# How can find this function by $x\in \mathbb{Q}^+$

Let $f:\mathbb{Q}^+\to \mathbb{Q}^+,f(x)+f(1/x)=1$ and $f(2x)=2f(f(x)),x\in \mathbb{Q}^+$, prove that $$f(x)=\dfrac{x}{x+1},x\in \mathbb{Q}^+$$ This Problem from my student.

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It might help to prove along the way that $f(x)=xf(1/x)$. – A Walker Mar 14 '13 at 1:36

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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