Weird Functional Equation problem on the irrationals

Question

This weird question was given by my professor as a part of my assignment :

$\textbf{Question :}$

“Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that it satisfies $f(x + y) = x f(\frac{1}{y}) + y f(\frac{1}{x})$, whenever $x$ and $y$ are both irrational numbers. Then prove that $f(0)$ is always $0$.”

$\textbf{My attempt :}$

Since $0$ is rational, we can’t let any of the variables to be $0$. So I tried the substitution $y = -x$ to get $f(0) = x f(\frac{1}{-x}) - x f(\frac{1}{x})$.

I tried proving f to be an even function after this, so that $f(0)$ becomes $0$. But I couldn’t do so or proceed further anyhow.

Can somebody kindly provide me hints or solutions for this problem ?

Answer 1

With $x = y = t$ (any irrational) you get $$ f(2t) = 2 t f(1/t) $$ and with $x = y = 1/(2t)$ you get $$f(1/t) = f(2t)/t$$ Substitute the first into the second and it says $$ f(1/t) = 2 f(1/t)$$ from which you conclude $f(1/t) = 0$, i.e. $f$ is $0$ on all irrationals. Then use $y = -x$ to get $f(0) = 0$.

Answer 2

Here is my solution, after taking suggestions from @RobertIsrael in the comments.

Let $x = y = \frac{1}{\sqrt{2}}$, in the original equation. After simple algebra, we get $f(\sqrt{2}) = 0$. Similarly we get $f(-\sqrt{2}) = 0$ by taking $x = y = \frac{-1}{\sqrt{2}}$ . Now let $x = \frac{1}{\sqrt{2}}$ and $y = −x$ to conclude.