Inverse functions

[PFuser1232]

Is there a way to formally prove that if $f$ and $g$ are multiplicative inverses of each other, then $f^{-1} (x) = g^{-1} (\frac{1}{x})$?

 

[Dr. Courtney]

I'd start by testing a number of special cases and see what approach the trickier special cases suggest.

Simple special cases almost never suggest approaches to proof, but the tricky special cases often do.

 

[pasmith]

Is there a way to formally prove that if $f$ and $g$ are multiplicative inverses of each other, then $f^{-1} (x) = g^{-1} (\frac{1}{x})$?

Let $h$ be the function which takes $x$ to $1/x$. Now if $f(x)g(x) = 1$ for all $x$ then $f = h \circ g$. Then $f^{-1} = g^{-1} \circ h^{-1}$. But $h = h^{-1}$ so $f^{-1} = g^{-1} \circ h$ as required.

 

[PFuser1232]

Let $h$ be the function which takes $x$ to $1/x$. Now if $f(x)g(x) = 1$ for all $x$ then $f = h \circ g$. Then $f^{-1} = g^{-1} \circ h^{-1}$. But $h = h^{-1}$ so $f^{-1} = g^{-1} \circ h$ as required.

Perfect. Thanks!