0

i have to find all functions $f: \mathbb{Q} \rightarrow \mathbb{R}$, such that $f(x+y)=f(x)+f(y)$.

So functions of the form $f(x) := ax, a \in \mathbb{R}$ satisfy the above condition: $$ f(x+y)=a(x+y)=ax+ay=f(x)+f(y) $$

But how do i proove that all functions that satisfy the above condition have the form $f(x) := ax, a \in \mathbb{R}$?

Thanks in advance!

Gaste
  • 1,339
  • Set $a=f(1)$ and draw consequences from the fact that you know $f(m)$ for all integers $m$ and that $f(n\frac{m}{n})=f(m)$. – egreg Oct 23 '13 at 08:33

1 Answers1

1

Let $f:\mathbb{Q}\rightarrow\mathbb{R}$ be such that $f(a+b)=f(a)+f(b)$ for all $a,b\in\mathbb{Q}$. It should be easy for you to prove the following:

  • for every $m\in\mathbb{Z}$, $p\in\mathbb{Q}$, $f(m\cdot p)=m\cdot f(p)$ (use induction on $m$).

From that, conclude that $f(m^{-1})=m^{-1}f(1)$ for every $m\in\mathbb{Z}\setminus\left\{0\right\}$.

Now, given $q\in\mathbb{Q}$, write $q=m/n$, with $m,n\in\mathbb{Z}$, $n\neq 0$. Then, $$f(q)=f(m/n)=m\cdot f(n^{-1})=m\cdot n^{-1}f(1)=q\cdot f(1).$$

Concluding, $f$ is of the form $f(x)=f(1)\cdot x$.

Luiz Cordeiro
  • 18,513