Problem - 3826
The function $f$ satisfies $f(0)=0$, $f(1)=1$, and $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$ for all $x,y\in\mathbb{R}$. Show that $f(x)=x$ for all rational numbers $x$.
Let $y=0$, then $f\left(\frac{x}{2}\right)=\frac{f\left(x\right)}{2}$, so \[f\left(x\right)+f\left(y\right)=2f\left(\frac{x+y}{2}\right)=f\left(x+y\right)\]for all $x,y\in\mathbb{R}$. Cauchy (see %%HREF%%3643%%) gives us that $f\left(x\right)=f\left(1\right)x$ for all $x\in\mathbb{Q}$, so since $f\left(1\right)=1$, we have that $f\left(x\right)=x$ for all rational numbers $x$.