$f: \mathbb{Z} \rightarrow \mathbb{Z}, f(x)=2x-3$ surjective I have a problem with a really easy task. The function $f: Z \rightarrow Z, f(x)=2x-3$ is given. I have to proof whether it is injective or surjective. 
So I basically thought that it is bijective but I have encountered some problems while solving proofing that it is surjective. I began :
I have to proof that $\forall y \in Z$ $\exists x \in Z: f(x)=y$.
Let $y \in Z$ and choose $x=(y+3)/2$. 
But then I am facing the problem that $x$ is only an element in $Z$ if $y$ is odd.
Now I do not know how to go on. Thanks for your help!

The function is not surjective, like you said all $2k-3$ are odd, therefore there is no $x\in \mathbb Z$ such that $y=2x-3$ if $y$ is even.
In your proof, you can't choose $x=\frac{y+3}{2}$ because it's not an integer.
