} {4 a^2} }} {{eqn | r = \frac c a }} {{end-eqn}} {{qed}}
From Cardano's Formula, the roots of $P$ are: :$(1): \quad x_1 = S + T - \dfrac b {3 a}$ :$(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \left({S - T}\right)$ :$(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \left({S - T}\right)$ where: : $S = \sqrt [3] {R + \sqrt{Q^3 + R^2}}$ : $T = \sqrt [3] {R - \sqrt{Q^3 + R^2}}$ === Zero Discriminant === First the easy case: $D = 0$. Hence $S = T = \sqrt [3] R$, and so $S + T = 2 \sqrt [3] R, S - T = 0$. From the above, this gives us: :$(1): \quad x_1 = 2 \sqrt [3] R - \dfrac b {3 a}$ :$(2): \quad x_2 = - \sqrt [3] R - \dfrac b {3 a}$ :$(3): \quad x_3 = - \sqrt [3] R - \dfrac b {3 a}$ Thus the roots $x_2$ and $x_3$ are equal, and all three roots are real. They are all equal when $R = 0$. {{qed|lemma}} === Positive Discriminant === Let $D = Q^3 + R^2 > 0$. Then $S = R + \sqrt{Q^3 + R^2}$ and $T = R - \sqrt{Q^3 + R^2}$ are wholly real and distinct. Therefore, so are $S + T$ and $S - T$. Hence: : $\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \left({S - T}\right)$ and : $\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \left({S - T}\right)$ are complex conjugates. {{qed|lemma}} === Negative Discriminant === Let $D = Q^3 + R^2 < 0$. Then $\sqrt D = \pm i \left|{Q^3 + R^2}\right| = \pm i E$, say, where $E > 0$. Thus $S^3 = R + i E, T^3 = R - i E$. Let $\sqrt [3] {R + i E} = p + i q$, and so $\sqrt [3] {R - i E} = p - i q$. Hence $S + T = 2 p, S - T = 2 i q$. So: {{begin-eqn}} {{eqn | l=y_1       | r=S + T       | c= }} {{eqn | r=2 p       | c }} {{eqn | l=y_2       | r=-\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \left({S - T}\right)       | c= }} {{eqn | r=- p - \sqrt 3 q       | c=(after some algebra) }} {{eqn | l=y_3       | r=-\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \left({S - T}\right)       | c= }} {{eqn | r=- p + \sqrt 3 q       | c=(after some algebra) }} {{end-eqn}} Subtracting $\dfrac b {3 a}$ from the above, we obtain the three distinct real solutions: :$(1): \quad x_1 = 2 p - \dfrac b {3 a}$ :$(2): \quad x_2 = -p - \sqrt 3 q - \dfrac b {3 a}$ :$(3): \quad x_3 = -p + \sqrt 3 q - \dfrac b {3 a}$ {{qed|lemma}}
From Cardano's Formula, the roots of $P$ are: :$(1): \quad x_1 = S + T - \dfrac b {3 a}$ :$(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$ :$(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$ where: :$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$ :$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$ Let $D = Q^3 + R^2 < 0$. Then $S^3 = R + i \sqrt {\size {Q^3 + R^2} }$. We can express this in polar form: :$S^3 = r \paren {\cos \theta + i \sin \theta}$ where: :$r = \sqrt {R^2 + \paren {\sqrt {Q^3 + R^2} }^2} = \sqrt {R^2 - \paren {Q^3 + R^2} } = \sqrt {-Q^3}$ :$\tan \theta = \dfrac {\sqrt {\size {Q^3 + R^2} } } R$ Then: :$\cos \theta = \dfrac R {\sqrt {-Q^3} }$ Similarly for $T^3$. The result: :$(1): \quad x_1 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3} - \dfrac b {3 a}$ :$(2): \quad x_2 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3 + \dfrac {2 \pi} 3} - \dfrac b {3 a}$ :$(3): \quad x_3 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3 + \dfrac {4 \pi} 3} - \dfrac b {3 a}$ follows after some algebra. {{qed}}
Note that: $\left({R, +, \circ}\right)$ is a ring by assumption. $\left({R, +}\right)$ is an abelian group by the definition of a ring. Let us verify the module axioms: {{begin-axiom}} {{axiom | n = 1         | q = \forall x, y, z \in R         | m = x \circ \left({y + z}\right) = \left({x \circ y}\right) + \left({x \circ z}\right) }} {{axiom | n = 2         | q = \forall x, y, z \in R         | m = \left({x + y}\right) \circ z = \left({x \circ z}\right) + \left({y \circ z}\right) }} {{axiom | n = 3         | q = \forall x, y, z \in R         | m = \left({x \circ y}\right) \circ z = x \circ \left({y \circ z}\right) }} {{end-axiom}} Axiom $(1)$ and $(2)$ follow from distributivity of $\circ$. Axiom $(3)$ follows from associativity of $\circ$.  {{qed|lemma}} Assume now that $\left({R, +, \circ}\right)$ has a unity, $1_R$.  For $\left({R, +, \circ}\right)_R$ to be unitary, it must satisfy the additional axiom: {{begin-axiom}} {{axiom | n = 4         | q = \forall x \in R         | m = 1_R \circ x = x }} {{end-axiom}} The axiom follows from the definition of a unity. {{qed}}
