Let, $\displaystyle f(x)$ be a real valued function definited for all reals. Proof (or disproof) that is there exists a function such as $\displaystyle g(f(x))=x$ is true for all reals. Then, $\displaystyle f(g(x))=x$ is always true.

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- Mar 9th 2006, 06:31 PMThePerfectHackerCommutativity of Inverse Functions.
Let, $\displaystyle f(x)$ be a real valued function definited for all reals. Proof (or disproof) that is there exists a function such as $\displaystyle g(f(x))=x$ is true for all reals. Then, $\displaystyle f(g(x))=x$ is always true.

- Mar 10th 2006, 03:46 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

normal functiom, then $\displaystyle \mathbb{R}$ is mapped one one onto

$\displaystyle (0,1)$, and so there is an inverse function $\displaystyle g(y)$ from

$\displaystyle (0,1)$ onto $\displaystyle \mathbb{R}$

Then $\displaystyle f(g(x))=x$ holds for all $\displaystyle x \in (0,1)$ but not for all $\displaystyle x \in \mathbb{R}$

That is the domains of $\displaystyle f\circ g$ and $\displaystyle g \circ f$ are not equal.

RonL