hcju/proofwikiqa huggingface dataset:
Construct the perpendicular from $E$ to $AC$, and call its foot $G$. Let $K$ be the intersection of $IE$ and $AC$. Let $L$ be the intersection of $EH$ and $AC$. :500px First we have: {{begin-eqn}} {{eqn | n = 1       | l = \angle CKI       | r = \angle EKG       | c = Two Straight Lines make Equal Opposite Angles }} {{eqn | l = \angle EGK       | r = \text {Right Angle}       | c = Tangent to Circle is Perpendicular to Radius }} {{eqn | l = \angle KIC       | r = \text {Right Angle}       | c = as $IF \perp CD$ }} {{eqn | n = 2       | ll= \therefore       | l = \angle EGK       | r = \angle KIC       | c = Euclid's Fourth Postulate }} {{eqn | l = IC       | r = EJ       | c = Opposite Sides and Angles of Parallelogram are Equal }} {{eqn | l = EJ       | r = EG       | c = as both are radii of the same circle }} {{eqn | n = 3       | ll= \therefore       | l = IC       | r = EG       | c = Euclid's First Common Notion }} {{eqn | ll= \therefore       | l = \Area \triangle IKC       | r = \Area \triangle GKE       | c = Triangle Angle-Angle-Side Equality: $(1)$, $(2)$ and $(3)$ }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | n = 4       | l = \angle HLA       | r = \angle GLE       | c = Two Straight Lines make Equal Opposite Angles }} {{eqn | l = \angle EGL       | r = \text {Right Angle}       | c = Tangent to Circle is Perpendicular to Radius }} {{eqn | l = \angle AHL       | r = \text {Right Angle}       | c = as $HJ \perp AD$ }} {{eqn | n = 5       | ll= \therefore       | l = \angle EGL       | r = \angle AHL       | c = Euclid's Fourth Postulate }} {{eqn | l = HA       | r = EF       | c = Opposite Sides and Angles of Parallelogram are Equal }} {{eqn | l = EF       | r = EG       | c = as both are radii of the same circle }} {{eqn | n = 6       | ll= \therefore       | l = HA       | r = EG       | c = Euclid's First Common Notion }} {{eqn | ll= \therefore       | l = \Area \triangle HAL       | r = \Area \triangle GEL       | c = Triangle Angle-Angle-Side Equality: $(4)$, $(5)$ and $(6)$ }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = \frac {\Area \Box ABCD} 2       | r = \frac {AD \cdot CD} 2       | c = Area of Parallelogram }} {{eqn | r = \Area \triangle ADC       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \Area \triangle HAL + \Area \triangle IKC + \Area \Box DHLKI }} {{eqn | r = \Area \triangle GEL + \Area \triangle GKE + \Area \Box DHLKI }} {{eqn | r = \Area \Box DHEI }} {{end-eqn}} {{qed}}
The crucial geometric truth to note is that: :$CJ = CG, AG = AF, BF = BJ$ This follows from the fact that: :$\triangle CEJ \cong \triangle CEG$, $\triangle AEF \cong \triangle AEG$ and $\triangle BEF \cong \triangle BEJ$ This is a direct consequence of the point $E$ being the center of the incircle of $\triangle ABC$. Then it is just a matter of algebra. Let $AF = a, FB = b, CJ = c$. {{begin-eqn}} {{eqn | l = \paren {a + b}^2 + \paren {b + c}^2       | r = \paren {a + c}^2       | c = Pythagoras's Theorem }} {{eqn | ll= \leadsto       | l = a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2       | r = a^2 + 2 a c + c^2       | c =  }} {{eqn | ll= \leadsto       | l = a b + b^2 + b c       | r = a c       | c =  }} {{eqn | ll= \leadsto       | l = a b + b^2 + b c + a c       | r = 2 a c       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a + b} \paren {b + c}       | r = 2 a c       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({c \mathbf u + \mathbf v}\right) \times \mathbf w       | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ c u_i + v_i & c u_j + v_j & c u_k + v_k \\ w_i & w_j & w_k \end{vmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ c u_i & c u_j & c u_k \\ w_i & w_j & w_k \end{vmatrix} + \begin{vmatrix} \mathbf i& \mathbf j & \mathbf k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{vmatrix}       | c = Determinant as Sum of Determinants }} {{eqn | r = c \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ u_i & u_j & u_k \\ w_i & w_j & w_k \end{vmatrix} + \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{vmatrix}       | c = Determinant with Row Multiplied by Constant }} {{eqn | r = c \left({\mathbf u \times \mathbf w}\right) + \mathbf v \times \mathbf w       | c = {{Defof|Vector Cross Product}} }} {{end-eqn}} {{qed}} Category:Vector Cross Product bhw03ywmrmsthq70e9dmqa9snur0fg6
