# Multiplicative function on rationals [duplicate]

Let $\Bbb Q^+$ be the set of positive rational numbers. Find all solutions $f:\Bbb Q^+ \to \Bbb R$ of the functional equation $$f(xy)=f(x)f(y), \quad x, y\in \Bbb Q.$$

Is $f(x)=x^a$ the only solution? If not, is it true if we assume that $f$ is continuous?

## marked as duplicate by user61527, Semsem, Claude Leibovici, Shuchang, user63181 Mar 13 '14 at 7:20

For example define $f(1)=1, f(p) = p$ for primes less than 100, and $f(p) = 1$ for primes $p > 100$, and extend it multiplicatively. This is in particular not a 1-1 function and satisfies your functional equation and is not of the form $f(x) = x^a$.
• Thank you very much for your perfect answer. Do you have an idea if we assume $f$ is continuous? – Chung. J Mar 13 '14 at 4:12