Q:

How to prove the multiples of $3 $ is denumerable?

Prove that the following set is denumerable.

$T$, the integer multiples of $3$.

A denumerable set is that the set is equivalent to the set of natural numbers.
Some of the multiples of 3 include 
$T = \{...,-3,0,3,6,9,12,...\}$
One must also consider the negative and positive integers multiples of 3. My question is how does one prove that this is one to one or that it is a bijection and onto?
So we know the $\Bbb Z \rightarrow T \rightarrow \Bbb N ,\text{ such that }. $ $f(z) =3z$
Then $f(x) = f(y) \Rightarrow 3x=3y \Rightarrow, x=y$, thus we can say it is one to one.   So by saying this can this be enough to say it is denumerable?

A:

There are two ways to prove X is denumerable.
1) Prove there is a bijection $\phi:X \rightarrow \mathbb N $.
Or
2) Take an Y that you know is denumerable and show there is a bijection $\rho:X\rightarrow Y $.  If need be you can show $\rho\circ\phi\rightarrow \mathbb N$ is bijection.
So....
We know $\mathbb Z $ is denumerable and if we bang our heads against a wall we'll remember the bijection was $\phi:\mathbb N\rightarrow \mathbb Z::2n+1\rightarrow n $ for odd naturals; $2n\rightarrow -n $ for even so {1,2,3,4,5,6,....} $\rightarrow ${0,-1,1,-2,2,-3...}.
Finding the bijection $\rho: \mathbb Z\rightarrow 3\mathbb Z$ is incredibly obvious.  $\rho:z\rightarrow 3*z $.  So the multiples of 3 and the integers are obviously 1-1.
Proof: if $\rho (x)=\rho (y)\implies 3x=3y\implies x=y $ so $\rho $ is injective.  If $x=3n\in 3\mathbb Z $ then $\rho (n)=3n=x $ so $\rho $ is surjective.
So $\mathbb N \leftarrow\rightarrow \mathbb Z \leftarrow\rightarrow 3\mathbb Z $.
If we want the exact bijection it is $\rho\circ\phi: 2n+1\rightarrow n\rightarrow 3n $; $2n\rightarrow -n\rightarrow -3n $.
I.e.  {1,2,3,4,5,6,7,8,9....} maps to {0,-3,3,-6,6,-9,9,-12,12......}.

