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The full question :

Find 2 functions $f,g: \mathbb R \to \mathbb R$

$f,g \neq 0 $

so that $fg = 0$

The confusing part is the way it is written, as far as I can tell there is no product of anything other than $0$ which yields $0$, is there??

Perhaps they meant $f(g(x))$ in this question?

Thanks in advance

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Maybe there are two function, each of them is sometimes zero sometimes not (which is what they may mean by $f \ne 0$). – GEdgar Feb 20 '14 at 17:58
$f\neq 0$ only means "$f$ is not always zero", i.e. "there exists $x$ such that $f(x)\neq 0$. – Clement C. Feb 20 '14 at 17:59
up vote 10 down vote accepted

Try this: $$f(x)=\begin{cases} 1& x>0 \\ 0 &x\le 0\end{cases}$$ $$g(x)=\begin{cases} 0& x>0 \\ 1 &x\le 0\end{cases}$$

Note: If they meant $f(g(x))$, then you would still need $f$ to be zero somewhere.

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I was going to suggest the same with rationals and irrationals. The idea is the same : split $\mathbb R$ into two sets. – user88595 Feb 20 '14 at 18:00

$$f(x)=\max(x,0)$$ $$g(x)=\min(x,0)$$

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Let X be the domain of f and g. Let f and g s.t. for any $x \in X$ we have $f(x)=0$ or $g(x)=0$. Than f(x)g(x)=0 in X.

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