# Proving that $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$ imply $f(x) = 0$ or $f(x) = x$ [duplicate]

Question: If $$f$$ is a function such that $$f(x+y) = f(x)+f(y) \qquad f(xy) = f(x)f(y)$$ for all $$x$$, then prove that $$f(x) = 0$$ or $$f(x) = x$$ for all $$x$$. (The fact that every positive number is a square of some number will be important.)

My attempt: I was able to show that this is satisfied for rational $$x$$ but we cant assume $$f$$ is continuous so can we prove this for all real $$x$$? Also, if I'm not mistaken, can I assume that $$f(x) \ne 0$$ or $$f(x) \ne x$$ for all $$x$$ and then prove that $$f(x) = x$$ or $$f(x) = 0$$ for all $$x$$ respectively?

• Consider $y=0$? Commented Sep 17, 2017 at 20:05
• If $f$ is a continuous function that it's obvious. Commented Sep 17, 2017 at 20:08

We have $f(1)=f(1)^2$, so $f(1)=1$ or $f(1)=0$. Moreover, $f(x)=f(1)x$ for rational $x$. Since $f(x^2)=f(x)^2\ge0$, we have $f(y)\ge0$ if $y\ge0$. But for $y>0$, $f(x+y)=f(x)+f(y)\ge f(x)$, so $f$ is monotone, i.e. $f(x)=f(1)x$ for all real $x$.