# Looking for a counterexample [duplicate]

Possible Duplicate:
Solution(s) to $f(x + y) = f(x) + f(y)$ (and miscellaneous questions…)

I am looking for a function $f:\mathbb{R}\rightarrow \mathbb{R}$ that for all $x$ and $y$ satisfies $f(x+y)=f(x)+f(y)$, but does not satisfy $f(\alpha x)=\alpha f(x)$ for al real $x$ and $\alpha$. I know that for rational $\alpha$ this property has to be satisfied, but can you provide an example where the property is not satisfied for irrational $\alpha$?

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## marked as duplicate by Haskell Curry, Chris Eagle, Ittay Weiss, Stefan Hansen, SeiriosJan 29 '13 at 8:07

Duplicate of math.stackexchange.com/q/115228/264 – Zev Chonoles Jan 29 '13 at 7:16
You will not find an explicit example, result cannot be proved without (some of) the Axiom of Choice. The Cauchy functional equation has been the subject of a number of MSE questions. – André Nicolas Jan 29 '13 at 7:17

This requires the axiom of choice in some form. Let $B$ be a Hamel basis for $\Bbb R$ considered as a vector space over the field $\Bbb Q$. Then for each $x\in\Bbb R$ there is a unique function $q_x:B\to\Bbb Q$ such that $$x=\sum_{b\in B}q_x(b)b$$ and $\{b\in B:q_x(b)\ne 0\}$ is finite. Now fix $b_0\in B$ and define
$$f:\Bbb R\to\Bbb R:x\mapsto q_x(b_0)\;.$$
I leave it to you to check that $f(x+y)=f(x)+f(y)$ for all $x,y\in\Bbb R$.
Now let $b\in B\setminus\{b_0\}$, and let $c=\dfrac{b}{b_0}\ne 0$; then
$$f(cb_0)=f(b)=0\ne c=cf(b_0)\;.$$