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Let $f:\mathbb{Q}^+\to \mathbb{Q}^+$ be a function such that $$f(x)+f\left(\frac1x\right)=1$$ and $$f(2x)=f\bigl(f(x)\bigr)$$ for all $x\in \mathbb{Q}^+$. Prove that $$f(x)=\frac{x}{x+1}$$ for all $x\in \mathbb{Q}^+$.

This problem is from my student.

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  • $\begingroup$ It might help to prove along the way that $f(x)=xf(1/x)$. $\endgroup$
    – awwalker
    Commented Mar 14, 2013 at 1:36

1 Answer 1

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Some ideas:

$$\text{I}\;\;\;\;x=1\Longrightarrow f(1)+f\left(\frac{1}{1}\right)=2f(1)=1\Longrightarrow \color{red}{f(1)=\frac{1}{2}}$$

$$\text{II}\;\;\;\;\;\;\;\;f(2)=2f(f(1))=2f\left(\frac{1}{2}\right)$$

But we also know that

$$f(2)+f\left(\frac{1}{2}\right) =1$$

so from II we get

$$3f\left(\frac{1}{2}\right)=1\Longrightarrow \color{red}{f\left(\frac{1}{2}\right)=\frac{1}{3}}\;,\;\;\color{red}{f(2)=\frac{2}{3}}$$

and also:

$$ \frac{1}{2}=f(1)=f\left(2\cdot \frac{1}{2}\right)=2f\left(f\left(\frac{1}{2}\right)\right)=2f\left(\frac{1}{3}\right)\Longrightarrow \color{red}{f\left(\frac{1}{3}\right)=\frac{1}{4}}\;,\;\color{red}{f(3)=\frac{3}{4}}$$

One more step:

$$f(4)=f(2\cdot2)=2f(f(2))=2f\left(\frac{2}{3}\right)=2\cdot 2f\left(f\left(\frac{1}{3}\right)\right)=4f\left(\frac{1}{4}\right)$$

and thus:

$$1=f(4)+f\left(\frac{1}{4}\right)=5f\left(\frac{1}{4}\right)\Longrightarrow \color{red}{f\left(\frac{1}{4}\right)=\frac{1}{5}}\;,\;\;\color{red}{f(4)=\frac{4}{5}}$$

...and etc.

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