Properties of solutions of the functional equation $f(kx) = kf(x)$

My problem is that I have a function: $f\colon\mathbb R\to\mathbb R$ with the property that $f(kx) = kf(x)$ for all $k,x \in\mathbb R$.

a) I shall show that $f(0)=0$

b) If $f$ is not the zero-function so that there is at least one $a\in\mathbb R$ with $f(a) \ne 0$, then $\forall x \in\mathbb R\colon f(x) = 0 \Leftrightarrow x=0$.

I don't really know how I shall proceed... Is it enough to say if: $k=0\Rightarrow kf(x)=0\cdot f(x)=0$ with $kf(x) = f(kx) = 0$ ? and how can I prove b) ?

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Plug in $1$ for $x$. –  rayradjr Nov 14 '12 at 18:09

a) You approach is correct. So you have $f(0) = f(0\cdot 0) = 0\cdot f(0) = 0$.

b) If there is $a\in \mathbb{R}$ such that $a\neq 0$ and $f(a)\neq 0$, then for any other $b\in \mathbb{R}$, $b\neq 0$ you have $$f(b) = f\left(\frac{b}{a}\cdot a\right) = \frac{b}{a}f(a) \neq 0.$$ So you have $f(x) = 0$ in and only if $x=0$ (by part (a)).

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That's kind of a complete answer for a homework problem... –  Thomas Andrews Nov 14 '12 at 18:13
@ThomasAndrews: oops I hadn't seen the homework tag. I got too exited.... At least the OP had shown some work already. –  Thomas Nov 14 '12 at 18:13
No worries, I've made that mistake many times. –  Thomas Andrews Nov 14 '12 at 18:15
Ok,now I've got it.Thank you :) –  phil Nov 14 '12 at 18:19

Yes, a) is really that easy.

For b) you only need to show that $x\ne0$ implies $f(x)\ne 0$. Hint: Consider $f(\frac ax\cdot x)$.

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