This is a special case of Module on Cartesian Product is Module: :$\struct {R^n, +, \circ}_R$ is an $R$-module where $n = 1$. {{qed}}
From the definition of division: :$a / b := a \times \paren {b^{-1} }$ where $b^{-1}$ is the inverse for rational multiplication. From Non-Zero Rational Numbers under Multiplication form Abelian Group, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group. From group axiom $G3$ it follows that every $b \in \Q_{\ne 0}$ has an inverse element $b^{-1} \in \Q$ under multiplication. From group axiom $G0$ it follows that $\Q_{\ne 0}$ is closed under multiplication. Hence the result: :$\forall a, b \in \Q_{\ne 0}: a / b \in \Q_{\ne 0}$ {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: : $\cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$ $P \left({1}\right)$ is trivially true: :$\cmod {z_1} = \cmod {z_1}$ === Basis for the Induction === $P \left({2}\right)$ is the case: : $\cmod {z_1 z_2} = \cmod {z_1} \cdot \cmod {z_2}$ which has been proved in Complex Modulus of Product of Complex Numbers. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: : $\cmod {z_1 z_2 \cdots z_k} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_k}$ Then we need to show: : $\cmod {z_1 z_2 \cdots z_{k + 1} } = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \cmod {z_1 z_2 \cdots z_{k + 1} }       | r = \cmod {\left({z_1 z_2 \cdots z_k}\right) z_{k + 1} }       | c =  }} {{eqn | r = \cmod {z_1 z_2 \cdots z_k} \cdot \cmod {z_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_k} \cdot \cmod {z_{k + 1} }       | c = Induction Hypothesis }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\forall n \in \N_{>0}: \cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$ {{qed}}
{{begin-eqn}} {{eqn | l = \cmod {\dfrac {z_1} {z_2} }       | r = \cmod {z_1 \times z_2^{-1} }       | c = {{Defof|Complex Division}} }} {{eqn | r = \cmod {z_1} \times \cmod {z_2^{-1} }       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \cmod {z_1} \times \cmod {z_2}^{-1}       | c = Complex Modulus of Reciprocal of Complex Number }} {{eqn | r = \dfrac {\cmod {z_1} } {\cmod {z_2} }       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
Let $z = a + i b$. {{begin-eqn}} {{eqn | l = \cmod {\frac 1 z}       | r = \cmod {\frac 1 {a + i b} }       | c =  }} {{eqn | r = \cmod {\frac {a - i b} {a^2 + b^2} }       | c = Inverse for Complex Multiplication }} {{eqn | r = \cmod {\frac a {a^2 + b^2} + i \frac {-b} {a^2 + b^2} }       | c =  }} {{eqn | r = \sqrt {\paren {\frac a {a^2 + b^2} }^2 + \paren {\frac {-b} {a^2 + b^2} }^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \frac {\sqrt {a^2 + b^2} } {a^2 + b^2}       | c =  }} {{eqn | r = \frac 1 {\sqrt {a^2 + b^2} }       | c =  }} {{eqn | r = \frac 1 {\cmod z}       | c = {{Defof|Complex Modulus}} }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$ $\map P 1$ is true by definition of the usual ordering on real numbers: :$\cmod {z_1} \le \cmod {z_1}$ === Basis for the Induction === $\map P 2$ is the case: :$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$ which has been proved in Triangle Inequality for Complex Numbers. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\cmod {z_1 + z_2 + \dotsb + z_k} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k}$ Then we need to show: :$\cmod {z_1 + z_2 + \dotsb + z_{k + 1} } \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2 + \dotsb + z_{k + 1} }       | r = \cmod {\paren {z_1 + z_2 + \dotsb + z_k} + z_{k + 1} }       | c = {{Defof|Indexed Summation}} }} {{eqn | o = \le       | r = \cmod {z_1 + z_2 + \dotsb + z_k} + \cmod {z_{k + 1} }       | c = Basis for the Induction }} {{eqn | o = \le       | r = \paren {\cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k} } + \cmod {z_{k + 