hcju/proofwikiqa huggingface dataset:
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 - a^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}       | r = \int \frac {\mathrm d z} {2 z^2}       | c = Integration by Substitution }} {{eqn | r = \frac {-1} {2 z} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 \left({x^2 - a^2}\right)} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({x^2 - a^2}\right)^2}       | r = \int \frac {x^2 - a^2 + a^2} {\left({x^2 - a^2}\right)^2} \ \mathrm d x       | c =  }} {{eqn | r = \int \frac {x^2 - a^2} {\left({x^2 - a^2}\right)^2} \ \mathrm d x + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\mathrm d x} {x^2 - a^2} + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2}       | c = simplification }} {{eqn | r = \frac 1 {2 a} \left({\frac {x - a} {x + a} }\right) + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2} + C       | c = Primitive of Reciprocal of $x^2 - a^2$ }} {{eqn | r = \frac 1 {2 a} \left({\frac {x - a} {x + a} }\right) + a^2 \left({\frac {-x} {2 a^2 \left({x^2 - a^2}\right)} - \frac 1 {4 a^3} \left({\frac {x - a} {x + a} }\right)}\right) + C       | c = Primitive of Reciprocal of $\left({x^2 - a^2}\right)^2$ }} {{eqn | r = \frac {-x} {2 \left({x^2 - a^2}\right)} + \frac 1 {4 a} \ln \left({\frac {x - a} {x + a} }\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({x^2 - a^2}\right)^2}       | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {\left({x^2 - a^2}\right)^2} \ \mathrm d x       | c =  }} {{eqn | r = \int \frac {x \left({x^2 - a^2}\right)} {\left({x^2 - a^2}\right)^2} \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {x \ \mathrm d x} {x^2 - a^2} + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}       | c = simplification }} {{eqn | r = \frac 1 2 \ln \left({x^2 - a^2}\right) + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2} + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac 1 2 \ln \left({x^2 - a^2}\right) + a^2 \left({-\frac 1 {2 \left({x^2 - a^2}\right)} }\right) + C       | c = Primitive of Reciprocal of $\dfrac x {\left({x^2 - a^2}\right)^2}$ }} {{eqn | r = \frac {-a^2} {2 \left({x^2 - a^2}\right)} + \frac 1 2 \ln \left({x^2 - a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2}^2}       | r = \int \left({\frac 1 {a^4 x} + \frac {-x} {a^4 \paren {x^2 - a^2} } + \frac x {a^2 \paren {x^2 - a^2}^2} }\right) \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\d x} x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \paren {\frac 1 2 \map \ln {x^2 - a^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac 1 {a^4} \ln \size x  + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C       | c = Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$ }} {{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C       | c = Logarithm of Power and $x^2 > 0$ }} {{eqn | r = \frac {-1} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 - a^2} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)^2}       | r = \int \left({\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \left({x + a}\right)} - \frac 3 {4 a^5 \left({x - a}\right)} + \frac 1 {4 a^4 \left({x + a}\right)^2} + \frac 1 {4 a^4 \left({x - a}\right)^2} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\mathrm d x} {x^2} + \frac 3 {4 a^5} \int \frac {\mathrm d x} {x + a} - \frac 3 {4 a^5} \int \frac {\mathrm d x} {x - a} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\mathrm d x} {x + a} - \frac 3 {4 a^5} \int \frac {\mathrm d x} {x - a} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{x + a}\right\vert - \frac 3 {4 a^5} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{x + a}\right\vert - \frac 3 {4 a^5} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^4} \frac {-1} {x + a} + \frac 1 {4 a^4} \frac {-1} {x - a} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{\frac {x + a} {x - a} }\right\vert + \frac {-1} {4 a^4 \left({ x + a}\right)} + \frac {-1} {4 a^4 \left({x - a}\right)} + C       | c = Difference of Logarithms }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left({\frac {x + a} {x - a} }\right) + \frac {-1} {4 a^4 \left({ x + a}\right)} + \frac {-1} {4 a^4 \left({x - a}\right)} + C       | c = Sign of Quotient of Factors of Difference of Squares }} {{eqn | r = \frac {-1} {a^4 x} - \frac x {2 a^4 \left({x^2 - a^2}\right)} + \frac 3 {4 a^5} \ln \left({\frac {x + a} {x - a} }\right) + C       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2}       | r = \int \paren {\frac 1 {a^4 x^3} + \frac 2 {a^6 x} - \frac {2 x} {a^6 \paren {x^2 - a^2} } + \frac x {a^4 \paren {x^2 - a^2}^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \frac {-1} {2 x^2} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \frac 1 2 \map \ln {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C       | c = Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$ }} {{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2} - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C       | c = Logarithm of Power and $x^2 > 0$ }} {{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
