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Question: Is there a real valued function $f:\mathbb{R}\to\mathbb{R}$ such that its set of discontinuities is $\mathbb{Q}$ and its set of zeros $\{x\in \mathbb{R}\mid f(x)=0\}$ is also $\mathbb{Q}$?

It's well known that the Thomae function has as discontinuities the rationals. However, its zero set is $\mathbb{R}\setminus\mathbb{Q}$. On the other hand, the characteristic function of the set of irrational numbers has the rationals as zero set but its set of discontinuities is $\mathbb{R}$. So none of the candidates, that come to mind first, work.

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There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $x$ on which $f$ doesn't vanish. Hence if the set of zeros of $f$ is dense, then the function has to be discontinuous at every value on which it doesn't vanish.

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