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I am given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ that has the property that $f(u+v)=f(u)+f(v)$ for all $u,v\in\mathbb{R}$. Then we define $m=f(1)$ and I am asked to prove that $f(x)=mx$ for all rational numbers $x$.

This should be pretty straightforward but I can't quite seem to nug this one out.

What I wrote out so far is as follows: $$f(x)=f(\frac{p}{q})=f(\frac{1_1}{q}+...+\frac{1_p}{q})=f(\frac{1_1}{q})+...+f(\frac{1_p}{q})=pf(\frac{1}{q}).$$ This is where I am stuck.

Walt
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    Well, then $m=f(1)=f\left( \frac qq \right)=qf\left( \frac 1q \right)\implies f\left( \frac 1q \right)=m\frac 1q$. – lulu Apr 12 '18 at 20:33

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Continue the hint in the comment: $qf(1/q) = \underbrace{f(1/q) + \cdots + f(1/q)}_{\mbox{$q$ times}} = f(q/q) = f(1) = m$

$f(1/q) = m/q \implies f(p/q) = \underbrace{f(1/q) + \cdots + f(1/q)}_{\mbox{$p$ times}} = pf(1/q) = pm/q = m \cdot p/q$

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