=== Necessary Condition === Suppose that $n = 0$ or $m = 0$. Then from Zero is Zero Element for Natural Number Multiplication: :$m \times n = 0$ {{qed|lemma}} === Sufficient Condition === Let $m \times n = 0$. Suppose WLOG that $n \ne 0$. {{begin-eqn}} {{eqn | l = n       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \le       | r = n       | c = Definition of One }} {{eqn | ll= \implies       | l = m \times n       | r = m \times \left({\left({n - 1}\right) + 1}\right)       | c = Definition of Difference }} {{eqn | r = m \times \left({n - 1}\right) + m       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | ll= \implies       | l = 0 \le m       | o = \le       | r = m \times \left({n - 1}\right) + m       | c =  }} {{end-eqn}} But as: : $m \times \left({n - 1}\right) \circ m = m \times n = 0$ it follows that: :$0 \le m \le 0$ and so as $\le$ is antisymmetric, it follows that $m = 0$. {{qed}}
{{WLOG}}, we will only prove $OB$ bisects $\angle AOC$. We have: {{begin-eqn}} {{eqn | l = OA       | r = OC       | c = {{Defof|Rhombus}} }} {{eqn | l = BA       | r = BC       | c = {{Defof|Rhombus}} }} {{eqn | l = OB       | r = OB       | c = Common Side }} {{eqn | ll = \leadsto       | l = \triangle OAB       | o = \cong       | r = \triangle OCB       | c = SSS }} {{end-eqn}} Comparing corresponding angles gives: :$\angle AOB = \angle COB$ hence $OB$ bisects $\angle AOC$. {{qed}}
{{WLOG}}, we will only prove $OB$ bisects $\angle AOC$. Let the position vector of $A$, $B$ and $C$ with respect to $O$ be $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively. By definition of rhombus, we have: {{begin-eqn}} {{eqn | n = a       | l = \mathbf a + \mathbf c       | r = \mathbf b       | c = Parallelogram Law }} {{eqn | n = b       | l = \norm {\mathbf a}       | r = \norm {\mathbf c}       | c =  }} {{end-eqn}} From the above we have: {{begin-eqn}} {{eqn | l = \cos \angle \mathbf a, \mathbf b       | r = \frac {\mathbf a \cdot \mathbf b} {\norm {\mathbf a} \norm {\mathbf b} }       | c = {{Defof|Dot Product|index = 2}} }} {{eqn | r = \frac {\mathbf a \cdot \paren {\mathbf a + \mathbf c} } {\norm {\mathbf a} \norm {\mathbf b} }       | c = from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$ }} {{eqn | r = \frac {\mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf c} {\norm {\mathbf a} \norm {\mathbf b} }       | c = Dot Product Distributes over Addition }} {{eqn | r = \frac { {\norm {\mathbf a} }^2 + \mathbf a \cdot \mathbf c} {\norm {\mathbf a} \norm {\mathbf b} }       | c = Dot Product of Vector with Itself }} {{eqn | r = \frac { {\norm {\mathbf c} }^2 + \mathbf a \cdot \mathbf c} {\norm {\mathbf c} \norm {\mathbf b} }       | c = from $(b)$ above: $\norm {\mathbf a} = \norm {\mathbf c}$ }} {{eqn | r = \frac {\mathbf c \cdot \mathbf c + \mathbf a \cdot \mathbf c} {\norm {\mathbf c} \norm {\mathbf b} }       | c = Dot Product of Vector with Itself }} {{eqn | r = \frac {\mathbf c \cdot \left({\mathbf a + \mathbf c}\right)} {\norm {\mathbf c} \norm {\mathbf b} }       | c = Dot Product Distributes over Addition }} {{eqn | r = \frac {\mathbf c \cdot \mathbf b} {\norm {\mathbf c} \norm {\mathbf b} }       | c = from $(a)$ above: $\mathbf b = \mathbf a + \mathbf c$ }} {{eqn | r = \cos \angle \mathbf c, \mathbf b       | c = {{Defof|Dot Product|index = 2}} }} {{end-eqn}} By definition of dot product, the angle between the vectors is between $0$ and $\pi$. From Shape of Cosine Function, cosine is injective on this interval. Hence: :$\angle \mathbf a, \mathbf b = \angle \mathbf c, \mathbf b$ The result follows. {{qed}}