Aiming for an expression in the form: : $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$ in order to use the technique of Integration by Parts, let: {{begin-eqn}} {{eqn | l = v       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d v} {\mathrm d x}       | r = 1       | c = Power Rule for Derivatives }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = u       | r = \frac 1 {\left({x^2 - a^2}\right)^{n - 1} }       | c =  }} {{eqn | r = \left({x^2 - a^2}\right)^{- \left({n - 1}\right)}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = - \left({n - 1}\right) \left({2 x}\right) \left({x^2 - a^2}\right)^{- \left({n - 1}\right) - 1}       | c = Power Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = - 2 \left({n - 1}\right) x \left({x^2 - a^2}\right)^{-n}       | c =  }} {{eqn | r = \frac {- \left({2 n - 2}\right) x} {\left({x^2 - a^2}\right)^n}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }       | r = \frac 1 {\left({x^2 - a^2}\right)^{n - 1} } x - \int x \frac {- \left({2 n - 2}\right) x} {\left({x^2 - a^2}\right)^n} \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x       | c = simplifying }} {{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 - a^2 + a^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x       | c =  }} {{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 - a^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}       | c = Linear Combination of Integrals }} {{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}        | c =  }} {{eqn | ll= \implies       | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }       | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}       | c =  }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}       | r = \frac {-x} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 - a^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int \frac {x \rd x} {\paren {x^2 - a^2}^n}       | r = \int \frac {\d z} {2 z^n}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int z^{-n} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \frac {z^{-n + 1} } {-n + 1} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 \paren {n - 1} z^{n - 1} } + C       | c = simplifying }} {{eqn | r = \frac {-1} {2 \paren {n - 1} \paren {x^2 - a^2}^{n - 1} } + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 - a^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n}       | r = \int \frac {\mathrm d z} {2 x^2 z^n}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z + a^2}\right) z^n}       | c = Primitive of Constant Multiple of Function and substituting for $x^2$ }} {{end-eqn}} From Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$: :$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$ Let: {{begin-eqn}} {{eqn | l = a       | o = :=       | r = 1       | c =  }} {{eqn | l = b       | o = :=       | r = a^2       | c =  }} {{eqn | l = x       | o = :=       | r = z       | c =  }} {{eqn | l = m       | o = :=       | r = -n       | c =  }} {{eqn | l = n       | o = :=       | r = -1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n}       | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z + a^2}\right) z^n}       | c =  }} {{eqn | r = \frac 1 2 \left({\frac {z^{-n+1} \left({z + a^2}\right)^{-1 + 1} } {\left({-n + 1}\right) a^2} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {\left({-n + 1}\right) a^2} \int z^{-n + 1} \left({z + a^2}\right)^{-1} \ \mathrm d z}\right)       | c =  }} {{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 z^{n-1} } - \frac 1 {2 a^2} \int \frac {\mathrm d z} {z^{n - 1} \left({z + a^2}\right)}       | c = simplification }} {{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n-1} } - \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({x^2 - a^2}\right)^{n - 1} }       | c = substituting for $z$ and $\mathrm d z$ }} {{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^{n - 1} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^m \rd x} {\paren {x^2 - a^2}^n}       | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {x^2 - a^2}^n}       | c =  }} {{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2 + a^2} \rd x} {\paren {x^2 - a^2}^n}       | c =  }} {{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2} \rd x} {\paren {x^2 - a^2}^n} + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^{n - 1} } + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }       | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{n - 1} \paren {x^2 - a^2} }       | c =  }} {{eqn | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{\paren {n - 1} + 1} }       | c =  }} {{eqn | r = \int \frac {x^2 \rd x} {x^m \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}        | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }       | c = changing sides }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}        | r = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \sgn {\frac {a - b} {a + b} }       | r = \map \sgn {a - b} \frac 1 {\map \sgn {a + b} }       | c = Signum Function is Completely Multiplicative }} {{eqn | r = \map \sgn {a - b} \map \sgn {a + b}       | c = Signum Function of Reciprocal }} {{eqn | r = \map \sgn {\paren {a - b} \paren {a + b} }       | c = Signum Function is Completely Multiplicative }} {{eqn | r = \map \sgn {a^2 - b^2}       | c = Difference of Two Squares }} {{eqn | r = \map \sgn {\paren {a + b} \paren {a - b} }       | c = Difference of Two Squares }} {{eqn | r = \map \sgn {a + b} \map \sgn {a - b}       | c = Signum Function is Completely Multiplicative }} {{eqn | r = \map \sgn {a + b} \frac 1 {\map \sgn {a - b} }       | c = Signum Function of Reciprocal }} {{eqn | r = \map \sgn {\frac {a + b} {a - b} }       | c = Signum Function is Completely Multiplicative }} {{end-eqn}} {{qed}} Category:Real Analysis Category:Signum Function ss7k87h3t6ty6ypqnd7cqi9f1xktmko
Let $x = 0$ or $y = 0$. Then: {{begin-eqn}} {{eqn | l = x y       | r = 0       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \map \sgn {x y}       | r = 0       | c =  }} {{end-eqn}} and either $\map \sgn x = 0$ or $\map \sgn y = 0$ and so: {{begin-eqn}} {{eqn | l = \map \sgn x \map \sgn y       | r = 0       | c =  }} {{eqn | r = \map \sgn {x y}       | c = from $(1)$ above }} {{end-eqn}} {{qed|lemma}} Let $x > 0$ and $y > 0$. Then: {{begin-eqn}} {{eqn | l = \map \sgn x       | r = 1       | c =  }} {{eqn | lo= \land       | l = \map \sgn y       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \map \sgn x \map \sgn y       | r = 1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = >       | r = 0       | c =  }} {{eqn | l = \map \sgn {x y}       | r = 1       | c ={{Defof|Signum Function}} }} {{eqn | r = \map \sgn x \map \sgn y       | c =  }} {{end-eqn}} {{qed|lemma}} Let $x < 0$ and $y < 0$. Then: {{begin-eqn}} {{eqn | l = \map \sgn x       | r = -1       | c =  }} {{eqn | lo= \land       | l = \map \sgn y       | r = -1       | c =  }} {{eqn | ll= \leadsto       | l = \map \sgn x \map \sgn y       | r = 1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = >       | r = 0       | c =  }} {{eqn | l = \map \sgn {x y}       | r = 1       | c = {{Defof|Signum Function}} }} {{eqn | r = \map \sgn x \map \sgn y       | c =  }} {{end-eqn}} {{qed|lemma}} Let $x < 0$ and $y > 0$. Then: {{begin-eqn}} {{eqn | l = \map \sgn x       | r = -1       | c =  }} {{eqn | lo= \land       | l = \map \sgn y       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \map \sgn x \map \sgn y       | r = -1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = <       | r = 0       | c =  }} {{eqn | l = \map \sgn {x y}       | r = -1       | c = {{Defof|Signum Function}} }} {{eqn | r = \map \sgn x \map \sgn y       | c =  }} {{end-eqn}} {{qed|lemma}} The same argument, mutatis mutandis, covers the case where $x > 0$ and $y < 0$. {{qed}} Category:Signum Function Category:Completely Multiplicative Functions 5e7i1rjnmnjqtw4i4x1qgjun1m0cife
