Proving the set $\{x ∈ \Bbb Z|x ≥ 3\}$ is countable 
Show that the set $\{x ∈ \Bbb Z|x ≥ 3\}$ is countable.

I'm really stuck on this problem, I have an exam really soon and I'm trying to solve it. So far I know that I have to show that it's onto and one-to-one.
Would I start by saying that $f(n)= 3 + n$ and then showing that it's onto and one-to-one?
 A: Yup, precisely!

Be sure to be clear about your definition for $f$, of course, in that you're using the definition $$f : \Bbb N \to \{x \in \Bbb Z | x \geq 3\}$$
Seems like a minor detail but it makes your argument clear since, usually, I like focusing these sorts of proofs by looking at the integers. Possibly just a personal quirk, but a lack of ambiguity never hurts.
From there, once you have shown $f$ is a bijection - that is, it is both onto (surjective) and one-to-one (injective) - then you can conclude its domain and codomain have the same cardinality. And thus,
$$|\Bbb N | = \aleph_0 = | \{x \in \Bbb Z | x \geq 3\} |$$
A: You already got the answer. A set is countable if there is a one-to-one and onto function between the set and the set of Integers or Natural numbers.
For the set you provided, we can see that there exits a one-to-one and onto function to the set of integers. which is the function
 $f(n) = n+3$ .
To prove its one-to-one just show that 
for any given $n_1$ and $n_2$ , if $f(n_1) = f(n_2)$
then $n_1$ = $n_2$.
To show its onto just show that
 given any $y$ you can find $x$ such that $y = f(x)$. 
Both are not difficult to prove.
