So I am given a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all real $x$ and $y$, is continuous at $x=0$, and $f(1)=1$. I need to show that $f(x)=x$ for all real $x$.
I have proven that $f$ is continuous for all real $x$ from the fact that it is continuous at one point, and I am not sure where to go next. (My prof gave a hint that I should prove $f(r)=r$ for all rational $r$, but I'm not sure how to do that.)
Because is continuous in $x=0$ $$\lim_{y\to 0}f(x+y)=f(0)+f(x)$$ so, $f(x)=f(0)+f(x)$ and from here $f(0)=0$
Now if $n \in \Bbb Z, f(n)=nf(1)=n $
For $n \in \Bbb Q,f(\frac1n+\frac1n+...\frac1n)=nf(\frac1n).$
Here you see that $f(1)=nf(\frac1n)\Rightarrow f(\frac1n)=\frac1n \Rightarrow f(q)=q,q \in \Bbb Q^*$
For $x \in \Bbb I$ you know that x is an infinite sum of $q \in \Bbb Q$
Now the conclusion that $f(x)=x$ for all real.