Proving Inverse Functions: Multiplicative Relationships

[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!