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Suppose you have: $$f:\Bbb R \rightarrow\Bbb Z \text{ where } f(x)= \lceil 2x-1 \rceil$$
Well, I know that this is not a one-to-one function, but I don't know how to show that it's onto. The reason being the ceiling constraints. What I thought of was: $$ f(x)= \lceil 2x-1 \rceil = \lceil 2x \rceil -1 \\ f(x)=y\\y+1 = \rceil = \lceil 2x \rceil $$
But I don't know what to do past this point.
As you pointed out, $f$ onto would mean that for each $y$, there is some $x$ such that $f(x) = y$, or, equivalently, that $$\lceil 2x-1\rceil = y.$$
Instead of trying to deal with the problem symbolically, which you said you don't know how to do, pick a specific value of $y$ and see if you can find an $x$ that maps to it.
The expression $\lceil 2x-1\rceil$ is rather simple, so perhaps if you tabulate a few values of $x$ and $y$, you will see a pattern that you can generalize.
