hcju/proofwikiqa huggingface dataset:
=== Lemma === {{:Mellin Transform of Heaviside Step Function/Lemma}} {{begin-eqn}} {{eqn | l = \mathcal M \left\{ {u_c \left({t}\right)}\right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s-1} u_c \left({t}\right) \ \mathrm d t       | c = Definition of Mellin Transform }} {{eqn | r = \int_c^{\to +\infty} t^{s-1} \ \mathrm d t       | c = Definition of Heaviside step function: integrand is elsewhere zero }} {{eqn | r = \left.{\dfrac {t^s} s}\right \vert_c^{+\infty}       | c = Primitive of Power }} {{eqn | r = 0 - \dfrac {c^s} s       | c = By Lemma }} {{eqn | r = - \dfrac {c^s} s       | c =  }} {{end-eqn}} {{qed}} Category:Mellin Transforms Category:Heaviside Step Function fo1by4t7vnl09obqnc6vzy2309b6z6q
{{begin-eqn}} {{eqn | l = \mathcal M \left\{ {u \left({c-t}\right)}\right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s - 1} u \left({c - t}\right) \mathrm d t       | c = Definition of Mellin transform }} {{eqn | r = \int_0^c t^{s-1} \mathrm d t       | c = Definition of Heaviside step function: integrand is elsewhere zero }} {{eqn | r = \left.{\dfrac {t^s} s}\right \vert_0^c       | c = Primitive of Power }} {{eqn | r = \dfrac {c^s} s - 0       | c =  }} {{eqn | r = \dfrac {c^s} s       | c =  }} {{end-eqn}} {{qed}} Category:Mellin Transforms Category:Heaviside Step Function 08zk1cdjldattlrkjqqbuxlsdtu0tdh
{{begin-eqn}} {{eqn | l = \mathcal M \left\{ {t^n f \left({t}\right)}\right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s - 1} t^n f \left({t}\right) \ \mathrm d t       | c = Definition of Mellin Transform }} {{eqn | r = \int_0^{\to +\infty} t^{\left({s + n}\right)-1} f \left({t}\right) \ \mathrm d t       | c = Exponent Combination Laws }} {{eqn | r = \mathcal M \left\{ {f \left({t}\right)} \right\} \left({s + n}\right)       | c = Definition of Mellin Transform }} {{end-eqn}} {{qed}} Category:Mellin Transforms p05uwj706t6ykmc70j1u6xvni60pgpn
{{begin-eqn}} {{eqn | l = \mathcal M \left\{ {\delta_c \left({t}\right) f \left({t}\right)} \right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s - 1} \delta_c \left({t}\right) f \left({t}\right) \ \mathrm d t       | c = {{Defof|Mellin Transform}} }} {{eqn | r = \int_{c^-}^{c^+} t^{s - 1} \delta_c \left({t}\right) f \left({t}\right) \ \mathrm d t       | c = {{Defof|Dirac Delta Function}}: integrand is elsewhere zero }} {{eqn | r = \int_{c^-}^{c^+} c^{s - 1} \delta_c \left({t}\right) f \left({c}\right) \ \mathrm d t       | c = $t$ is constant in interval $\left[{c^-\,.\,.\,c^+}\right]$ }} {{eqn | r = c^{s - 1} f \left({c}\right) \int_{c^-}^{c^+} \delta_c \left({t}\right) \ \mathrm d t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = c^{s - 1} f \left({c}\right)       | c = {{Defof|Dirac Delta Function}} }} {{end-eqn}} {{qed}} Category:Mellin Transforms Category:Dirac Delta Function 9xmrwrtkjvm7an8dlpo9l4778l3rskm
By definition of ring with unity, $\left({R, \circ}\right)$ is a monoid whose identity element is $1_R$. From Identity Element is Idempotent (applied to $1_R$): :$1_R \circ 1_R = 1_R$ which was to be proven.  {{qed}} Category:Ring Theory sykxawsy95ekov7hb6d6eq70yxiv8oq
