Build a bijection $f: \mathbb{R} \to \mathbb{R}\setminus \mathbb{N}$.
What about $f(x)=\pi \cdot x?$
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5Fails for $x=\frac{1}{\pi}$ – barak manos Dec 26 '14 at 08:36
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See also: Find a bijection from $\mathbb R$ to $\mathbb R-\mathbb N$ – Martin Sleziak Jul 15 '15 at 13:24
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Map $\mathbb{Z}$ bijectively to $\{...,-3,-2,-1,0\}$ and leave the other points from $\mathbb{R}$ untouched.
C_M
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Hint: do you know how to build a bijection $[0,1]\to (0,1]$? You "hide" $0$ somewhere inside an interval (in some infinite sequence). In your case you can hide all positive integers in some sequence too.
Update: you can choose any sequence $\{a_i\}$ and for all $n\in\mathbb{N} \quad f(n) = a_{2n},$ for elements from sequence $f(a_i) = a_{2i+1},$ and otherwise $f(x) = x$.
sas
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Shouldn't a bijection cover the entire range of $\mathbb{R}$ \ $\mathbb{N}$ in the other direction? – barak manos Dec 26 '14 at 08:44
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@barakmanos, excuse me, I do not completely understand your remark. I've updated my post, though. – sas Dec 26 '14 at 08:56