# If $f$ is a continuous function and $f(a + b) = f(a) + f(b)$, how do I prove that $f(x) = mx$ where $m=f(1)$? [duplicate]

If $$f$$ is a continuous function and $$f ( a + b ) = f ( a ) + f ( b )$$, how do I prove that $$f ( x ) = m x$$ for any $$x$$ in real numbers, where $$m = f ( 1 )$$?

I know that I have to start by showing $$f ( x ) = m x$$ for any rational $$x$$ and then extend that to any real number with continuity. However, I do not know how to go about it.

• Commented Jul 12, 2020 at 23:59

First use induction to show $$f(n)=nm$$ for $$n\in\Bbb N$$. Next, $$f(0)=f(0)+f(0)$$ implies $$f(0)=0$$, so $$0=f(1-1)=m+f(-1)$$. This tells us $$f(-1)=-m$$, and now we can show by induction $$f(n)=mn$$ for any $$n\in\Bbb Z$$.
Next, we claim that, for any rational number $$p/q$$, $$f(p/q)=mp/q$$. We can assume $$q$$ is a positive integer, and then
$$mp=f(p)=f(qp/q)=f(p/q+p/q+\cdots+p/q)=qf(p/q),$$
which gives the result. Now use the fact that $$f$$ is continuous to extend the result to every real number.