= Rational homogeneity of an additive function
{title2=$f(qx)=qf(x)\quad(q\in\mathbb Q)$}
For every <additive function> $f:\mathbb R\to\mathbb R$, every <real number> $x$, and every <rational number> $q$, one has $f(qx)=qf(x)$. This is the <scalar multiplication> part of being a <linear map> over the <rational numbers>.
First $f(0)=f(0)+f(0)$ gives $f(0)=0$, and $f(x)+f(-x)=f(0)$ gives $f(-x)=-f(x)$. Repeated <addition> gives $f(nx)=nf(x)$ for each positive <integer> $n$, and the negative case follows from the sign identity. If $q=m/n$ with $m\in\mathbb Z$ and $n>0$, then
$$
nf\left(\frac mn x\right)=f(mx)=mf(x),
\qquad
f\left(\frac mn x\right)=\frac mn f(x).
$$
No <continuity> was used. In particular $f(1)=3$ fixes $f(-2/3)=-2$, $f(5/6)=5/2$, and every other rational value, even for a <discontinuous additive function>.
Back to article page
