How to find a bijective mapping such that a) $f: \mathbb{N} \rightarrow \mathbb{N} \cup \{0\}$
I'm thinking $f(x) = x-1$, because in order to get 0, it would just be 1-1. Is this correct? I can show this is a bijection because it is surjective, i.e.: $ y = x - 1 \implies x = y+ 1$ and $f(x) = f(y+1) = y+1-1 = y$ and it is injective because $f(a) = a -1 = f(b) = b-1$ and adding one to both sides yields $a = b$.
b) $f: \mathbb{N} \rightarrow \mathbb{N} \backslash \{1,2,3,...2017\}$
I'm a bit confused on this, if I do $f(x) = x + 2017$ and I can show this is a bijection the same way. Is this correct? 

That would be correct.  This is a variant of the Hilbert hotel "technique", whereby you can essentially lose an extra element in (and only in) an infinite set.
As Cantor said,  "i see it but I can't believe it. "
