$f: \mathbb{Z} \rightarrow \mathbb{Z}, f(x)=2x-3$ 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!
 A: 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.
A: Yes, it's good reasoning. Function is surjective("onto") on recticted set $2Z+1$ not on Z
It's injective on all $Z$ because f(a)=f(b) =>2a+3=2b+3=>a=b. Cleraly, f is bijection on $2Z+1$.
A: To show that it is not surjective, it suffices to give an example of $y\in\mathbb Z$ such that $f(x)=y$ has no solutions. Just pick $y=-2$ and derive that
$$
f(x)=-2\iff 2x-3=-2\iff 2x=1
$$
and it should be quite clear that this has the unique solution $x=0.5$ over $\mathbb Q$ so no solutions over $\mathbb Z$.
A: You suspect right that the function is not surjective. To prove it correctly, you have to show an integer $y$ so that $y\ne f(x)$ for any integer $x$. Since $2x-3$ seems to be odd, try $y=0$:

for every integer $x$, $2x$ is even, so $2x\ne3$ and therefore $0\ne2x-3$, for every integer $x$

