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Define $f : \mathbb{R} \to \mathbb{R}$ such that $f(x)\leq f(y)$ whenever $x\leq y$ and $f^{2018} (z) \in \mathbb{Z}$ $\forall z \in \mathbb{R}$. Does there exist a function $f$ such that $f(x)$ is an integer for only finitely many values of $x$?

I think that the condition of the problem forces $f^n(x) \in \mathbb{Z}$ for all integer $n\geq 2018$. But this doesn't help I guess.

And I'm somewhat sure that the answer is NO.

Side-Note : $f^n (x)$ denotes $n^\text{th}$ composition of $f$

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