We have that: $P \left({k, n}\right) = \begin{cases} 0 & : n = 0 \\ P \left({k, n - 1}\right) + \left({k - 2}\right) \left({n - 1}\right) + 1 & : n > 0 \end{cases}$ Proof by induction: For all $n \in \N_{>0}$, let $\Pi \left({n}\right)$ be the proposition: : $\displaystyle P \left({k, n}\right) = \sum_{j \mathop = 1}^n \left({\left({k - 2}\right) \left({j - 1}\right) + 1}\right)$ === Basis for the Induction === $\Pi(1)$ is the statement that $P \left({k, 1}\right) = 1$. This follows directly from: {{begin-eqn}} {{eqn | l = P \left({k, 1}\right)       | r = P \left({k, 0}\right) + \left({k - 2}\right) \left({0}\right) + 1 }} {{eqn | r = 1 }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\Pi \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $\Pi \left({r + 1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle P \left({k, r}\right) = \sum_{j \mathop = 1}^r \left({\left({k - 2}\right) \left({j - 1}\right) + 1}\right)$ Then we need to show: :$\displaystyle P \left({k, r + 1}\right) = \sum_{j \mathop = 1}^{r + 1} \left({\left({k - 2}\right) \left({j - 1}\right) + 1}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = P \left({k, r + 1}\right)       | r = P \left({k, r}\right) + \left({k - 2}\right) r + 1 }} {{eqn | r = \sum_{j \mathop = 1}^r \left({\left({k - 2}\right) \left({j - 1}\right) + 1}\right) + \left({k - 2}\right) r + 1       | c = from the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{r + 1} \left({\left({k - 2}\right)\left({j - 1}\right) + 1}\right) }} {{end-eqn}} So $\Pi \left({r}\right) \implies \Pi \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \in \N: P \left({k, n}\right) = \sum_{j \mathop = 1}^n \left({\left({k - 2}\right)\left({j - 1}\right) + 1}\right)$ {{qed}} Category:Polygonal Numbers p2eohhczq1q05egn2ayjve0pcuzimhw
By Bézout's Lemma there exists $a, b \in \Z$ such that $a p + b q = 1$. Then for $s \otimes_\Z t \in \Z / p \Z \otimes \Z / q \Z$: {{begin-eqn}} {{eqn | l = s \otimes t       | r = (s \left({a p + b q}\right)) \otimes t       | c = $s = s \cdot 1$ }} {{eqn | r = (s a p + s b q) \otimes t       | c = By module axiom 2 }} {{eqn | r = s b q \otimes t + s a p \otimes t       | c = By equality in tensor product }} {{eqn | r = s b \otimes q t + s a p \otimes t       | c = By equality in tensor product }} {{eqn | r = 0       | c = by Tensor with Zero Element is Zero in Tensor and the fact that $qt = 0$ in $\Z_q$ and $sap=0$ in $\Z_p$ }} {{end-eqn}} {{qed}} {{explain|Notation needs to be tightened up. When the fundamental stuff like this is being addressed, it is important to explain exactly which operation of which component of the module is being used: the scalar product, or the group operation, or whichever of the ring operations. It is inadequate to use mere concatenation, as it is never completely clear which elements of which structure each one is. As for the rest of this proof, it is still impenetrably vague. The reason for the last line is a complete mystery -- the actual definition of $0$ also needs to be stated.}}
=== Axiom 1 === Let $x, y \in T$ with $x = (s_i)_{i\in I}$ and $y = (t_i)_{i\in I}$. Let $\lambda\in R$. Then: {{begin-eqn}} {{eqn | l = \lambda \circ (x + y)       | r = \lambda \circ ((s_i)_{i\in I} + (t_i)_{i\in I})       | c = By definition of elements in direct sum }} {{eqn | r = \lambda \circ (s_i + t_i)_{i\in I}       | c = By addition in direct sum }} {{eqn | r = (\lambda \circ s_i + \lambda \circ t_i)_{i\in I}       | c = By $R$-action in direct sum }} {{eqn | r = (\lambda \circ s_i)_{i\in I} + (\lambda \circ t_i)_{i\in I}       | c = By addition in direct sum }} {{eqn | r = \lambda \circ (s_i)_{i\in I} + \lambda \circ (t_i)_{i\in I}       | c = By $R$-action in direct sum }} {{eqn | r = \lambda \circ x + \lambda \circ y       | c =  By definition of elements in direct sum }} {{end-eqn}} {{qed|lemma}} === Axiom 2 === Let $x \in T$ with $x = (s_i)_{i\in I}$ Let $\lambda, \mu \in R$. Then: {{begin-eqn}} {{eqn | l = \left({\lambda + \mu}\right) \circ x       | r = (\lambda+\mu) \circ (s_i)_{i\in I}        | c = By definition of elements in direct sum }} {{eqn | r = ((\lambda + \mu) \circ s_i)_{i\in I}       | c = By definition or $R$-action in direct sum }} {{eqn | r = (\lambda \circ s_i + \mu\circ s_i)_{i\in I}       | c = By definition or $R$-action in modules }} {{eqn | r = (\lambda \circ s_i)_{i\in I} + (\mu\circ s_i)_{i\in I}       | c = By definition or sum in direct sum }} {{eqn | r = \lambda \circ (s_i)_{i\in I} + \mu\circ (s_i)_{i\in I}       | c = By definition of $R$-action on direct sum }} {{eqn | r = \lambda \circ x + \mu\circ x       | c = By original equality }} {{end-eqn}} {{qed|lemma}} === Axiom 3 === Let $x\in T$ with $x = (s_i)_{i\in I}$. Let $\lambda, \mu \in R$. Then: {{begin-eqn}} {{eqn | l = (\lambda \times \mu) \circ x       | r = (\lambda \times \mu) \circ (s_i)_{i\in I}       | c = By original equality }} {{eqn | r = ((\lambda \times\mu) \circ s_i)_{i\in I}       | c = By Definition of $R$-action on direct sum }} {{eqn | r = (\lambda \circ (\mu\circ s_i))_{i\in I}       | c = By definition of modules }} {{eqn | r = \lambda \circ (\mu \circ s_i)_{i\in I}       | c = Definition of $R$-action on direct sum }} {{eqn | r = \lambda \circ \left({\mu \circ x}\right)       | c = By original equality }} {{end-eqn}} {{qed}} Category:Tensor Algebra Category:Module Theory dzo7uolnglmq24k5wog3smx85vt5h3o
Note that $A$ is non-empty as the empty set does not admit a supremum (in $\R$). Suppose $\epsilon \in \R_{>0}$ such that: :$\forall x \in A: b − x \ge \epsilon$ Then: :$\forall x \in A: b − \epsilon \ge x$ and so $b − \epsilon$ would be an upper bound of $A$ which is less than $b$. But since $b$ is a supremum of $A$ there can be no such $b − \epsilon$. From that contradiction it follows that: :$\exists x \in A: b − x < \epsilon$ {{qed}}
Let $m \in M$ and $n \in N$ Then {{begin-eqn}} {{eqn | l = m \otimes_R n       | r = \paren {m + 0} \otimes_R n       | c = {{GroupAxiom|2}} }} {{eqn | r = m \otimes_R n + 0 \otimes_R n       | c = {{Defof|Tensor Equality}} }} {{eqn | r = m \otimes_R \paren {n + 0}       | c = {{GroupAxiom|2}} }} {{eqn | r = m \otimes_R n + m\otimes_R 0       | c = {{Defof|Tensor Equality}} }} {{eqn | r = m \otimes_R n + 0 \otimes_R \paren {n + 0}       | c = {{GroupAxiom|2}} }} {{eqn | r = m \otimes_R n + m \otimes_R 0 + 0 \otimes_R n + 0 \otimes_R 0       | c = {{Defof|Tensor Equality}} }} {{eqn | r = m \otimes_R n + 0 \otimes_R 0       | c = {{Defof|Tensor Equality}} }} {{end-eqn}} Hence $0 \otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module. {{qed}} Category:Tensor Algebra Category:Homological Algebra 3plttq0y0vknwtgupk53xvo8d8yew9f
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arcsec} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}       | c = Derivative of $\operatorname{arcsec} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = 1       | c =  }} {{eqn | ll= \implies       | l = v       | r = x       | c = Primitive of Constant }} {{end-eqn}} First let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$. Then: {{begin-eqn}} {{eqn | l = \int \operatorname{arcsec} \frac x a \ \mathrm d x       | r = x \operatorname{arcsec} \frac x a - \int x \left({\frac a {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = x \operatorname{arcsec} \frac x a - a \int \frac {\mathrm d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \operatorname{arcsec} \frac x a - a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ }} {{end-eqn}} Similarly, let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$. Then: {{begin-eqn}} {{eqn | l = \int \operatorname{arcsec} \frac x a \ \mathrm d x       | r = x \operatorname{arcsec} \frac x a - \int x \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = x \operatorname{arcsec} \frac x a + a \int \frac {\mathrm d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \operatorname{arcsec} \frac