{{begin-eqn}} {{eqn | l = \map {\MM \set {e^{-a t^n} } } s       | r = \int_0^{\to +\infty} t^{s-1} e^{-a t^n} \rd t       | c = {{Defof|Mellin Transform}} }} {{eqn | r = \int_0^{\to +\infty} \paren {\dfrac t {\sqrt[n] a} }^{s - 1} e^{-a \paren {\dfrac t {\sqrt[n] {a} } }^n} \dfrac {\d t} {\sqrt[n] a}       | c = Integration by Substitution, $t \mapsto \dfrac t {\sqrt[n] a}$, $\d t \mapsto \dfrac {\d t} {\sqrt[n] {a} }$ }} {{eqn | r = a^{-s/n} \int_0^{\to +\infty} t^{s - 1} e^{-t^n} \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = a^{-s/n} \int_0^{\to +\infty} t^{\paren {s - 1}/n} e^{-t} \dfrac {t^{1 / n - 1} \rd t} n       | c = Integration by Substitution, $t \mapsto \sqrt[n] t$, $\d t \mapsto \dfrac {t^{1 / n - 1} \rd t} n$ }} {{eqn | r = \dfrac{a^{-s/n} } n \int_0^{\to +\infty} t^{s/n-1} e^{-t} \rd t       | c = Primitive of Constant Multiple of Function, Exponent Combination Laws }} {{eqn | r = \dfrac{a^{-s/n} } n \map \Gamma {\dfrac s n}       | c = {{Defof|Gamma Function}} }} {{end-eqn}} {{qed}} Category:Mellin Transforms 2yxilpge4j3790l4fznsrij0k2le73c
{{begin-eqn}} {{eqn | l = \mathcal M \left\{ {\lambda f \left({t}\right) + g \left({t}\right) }\right\} \left({s}\right)       | r = \int_0^{\to +\infty} t^{s - 1} \left({\lambda f \left({t}\right) + g \left({t}\right)}\right) \mathrm d t       | c = Definition of Mellin Transform }} {{eqn | r = \lambda \int_0^{\to +\infty} t^{s - 1} f \left({t}\right) \mathrm d t + \int_0^{\to +\infty} t^{s - 1} g \left({t}\right) \mathrm d t       | c = distributing $t^{s - 1}$, Linear Combination of Complex Integrals }} {{eqn | r = \lambda \mathcal M \left\{ {f \left({t}\right)}\right\} \left({s}\right) + \mathcal M \left\{ {g \left({t}\right)}\right\} \left({s}\right)       | c = Definition of Mellin Transform }} {{end-eqn}} {{qed}} Category:Mellin Transforms 9apkv5zjrkq9lazbsm2tvhhcab6gxin
Let $\phi: \C_{\ne 0} \to \R_{> 0} \times K$ be the mapping: :$\map \phi {r e^{i \theta} } = \paren {r, e^{i \theta} }$ $\forall \tuple {a, b} \in \R_{> 0} \times K:\exists z = a \times b \in \C$ such that: :$\map \phi z = \tuple {a, b}$ by Complex Multiplication is Closed and $\R \subset \C$. So $\phi$ is surjective. To prove $\phi$ is injective, let $\map \phi {r_1 e^{i \theta_1} } = \map \phi {r_2 e^{i \theta_2} }$. {{begin-eqn}} {{eqn | l = \map \phi {r_1 e^{i \theta_1} }       | r = \map \phi {r_2 e^{i \theta_2} }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {r_1, e^{i \theta_1} }       | r = \paren {r_2, e^{i \theta_2} }       | c = Definition of $\phi$ }} {{eqn | ll= \leadsto       | l = r_1 = r_2       | o = \land       | r = e^{i \theta_1} = e^{i \theta_2}       | c =  }} {{eqn | ll= \leadsto       | l = r_1 e^{i \theta_1}       | r = r_2 e^{i \theta_2}       | c =  }} {{end-eqn}} So $\phi$ is injective, thus bijective. Also: {{begin-eqn}} {{eqn | l = \map \phi {r_1 e^{i \theta_1} \times r_2 e^{i \theta_2} }       | r = \map \phi {r_1 r_2 e^{i \theta_1 + i \theta_2} }       | c = Product of Complex Numbers in Exponential Form }} {{eqn | r = \paren {r_1 r_2, e^{i \theta_1 + i \theta_2} }       | c = Definition of $\phi$ }} {{eqn | r = \paren {r_1 \times r_2, e^{i \theta_1} \times e^{i \theta_2} }       | c = Exponential of Sum }} {{eqn | r = \paren {r_1, e^{i\theta_1} } \times \paren {r_2, e^{i\theta_2} }       | c = {{Defof|Operation Induced by Direct Product|subdef = General Definition}} }} {{eqn | r = \map \phi {r_1 e^{i \theta_1} } \times \map \phi {r_2 e^{i \theta_2} }       | c = Definition of $\phi$ }} {{end-eqn}} {{explain|It is not clear how $\paren {r_1 \times r_2, e^{i \theta_1} \times e^{i \theta_2} } {{=}} \paren {r_1, e^{\theta_1} } \times \paren {r_2, e^{\theta_2} }$. The first $\times$ is the times of multiplication, real and complex, and the second appears to be the $\times$ of cartesian product. In any case, it needs to be established that $paren {a \times b, c \times d} {{=}} \tuple {a, c} \times \tuple {b, d}$ and in what context this holds -- and indeed, what it actually means.