real analysis - Continuous and additive implies linear

The following problem is from Golan's linear algebra book. I have posted a solution in the comments.

Problem: Let $f(x):\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function satisfying $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}$. Show $f$ is a linear transformation.

Answer

The only property of linear transformations that we still need to verify is that $f(xt)=tf(x)$ for all $x,y\in \mathbb{R}.$ It is enough to establish this result just for rational numbers. If $j$ is irrational and $x\in \mathbb{R}$,, we can find a rational $r$ with $|jx-rx|<\delta$ for any positive real $\delta$. By continuity, for every $\epsilon>0$, we can choose $\delta$ so that $|f(rx)-f(jx)|<\epsilon$. This condition also gives $|r-j|<\delta/|j|$, and choosing $\delta$ to be even smaller if necessary gives $|f(x)-f(r)|<\epsilon$, too. Putting this all together gives

$|jf(x)-f(jx)|<|j|f(x)-f(r)| + |f(jx)-f(rx)|<(|j|+1)\epsilon$

and we can make this arbitrarily small, giving the desired result.

To verify the property for rationals, we first verify it for integers. If $n\in \mathbb{N}$, then

$nf(x)=f(x)+f(x)+...+f(x)=f(nx)$

by hypothesis. Also,

$f(x)=f(x/n)+f(x/n)+\cdots f(x/n)=nf(x/n)$

so $\frac{1}{n}f(x)=f(\frac{x}{n})$. Combining the above shows we have scalar multiplication for all positive rationals.

Noting that $f(0)=f(0)+f(0)$ gives $f(0)=0$, and

$f(0)=f(-x)+f(x)\Rightarrow -f(-x)=f(x)$. Using this allows us to extend scalar multiplication to negative rationals and completes the proof.