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I have a hard time to prove if a function is injective and/or surjective when there's ceiling, floor or absolute functions in it like this:

$$ \ u: \mathbb{R} \rightarrow \mathbb{R}, \ u(x) = 3\lfloor x \rfloor + 1 $$

$$ \ v: \mathbb{Z} \rightarrow \mathbb{Z}, \ v(x) = \lceil\dfrac{x}{2} \rceil $$

My teacher only told me that the v(x) is surjective but without further explanation. I don't understand how to handle ceil and floor in algebra, like when I try to prove v(x) or u(x) is injective and/or surjective.

Any help would be very appreciate.

Thanks.

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    Are you sure it's $\frac{\lceil x \rceil}{2}$ instead of $\lceil \frac{x}{2} \rceil$? The former is not always an integer. – 2017-02-04
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    @FabioSomenzi yes you are right I made a mistake – 2017-02-04
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    You apply the definitions. If, say, $v$ is surjective, then for every $y \in \mathbb{Z}$, there is (at least) an $x$ such that $v(x) = y$. Can you see how to get $x$ as a function of $y$? If you have trouble, try a few numerical examples. If $y=5$, both $x=9$ and $x=10$ work. – 2017-02-04
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    @FabioSomenzi That's what I can't do I don't know how to get x as a function y. I know if y = 5 i can find that x can be 9 or 10 by trying number near 2y but I can't figure out how to get a general function to express this equation. It seems to me that x can be 2y or 2y-1 but I don't know where to start after that. – 2017-02-04
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    You are effectively done. You can say that for every $y$, $\lceil \frac{2y}{2} \rceil = y$, which proves that $v$ is surjective. You could also choose $2y-1$, or any function of $y$ that picks between $2y$ and $2y-1$ in some fashion that depends on $y$, but you don't have to. You already proved that for every element of the codomain, there's an element of the domain mapped to it. – 2017-02-04

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