Let f(x+y)=f(x)+f(y) where x, y, f(x) are all rational numbers. Show that f(qx)=qf(x) for all q, x rational.
The Attempt at a Solution
It's easy to get to f(nx)=nf(x). Because f(x+...+x)=f(x)+...+f(x)=nf(x). But going from there to rational is hard. I think I've done it before, but I can't get it. I've been stumped.