1} }       | c = Induction Hypothesis }} {{eqn | o = \le       | r = \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k} + \cmod {z_{k + 1} }       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Let $z_3 := z_1 + z_2$. Then: {{begin-eqn}} {{eqn | l = \cmod {z_3} + \cmod {\paren {-z_2} }       | o = \ge       | r = \cmod {z_3 + \paren {-z_2} }       | c = Triangle Inequality for Complex Numbers }} {{eqn | ll= \leadsto       | l = \cmod {z_3} + \cmod {z_2}       | o = \ge       | r = \cmod {z_3 - z_2}       | c = Complex Modulus of Additive Inverse }} {{eqn | ll= \leadsto       | l = \cmod {z_1 + z_2} + \cmod {z_2}       | o = \ge       | r = \cmod {z_1}       | c = substituting $z_3 = z_1 + z_2$ }} {{eqn | ll= \leadsto       | l = \cmod {z_1 + z_2}       | o = \ge       | r = \cmod {z_1} - \cmod {z_2}       | c =  }} {{end-eqn}} {{qed}}
Let $z = a + i b$. {{begin-eqn}} {{eqn | l = \cmod {\paren {-z} }       | r = \cmod {\paren {-a - i b} }       | c = {{Defof|Negative of Complex Number}} }} {{eqn | r = \sqrt {\paren {-a}^2 + \paren {-b}^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt {a^2 + b^2}       | c = Even Power of Negative Real Number }} {{eqn | r = \cmod {a + i b}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \cmod z       | c = Definition of $z$ }} {{end-eqn}} {{qed}} Category:Complex Modulus 6wt6y75bct78sw0c4q7wfse098r4xw7
From Real Numbers form Ordered Field, $\R$ is a field. By definition, $\R$ is therefore a ring. The result follows from Power of Ring Negative. {{qed}} Category:Real Numbers qg7zw1ow21dyrvq4hqvq50ymhvolfoy
This is an instance of Cardinality of Cartesian Product: General Result, where each set is equal.  {{qed}} Category:Cardinality of Cartesian Product bqiirzq13psf5dl8si523so5oqtu0c3
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$ Let this be expressed as: :$\displaystyle \prod_{j \mathop = 1}^n z_j = \prod_{j \mathop = 1}^n r_j \sum_{j \mathop = 1}^n \paren {\cos \theta_j + i \sin \theta_j}$ $\map P 1$ is the case: :$r_1 \paren {\cos x + i \sin x} = r_1 \paren {\cos x + i \sin x }$ which is trivially true. === Basis for the Induction === $\map P 2$ is the case: :$z_1 z_2 = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} }$ which is proved in Product of Complex Numbers in Polar Form. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k z_j = \prod_{j \mathop = 1}^k r_j \sum_{j \mathop = 1}^k \paren {\cos \theta_j + i \sin \theta_j}$ Then we need to show: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} z_j = \prod_{j \mathop = 1}^{k + 1} r_j \sum_{j \mathop = 1}^{k + 1} \paren {\cos \theta_j + i \sin \theta_j}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn| l = \prod_{j \mathop = 1}^{k + 1} z_j      | r = \paren {\prod_{j \mathop = 1}^k z_j} z_{k + 1} }} {{eqn| r = \paren {\prod_{j \mathop = 1}^k r_j \sum_{j \mathop = 1}^k \paren {\cos \theta_j + i \sin \theta_j} } \paren {r_{k + 1} \paren {\cos \theta_{k + 1} + i \sin \theta_{k + 1} } }      | c = Induction Hypothesis }} {{eqn| r = \paren {\prod_{j \mathop = 1}^k r_j} r_{k + 1} \paren {\sum_{j \mathop = 1}^k \paren {\cos \theta_j + i \sin \theta_j} + \cos \theta_{k + 1} + i \sin \theta_{k + 1} }      | c = Basis for the Induction }} {{eqn| r = \prod_{j \mathop = 1}^{k + 1} z_j = \prod_{j \mathop = 1}^{k + 1} r_j \sum_{j \mathop = 1}^{k + 1} \paren {\cos \theta_j + i \sin \theta_j}      | c =  }} {{end-eqn}} Hence, by induction, for all $n \in \N_{>0}$: :$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$ {{qed}}
From Product of Complex Numbers in Polar Form: General Result: :$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$ Setting $z_1 = z_2 = \cdots = z_n = r \paren {\cos x + i \sin x}$ gives the result.
600px The above diagram illustrates the $7$th roots of unity. {{ProofWanted}}
From Equation of Unit Circle, the unit circle whose center is at the origin of the Cartesian $xy$ plane has the equation: :$x^2 + y^2 = 1$ Identifying the Cartesian $xy$ plane with the complex plane: {{ProofWanted|Can't think of the precise words I need for this at the moment}}