I want to show that $f(x)=x.f(1)$ where $f:R\to R$ is additive. [duplicate]
HINTS:
Look at $0$ first: $f(0)=f(0+0)=f(0)+f(0)$, so $f(0)=0=0\cdot f(1)$.
Use induction to prove that $f(n)=nf(1)$ for every positive integer $n$, and use $f(0)=0$ to show that $f(n)=nf(1)$ for every negative integer as well.
$f(1)=f\left(\frac12+\frac12\right)=f\left(\frac13+\frac13+\frac13\right)=\dots\;$.
Once you’ve got it for $f\left(\frac1n\right)$, use the idea of (2) to get it for all rationals.
Then use continuity at a point.