}} So $\phi$ is a group homomorphism. Since it is bijective, it is a group isomorphism. {{qed}} Category:Complex Numbers Category:External Direct Products 3gg7vhxarvgm66j64kqwotvpxn57vdq
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$. Then: {{begin-eqn}} {{eqn | l = \mathbf x \overline{\mathbf x}       | r = \left({a^2+b^2+c^2+d^2}\right)\mathbf 1       | c = Product of Quaternion with Conjugate }} {{eqn | r = \left\vert \mathbf x \right\vert ^2 \mathbf 1       | c = Definition of Quaternion Modulus }} {{end-eqn}} {{qed}} Category:Quaternion Modulus hjk6fftadnyj046lmgg4czh6x3agskf
Here we adopt the definition of $\ln x$ to be: :$\displaystyle \ln x := \int_1^x \dfrac {\mathrm d t} t$ {{begin-eqn}} {{eqn | l = \ln \left({x^r}\right)       | r = \int_1^{x^r} \dfrac {\mathrm d t} t       | c = {{Defof|Natural Logarithm}} }} {{eqn | r = \int_1^x \dfrac {r t^{r - 1} \mathrm d t} {t^r}       | c = Integration by Substitution: $t\mapsto t^r$, $\mathrm d t \mapsto r t^{r - 1} \mathrm d t$, $1 \mapsto 1$, $x^r\mapsto x$ }} {{eqn | r = r \int_1^x \dfrac {\mathrm d t} t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = r \ln x       | c = {{Defof|Natural Logarithm}} }} {{end-eqn}} {{qed}}
=== Proof of $\paren 1$ === It follows directly from Complex Numbers form Subfield of Quaternions and Complex Multiplication is Commutative. {{qed|lemma}} === Proof of $\paren 2$ === Let $\mathbf x \in \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a\in \R}$. Let $\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k: e, f, g, h \in \R$. Then: {{begin-eqn}} {{eqn | l = \mathbf x \times \mathbf y        | r = a e \mathbf 1 + a f \mathbf i + a g \mathbf j + a h \mathbf k       | c = {{Defof|Quaternion Multiplication}}; the other terms are $0$ }} {{eqn | r = \mathbf y \times \mathbf x       | c = {{Defof|Quaternion Multiplication}} }} {{end-eqn}} The above is the proof of $\paren {2a}$, and the proof of $\paren {2b}$ is similar. {{qed|lemma}} === Proof of $\paren 3$ === {{begin-eqn}} {{eqn | l = \mathbf x \times \mathbf y       | r = \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y} \times \mathbf y       | c =  }} {{eqn | r = \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y} \times \overline {\paren {\overline {\mathbf y} } }       | c = Quaternion Conjugation is Involution }} {{eqn | r = \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \cmod {\overline{\mathbf y} }^2 \mathbf 1       | c = Quaternion Modulus in Terms of Conjugate }} {{eqn | r = \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times  \cmod {\mathbf y}^2 \mathbf 1       | c = Quaternion Modulus of Conjugate }} {{eqn | r = \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \mathbf y \times \overline {\mathbf y}       | c = Quaternion Modulus in Terms of Conjugate }} {{eqn | r = \mathbf y \times \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y}       | c = from $\paren 2$ }} {{eqn | r = \mathbf y \times \mathbf x       | c =  }} {{end-eqn}} {{qed}} Category:Quaternions bcp2yc29muagbt0j5sjmkgnju4nqbpm
