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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

Prove that the only function $f(x)$ satisfying:

(i) $f(x)$ is strictly increasing.

(ii) $f(x)=f^{-1}(x)$

is $f(x)=x$.

$f(x)=f^{-1}(x)$

$f(f(x))=x$

Since $f(x)$ is strictly increasing,

(for any $a$)

$a<f(a)\implies f(a)<f(f(a))=a$ (Contradiction)

$a>f(a)\implies f(a)>f(f(a))=a$ (Contradiction)

So, $f(a)=a$ for all $a$.

ie $f(x)=x$ is the only such function.

(Got it, never mind.)

(i) $f(x)$ is strictly increasing.

(ii) $f(x)=f^{-1}(x)$

is $f(x)=x$.

**My attempt:**$f(x)=f^{-1}(x)$

$f(f(x))=x$

Since $f(x)$ is strictly increasing,

(for any $a$)

$a<f(a)\implies f(a)<f(f(a))=a$ (Contradiction)

$a>f(a)\implies f(a)>f(f(a))=a$ (Contradiction)

So, $f(a)=a$ for all $a$.

ie $f(x)=x$ is the only such function.

(Got it, never mind.)

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