x a + a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \arcsec \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \dfrac a {\size x \sqrt {x^2 - a^2} }       | c = Derivative of $\arcsec \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = x       | c = Primitive of Constant }} {{end-eqn}} We then have: {{begin-eqn}} {{eqn | l = \int \arcsec \frac x a \rd x       | r = x \arcsec \frac x a - \int x \paren {\dfrac a {\size x \sqrt {x^2 - a^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | n = 1       | r = x \arcsec \frac x a - \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x + C       | c = rearrangement }} {{end-eqn}} Let $x > 1$. Then: {{begin-eqn}} {{eqn | l = \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x       | r = \int \frac a { {\sqrt {x^2 - a^2} } } \rd x       | c = {{Defof|Absolute Value}} }} {{eqn | r = a \int \frac 1 { {\sqrt {x^2 - a^2} } } \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C       | c = Integral of One Over Square Root of Binomial }} {{eqn | r = a \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as argument of logarithm is positive }} {{end-eqn}} Similarly, let $x < -1$. Then:  {{begin-eqn}} {{eqn | l = \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x       | r = \int \paren {-1} \frac a { {\sqrt{x^2 - a^2} } } \rd x       | c = {{Defof|Absolute Value}} }} {{eqn | r = -a \int \frac 1 { {\sqrt{x^2 - a^2} } } \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C       | c = Integral of One Over Square Root of Binomial }} {{eqn | r = -a \paren {-\ln \size {x + \sqrt {x^2 - a^2} } } + C       | c = as argument of logarithm is negative }} {{eqn | r = a \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as argument of logarithm is negative }} {{end-eqn}} {{qed}} Category:Primitives involving Inverse Secant Function lyfdo47tn7b8vs8a82zr6p4vqsqu1eh
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }       | r = \cosh^{-1} {\frac x a} + C'       | c = Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: $\cosh^{-1}$ form }} {{eqn | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - 1} } + C'       | c = {{Defof|Real Inverse Hyperbolic Cosine}} }} {{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C'       | c = rearrangement }} {{eqn | r = \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C'       | c = rearrangement }} {{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a + C'       | c = Difference of Logarithms }} {{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } + C       | c = putting $C = -\ln a + C'$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf x \times \mathbf x       | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ x_i & x_j & x_k \\ x_i & x_j & x_k \\ \end{vmatrix}      | c = {{Defof|Vector Cross Product}} }} {{eqn | r = \mathbf 0       | c = Square Matrix with Duplicate Rows has Zero Determinant }} {{end-eqn}} {{qed}} Category:Vector Cross Product 1bl292w88tnbfskcppb6mc83yd4zye2
By definition of the Riemann zeta function: {{begin-eqn}} {{eqn | l = \frac 1 {\zeta \left({z}\right)}       | r = \prod_{p \text{ prime} } \left({1 - p^{-z} }\right)       | c =  }} {{eqn | r = \left({1 - \frac 1 {2^z} }\right) \left({1 - \frac 1 {3^z} }\right) \left({1 - \frac 1 {5^z} }\right) \left({1 - \frac 1 {7^z} }\right) \left({1 - \frac 1 {11^z} }\right) \cdots       | c =  }} {{end-eqn}} The expansion of this product will be: :$\displaystyle 1 + \sum_{n \text{ prime}} \left({\frac{-1} {n^z} }\right) + \sum_{n \mathop = p_1 p_2} \left({ \frac{-1}{p_1^z} \frac{-1} {p_2^z} }\right) + \sum_{n \mathop = p_1 p_2 p_3} \left({ \frac {-1} {p_1^z} \frac {-1} {p_2^z} \frac{-1}{p_3^z} }\right) + \cdots$ which is precisely: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mu \left({n}\right)} {n^z}$ as desired. {{handwaving|get this from Dirichlet Series of Inverse of Arithmetic Function instead}} {{qed}} Category:Riemann Zeta Function 5s8vvcoogyqnqlpso596vxiumn1kgd1