Let $y \in \R_{>0}$ be fixed. Consider the function: :$\map f x = \ln x y - \ln x$ From the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule for Derivatives: :$\forall x > 0: \map {f'} x = \dfrac 1 {x y} y - \dfrac 1 x = \dfrac 1 x - \dfrac 1 x = 0$ Thus from Zero Derivative implies Constant Function, $f$ is constant: :$\forall x > 0: \ln x y - \ln x = c$ To determine the value of $c$, put $x = 1$. From Logarithm of 1 is 0: :$\ln 1 = 0$ Thus: :$c = \ln y - \ln 1 = \ln y$ and hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \ln x + \ln y       | r = \int_1^x \dfrac {\d t} t + \int_1^y \dfrac {\d s} s       | c = {{Defof|Natural Logarithm}} }} {{eqn | r = \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t / x} {t / x}       | c = Integration by Substitution: $s \mapsto t / x$, $\d s \mapsto \d t / x$, $1 \mapsto x$, $y \mapsto x y$ }} {{eqn | r = \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t} t       | c = simplifying }} {{eqn | r = \int_1^{x y} \dfrac {\d t} t       | c = Sum of Integrals on Adjacent Intervals for Continuous Functions }} {{eqn | r = \ln x y       | c = {{Defof|Natural Logarithm}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\sinh x}       | r = \map {\frac \d {\d x} } {\dfrac {e^x - e ^{-x} } 2}       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac 1 2 \paren {\map {\frac \d {\d x} } {e^x} - \map {\frac \d {\d x} } {e^{-x} } }       | c = Linear Combination of Derivatives }} {{eqn | r = \frac 1 2 \paren {e^x - \paren {-e^{-x} } }       | c = Derivative of Exponential Function, Chain Rule for Derivatives }} {{eqn | r = \frac {e^x + e^{-x} } 2       | c = simplification }} {{eqn | r = \cosh x       | c = {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\sinh x}       | r = \lim_{h \mathop \to 0} \frac {\map \sinh {x + h} - \sinh x} h       | c = {{Defof|Derivative}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cosh {\frac {x + h + x} 2} \map \sinh {\frac {x + h - x} 2} } h       | c = Hyperbolic Sine minus Hyperbolic Sine }} {{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cosh {x + \frac h 2} \map \sinh {\frac h 2} } h       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map \cosh {x + \frac h 2} \map \sinh {\frac h 2} } {\frac h 2}       | c =  }} {{eqn | r = \lim_{2 d \mathop \to 0} \frac {\map \cosh {x + d} \map \sinh d} d       | c = where $d = \dfrac h 2$ }} {{eqn | r = \lim_{d \mathop \to 0} \frac {\map \cosh {x + d} \map \sinh d} d       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {\map \sinh d} d       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {e^d - e^{-d} } {2 d}       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {e^{2 d} - 1 } {2 d e^d}       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \frac {e^{2 d} - 1} {2 d}       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \lim_{d \mathop \to 0} \frac {e^{2 d} - 1} {2 d}       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \lim_{2 d \mathop \to 0} \frac {e^{2 d} - 1} {2 d}       | c =  }} {{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d}       | c = Derivative of Exponential at Zero }} {{eqn | r = \cosh x       | c =  }} {{end-eqn}} {{qed}}
