=== Definition 1 implies Definition 2 === Let $x = \cosh y$, where $y > 0$. Let $z = e^y$. Then: {{begin-eqn}} {{eqn | l = x       | r = \frac {e^y + e^{-y} } 2       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | ll= \leadsto       | l = 2 x       | r = e^y + e^{-y}       | c =  }} {{eqn | ll= \leadsto       | l = 2 x e^y       | r = e^{2 y} + 1       | c =  }} {{eqn | ll= \leadsto       | l = z^2 - 2 x z + 1       | r = 0       | c = Power of Power }} {{eqn | ll= \leadsto       | l = z       | r = \frac {2 x \pm \sqrt {\paren {-2 x}^2 - 4} } 2       | c = Quadratic Formula }} {{eqn | r = x \pm \sqrt{x^2 - 1}       | c =  }} {{eqn | ll= \leadsto       | l = e^y       | r = x \pm \sqrt{x^2 - 1}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \map \ln {x \pm \sqrt {x^2 - 1} }       | c =  }} {{end-eqn}} Also, from Minimum of Real Hyperbolic Cosine Function: :$x = \cosh y \ge 1$ Also: {{begin-eqn}} {{eqn | l = x       | r = \sqrt {x^2}       | c = $x$ is positive }} {{eqn | o = >       | r = \sqrt {x^2 - 1}       | c = Square Root is Strictly Increasing }} {{eqn | ll= \leadsto       | l = x - \sqrt {x^2 - 1}       | o = >       | r = 0 }} {{end-eqn}} Thus $x - \sqrt {x^2 - 1}$ is (strictly) positive. {{AimForCont}} $x - \sqrt {x^2 - 1} > 1$. Then: {{begin-eqn}} {{eqn | l = x - \sqrt {x^2 - 1}       | o = >       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x - 1       | o = >       | r = \sqrt {x^2 - 1}       | c = Both sides are (strictly) positive }} {{eqn | ll= \leadsto       | l = \paren {x - 1}^2       | o = >       | r = x^2 - 1       | c = {{RHS}} is (strictly) positive because $x \ge 1$ }} {{eqn | ll= \leadsto       | l = x^2 - 2 x + 1       | o = >       | r = x^2 - 1       | c =  }} {{eqn | ll= \leadsto       | l = 2       | o = >       | r = 2 x       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = <       | r = 1       | c =  }} {{end-eqn}} and a contradiction is deduced. Therefore: :$x - \sqrt {x^2 - 1} < 1$ From Logarithm is Strictly Increasing: :$y = \map \ln {x - \sqrt {x^2 - 1} } < \ln 1 = 0$ Since $y$ is (strictly) positive from the first definition of real inverse hyperbolic cosine: :$y = \map \ln {x + \sqrt {x^2 - 1} }$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $z = x + \sqrt {x^2 - 1}$. Then: :$y = \ln z$ {{begin-eqn}} {{eqn | l = \map \cosh {\map \ln {x + \sqrt {x^2 - 1} } }       | r = \map \cosh {\ln z}       | c =  }} {{eqn | r = \frac {e^{\ln z} + e^{-\ln z} } 2       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac {z + \frac 1 z} 2       | c =  }} {{eqn | r = \frac {z^2 + 1} {2 z}       | c =  }} {{eqn | r = \frac {\paren {x + \sqrt {x^2 - 1} }^2 + 1} {2 x + 2 \sqrt {x^2 - 1} }       | c =  }} {{eqn | r = \frac {x^2 + 2 x \sqrt {x^2 - 1} + \paren {x^2 - 1} + 1} {2 x + 2 \sqrt {x^2 - 1} }       | c =  }} {{eqn | r = \frac {2 x^2 + 2 x \sqrt {x^2 - 1} } {2 x + 2 \sqrt {x^2 - 1} }       | c =  }} {{eqn | r = \frac {\paren {2 x + 2 \sqrt {x^2 - 1} } x} {2 x + 2 \sqrt {x^2 - 1} }       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} If $-1 < x < 1$, $z$ is not defined. If $x \le -1$: {{begin-eqn}} {{eqn | l = \sqrt {x^2 - 1}       | o = <       | r = \sqrt {x^2}       | c = Square Root is Strictly Increasing }} {{eqn | r = -x       | c = $x$ is negative }} {{eqn | ll= \leadsto       | l = x + \sqrt{x^2 - 1}       | o = <       | r = 0       | c =  }} {{end-eqn}} If $x \ge 1$, $z \ge 1$. Therefore, $y = \ln z \ge \ln 1 = 0$. {{qed|lemma}} Therefore: {{begin-eqn}} {{eqn | n = 1       | l = y > 0 \land x = \cosh y       | o = \implies       | r = y = \map \ln {x + \sqrt {x^2 + 1} }       | c = Definition 1 implies Definition 2 }} {{eqn | n = 2       | l = y = \map \ln {x + \sqrt {x^2 + 1} }       | o = \implies       | r = x = \cosh y \land y > 0       | c = Definition 2 implies Definition 1 }} {{eqn | ll= \leadsto       | l = y > 0 \land x = \cosh y       | o = \iff       | r = y = \map \ln {x + \sqrt {x^2 + 1} }       | c =  }} {{end-eqn}} {{qed}}