Let $x \in \R_{\ne 0}$. Then either $x > 0$ or $x < 0$. Let $x > 0$. Then: {{begin-eqn}} {{eqn | l = \frac x {\left\vert{x}\right\vert}       | r = \frac x x       | c = Definition of Absolute Value, as $x > 0$ }} {{eqn | r = 1       | c =  }} {{eqn | r = \operatorname{sgn} \left({x}\right)       | c = Definition of Signum Function, as $x > 0$ }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \frac {\left\vert{x}\right\vert} x       | r = \frac x x       | c = Definition of Absolute Value, as $x > 0$ }} {{eqn | r = 1       | c =  }} {{eqn | r = \operatorname{sgn} \left({x}\right)       | c = Definition of Signum Function, as $x > 0$ }} {{end-eqn}} {{qed|lemma}} Let $x < 0$. Then: {{begin-eqn}} {{eqn | l = \frac x {\left\vert{x}\right\vert}       | r = \frac x {-x}       | c = Definition of Absolute Value, as $x < 0$ }} {{eqn | r = -1       | c =  }} {{eqn | r = \operatorname{sgn} \left({x}\right)       | c = Definition of Signum Function, as $x < 0$ }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \frac {\left\vert{x}\right\vert} x       | r = \frac {-x} x       | c = Definition of Absolute Value, as $x < 0$ }} {{eqn | r = -1       | c =  }} {{eqn | r = \operatorname{sgn} \left({x}\right)       | c = Definition of Signum Function, as $x < 0$ }} {{end-eqn}} {{qed}} Category:Signum Function Category:Absolute Value Function ijvb82dxkam36lfkeikjdgawoymdy7y
Let $x = 0$ or $y = 0$. Then: {{begin-eqn}} {{eqn | l = x y       | r = 0       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \left\vert{x y}\right\vert       | r = 0       | c =  }} {{end-eqn}} and either $\left\vert{x}\right\vert = 0$ or $\left\vert{y}\right\vert = 0$ and so: {{begin-eqn}} {{eqn | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | r = 0       | c =  }} {{eqn | r = \left\vert{x y}\right\vert       | c = from $(1)$ above }} {{end-eqn}} {{qed|lemma}} Let $x > 0$ and $y > 0$. Then: {{begin-eqn}} {{eqn | l = \left\vert{x}\right\vert       | r = x       | c =  }} {{eqn | lo= \land       | l = \left\vert{y}\right\vert       | r = y       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | r = x y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = >       | r = 0       | c =  }} {{eqn | l = \left\vert{x y}\right\vert       | r = x y       | c = Definition of Absolute Value }} {{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | c =  }} {{end-eqn}} {{qed|lemma}} Let $x < 0$ and $y < 0$. Then: {{begin-eqn}} {{eqn | l = \left\vert{x}\right\vert       | r = -x       | c =  }} {{eqn | lo= \land       | l = \left\vert{y}\right\vert       | r = -y       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | r = \left({-x}\right) \left({-y}\right)       | c =  }} {{eqn | r = x y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = >       | r = 0       | c =  }} {{eqn | l = \left\vert{x y}\right\vert       | r = x y       | c = {{Defof|Absolute Value}} }} {{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | c =  }} {{end-eqn}} {{qed|lemma}} Let $x < 0$ and $y > 0$. Then: {{begin-eqn}} {{eqn | l = \left\vert{x}\right\vert       | r = - x       | c =  }} {{eqn | lo= \land       | l = \left\vert{y}\right\vert       | r = y       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | r = - x y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x y       | o = <       | r = 0       | c =  }} {{eqn | l = \left\vert{x y}\right\vert       | r = - x y       | c = {{Defof|Absolute Value}} }} {{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert       | c =  }} {{end-eqn}} {{qed|lemma}} The same argument, mutatis mutandis, covers the case where $x > 0$ and $y < 0$. {{qed}}
Follows directly from: : Real Numbers form Ordered Integral Domain : Product of Absolute Values on Ordered Integral Domain. {{qed}}
We have the result Multiplicative Group of Reals is Subgroup of Complex. Therefore, any result applying to all complex numbers will also hold for all real numbers. The result follows from Complex Modulus of Product of Complex Numbers. {{qed}}
Let $z = \dfrac x y$. Then: {{begin-eqn}} {{eqn | l = z       | r = \dfrac x y       | c =  }} {{eqn | ll= \implies       | l = x       | r = y z       | c =  }} {{eqn | ll= \implies       | l = f \left({x}\right)       | r = f \left({y}\right) f \left({z}\right)       | c = as $f$ is completely multiplicative }} {{eqn | ll= \implies       | l = \frac {f \left({x}\right)} {f \left({y}\right)}       | r = f \left({z}\right)       | c = dividing by $f \left({y}\right)$ }} {{eqn | ll= \implies       | l = \frac {f \left({x}\right)} {f \left({y}\right)}       | r = f \left({\frac x y}\right)       | c = by definition of $z$ }} {{end-eqn}} {{qed}} Category:Multiplicative Functions qkds76zo6dwiwedvaofz6vsj1h1kkca
{{begin-eqn}} {{eqn | l = \map \sgn x       | r = 1       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = >       | r = 0       | c = {{Defof|Signum Function}} }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | o = >       | r = 0       | c = Reciprocal of Strictly Positive Real Number is Strictly Positive }} {{eqn | ll= \leadstoandfrom       | l = \map \sgn {\dfrac 1 x}       | r = 1       | c = {{Defof|Signum Function}} }} {{end-eqn}} {{qed}} Category:Signum Function 3fktxynn8ed1viz3c18ts1pma7lwj8f