For each $n \in P$, define $\map A n$ as: :$\map A n = \set {h: P \to T \mid \map h 0 = a \land \forall m < n: \map h {\map s n} = \map g {\map h m} }$ {{MissingLinks|Ordering $<$ on Peano Structure}} First, we prove for all $n \in P$ that $\map A n$ is not empty. More formally, we prove that $A = \set {n \in P: \map A n \ne \O} = P$. For $n = 0$, there are no $m < n$, so any mapping $h: P \to T$ with $\map h 0 = a$ is an element of $\map A 0$. Since the constant mapping $\map h n = a$ satisfies this condition, it follows that $0 \in A$. Now suppose that $n \in A$, and let $h \in \map A n$. Define $h': P \to T$ by, for all $n' \in P$: :$\map {h'} {n'} = \begin {cases} \map h {n'} & : n' \le n \\ \map g {\map h n} & : n' > n \end {cases}$ To check that $h' \in \map A {\map s n}$, we have to verify that: :$\forall m < \map s n: \map {h'} {\map s m} = \map g {\map {h'} m}$ Since $h \in \map A n$, only the case $m = n$ needs to be verified: {{begin-eqn}} {{eqn | l = \map {h'} {\map s n}       | r = \map g {\map h n}       | c = as $\map s n > n$ }} {{eqn | r = \map g {\map {h'} n}       | c = Definition of $h'$ }} {{end-eqn}} Therefore $h' \in \map A {\map s n}$, and so $\map A {\map s n} \ne \O$. That means $\map s n \in A$, and by Axiom $(\text P 5)$, we conclude $A = P$. Now let $A' = \set {n \in P: \forall h, h' \in \map A n: \map h n = \map {h'} n}$. Then by definition of $\map A )$, it follows that $0 \in A'$. Suppose now that $n \in A'$, and let $h, h' \in \map A {\map s n}$. Then: {{begin-eqn}} {{eqn | l = \map h {\map s n}       | r = \map g {\map h n}       | c = as $h \in \map A {\map s n}$ }} {{eqn | r = \map g {\map {h'} n}       | c = as $h, h' \in \map A n$ }} {{eqn | r = \map {h'} {\map s n}       | c = as $h \in \map A {\map s n}$ }} {{end-eqn}} Hence $\map s n \in A'$, and by Axiom $(\text P 5)$, we conclude $A' = P$. Because any $f: \N \to T$ as in the theorem statement needs to be in all $\map A n$, it follows that such an $f$ is necessarily unique. Finally, we can define $f: P \to T$ by: :$\map f n = \map {h_n} n$ where $h_n \in \map A n$. It is immediate from the definition of the $\map A n$ that: :$\forall m, n, m < n: \map A n \subseteq \map A m$ Hence, for every $m, n$ such that $m < n$: :$\map {h_m} m = \map {h_n} n$ Thus, for all $m < n$: :$\map f m = \map {h_n} m$ Since $h_n \in \map A n$, it follows that also: :$f \in \map A n$ Thus, since $n$ was arbitrary, it follows that for all $n \in P$: :$\map f {\map s n} = \map g {\map f n}$ as desired. {{qed}} Category:Principle of Recursive Definition Category:Mapping Theory Category:Peano's Axioms 08ideooijeeu89e0f7hx1sewzyu595e
Take the function $F$ generated in the second principle of transfinite recursion. Set $f = F {\restriction_\omega}$. {{begin-eqn}} {{eqn | l = \map f \O       | r = \map F \O       | c = $\O \in \omega$ }} {{eqn | l = \map f {n^+}       | r = \map F {n^+}       | c = $n^+ \in \omega$ }} {{eqn | r = \map g {\map F n}       | c = Second Principle of Transfinite Recursion }} {{eqn | r = \map g {\map f n}       | c = $n \in \omega$ and the definition of $f$ }} {{end-eqn}} Therefore, such a function exists. Now, suppose there are two functions $f$ and $f'$ that satisfy this: :$\map f \O = \map {f'} \O$ Then: {{begin-eqn}} {{eqn | l = \map f {n^+}       | r = \map g {\map f n}       | c = By Hypothesis }} {{eqn | r = \map g {\map {f'} n}       | c = Inductive Hypothesis }} {{eqn | r = \map {f'} {n^+}       | c = By Hypothesis }} {{end-eqn}} This completes the proof. {{qed}} </onlyinclude>
