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**multiplicative inverses**of each other, then ##f^{-1} (x) = g^{-1} (\frac{1}{x})##?

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

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pasmith

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Let [itex]h[/itex] be the function which takes [itex]x[/itex] to [itex]1/x[/itex]. Now if [itex]f(x)g(x) = 1[/itex] for all [itex]x[/itex] then [itex]f = h \circ g[/itex]. Then [itex]f^{-1} = g^{-1} \circ h^{-1}[/itex]. But [itex]h = h^{-1}[/itex] so [itex]f^{-1} = g^{-1} \circ h[/itex] as required.multiplicative inversesof each other, then ##f^{-1} (x) = g^{-1} (\frac{1}{x})##?

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

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