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Axiom $E$ states: :$\forall a, b \in \N_{>0}$, either: ::$a = b$, in which case $(1)$ holds ::$\exists x \in \N_{> 0}: a = b + x$, in which case, by definition of the ordering defined, $a > b$, in which case $(2)$ holds ::$\exists x \in \N_{> 0}: a + x = b$, in which case, by definition of the ordering defined, $a < b$, in which case $(3)$ holds. Hence the result. {{qed}}
Consider the matrix form of $\mathbf x$ and $\mathbf y$: :$\mathbf x = \begin{bmatrix} a & b \\ -\overline b & \overline a \end{bmatrix}$ :$\mathbf y = \begin{bmatrix} c & d \\ -\overline d & \overline c \end{bmatrix}$ where $a, b, c, d \in \C$. {{begin-eqn}} {{eqn | l = \overline{\mathbf x \times \mathbf y}       | r = \overline{\begin{bmatrix} a & b \\ -\overline b & \overline a \end{bmatrix} \begin{bmatrix} c & d \\ -\overline d & \overline c \end{bmatrix} }       | c =  }} {{eqn | r = \overline{\begin{bmatrix} a c - b \overline d & a d + b \overline c \\ - \overline b c - \overline a \overline d & - \overline b d + \overline a \overline c \end{bmatrix} }       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} \overline{a c - b \overline d} & -\left({a d + b \overline c}\right) \\ \overline b c + \overline a \overline d & \overline{- \overline b d + \overline a \overline c} \end{bmatrix}       | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}} }} {{eqn | r = \begin{bmatrix} \overline a \overline c - \overline b d & - a d - b \overline c \\ \overline {a d} + \overline b c & a c - b \overline d \end{bmatrix}       | c = Complex Conjugation is Automorphism, Complex Conjugation is Involution }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \overline{\mathbf y} \times \overline{\mathbf x}       | r = \overline{\begin{bmatrix} c & d \\ -\overline d & \overline c \end{bmatrix} } \ \overline{\begin{bmatrix} a & b \\ -\overline b & \overline a \end{bmatrix} }        | c =  }} {{eqn | r = \begin{bmatrix} \overline c & - d \\ \overline d & c \end{bmatrix} \begin{bmatrix} \overline a & - b \\ \overline b & a \end{bmatrix}       | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}} }} {{eqn | r = \begin{bmatrix} \overline a \overline c - \overline b d & - a d - b \overline c \\ \overline {a d} + \overline b c & a c - b \overline d \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \overline{\mathbf x \times \mathbf y}       | c = from above }} {{end-eqn}} but: {{begin-eqn}} {{eqn | l = \overline{\mathbf x} \times \overline{\mathbf y}       | r = \overline{\begin{bmatrix} a & b \\ -\overline b & \overline a \end{bmatrix} } \ \overline{\begin{bmatrix} c & d \\ -\overline d & \overline c \end{bmatrix} }       | c =  }} {{eqn | r = \begin{bmatrix} \overline a & - b \\ \overline b & a \end{bmatrix} \times \begin{bmatrix} \overline c & - d \\ \overline d & c \end{bmatrix}       | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}} }} {{eqn | r = \begin{bmatrix} \overline a \overline c - b \overline d & - \overline a d - b c \\ a \overline d + \overline {b c} & a c - \overline b d \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | o = \ne       | r = \overline{\mathbf x \times \mathbf y}       | c = from above }} {{end-eqn}} {{qed}}
Let $\mathbf x, \mathbf y$ be in their matrix form. Then: {{begin-eqn}} {{eqn | l = \left\vert{\mathbf x \mathbf y}\right\vert       | r = \sqrt{\det \left({\mathbf x \mathbf y}\right)}       | c = Definition of Modulus }} {{eqn | r = \sqrt{\det \left({\mathbf x}\right) \det \left({\mathbf y}\right)}       | c = Determinant of Matrix Product }} {{eqn | r = \sqrt{\det \left({\mathbf x}\right)} \sqrt{\det \left({\mathbf y}\right)}       | c = Exponent Combination Laws }} {{eqn | r = \left\vert{\mathbf x}\right\vert \left\vert{\mathbf y}\right\vert       | c = Definition of Modulus }} {{end-eqn}} {{qed}} Category:Quaternion Modulus j6tnwbfmkvn4zri49y8437bsd1g3eya
{{begin-eqn}} {{eqn | o =       | r = 2 \, \map \sinh {\frac {x + y} 2} \, \map \sinh {\frac {x - y} 2}       | c =  }} {{eqn | r = 2 \, \frac {\map \cosh {\dfrac {x + y} 2 + \dfrac {x - y} 2} - \map \cosh {\dfrac {x + y} 2 - \dfrac {x - y} 2} } 2       | c = Simpson's Formula for Hyperbolic Sine by Hyperbolic Sine }} {{eqn | r = \cosh \frac {2 x} 2 - \cosh \frac {2 y} 2       | c =  }} {{eqn | r = \cosh x - \cosh y       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sinh x \sinh y       | r = i^2 \map \sin {\frac x i} \map \sin {\frac y i}       | c = Sine in terms of Hyperbolic Sine }} {{eqn | r = -\map \sin {\frac x i} \map \sin {\frac y i}       | c = $i^2 = -1$ }} {{eqn | r = -\frac {\map \cos {\frac x i - \frac y i} - \map \cos {\frac x i + \frac y i} } 2       | c = Simpson's Formula for Sine by Sine }} {{eqn | r = \frac {\map \cos {\frac x i + \frac y i} - \map \cos {\frac x i - \frac y i} } 2       | c = simplifying }} {{eqn | r = \frac {\map \cosh {x + y} - \map \cosh {x - y} } 2       | c = Cosine in terms of Hyperbolic Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \frac {\map \cosh {x + y} - \map \cosh {x - y} } 2 }} {{eqn | r = \frac {\paren {\cosh x \cosh y + \sinh x \sinh y} - \map \cosh {x - y} } 2       | c = Hyperbolic Cosine of Sum }} {{eqn | r = \frac {\paren {\cosh x \cosh y + \sinh x \sinh y} - \paren {\cosh x \cosh y - \sinh x \sinh y} } 2       | c = Hyperbolic Cosine of Difference }} {{eqn | r = \frac {2 \sinh x \sinh y} 2 }} {{eqn | r = \sinh x \sinh y }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \overline {\paren {\overline x} }       | r = \overline {\paren {\overline {\tuple {a, b} } } }       | c = {{Defof|Octonion}} }} {{eqn | r = \overline {\tuple {\overline a, -b} }       | c = {{Defof|Conjugate of Octonion}} }} {{eqn | r = \tuple {\overline {\paren {\overline a} }, -\paren {-b} }       | c = {{Defof|Conjugate of Octonion}} }} {{eqn | r = \tuple {a, b}       | c = Quaternion Conjugation is Involution }} {{eqn | r = x       | c = {{Defof|Octonion}} }} {{end-eqn}} {{qed}} Category:Octonions Category:Involutions sr6an7ebrb6i8b07h018s11isrpfhg5
{{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \implies       | l = \exists c \in \N_{>0}: a       | r = b + c       | c = Definition of Ordering on $1$-Based Natural Numbers }} {{eqn | ll= \implies       | l = a + n       | r = \left({b + c}\right) + n       | c = Definition of binary operation }} {{eqn | r = b + \left({c + n}\right)       | c = Natural Number Addition is Associative }} {{eqn | r = b + \left({n + c}\right)       | c = Natural Number Addition is Commutative }} {{eqn | r = \left({b + n}\right) + c       | c = Natural Number Addition is Associative }} {{eqn | ll= \implies       | l = a + n       | o = <        | r = b + n       | c = Definition of Ordering on $1$-Based Natural Numbers }} {{end-eqn}} {{qed}}