Few questions about the basics of Cardinality I am looking for some help to either conform that my reasoning is sound, or to please elaborate to me more on the subject so I can gain a better understanding. I am studying some from my class notes, and some from other sources.
Suppose we are able to use the Cantor-Bernstien theorem i.e. that if $|A| \le |B|$ and $|B| \le |A|$ then $|A|=|B|$ or in other words if there is an injection from A to B and an injection from B to A, then there exists a bijection from A to B ( and hence a bijection from B to A as well), without requiring proof, and where $|A| \le |B|$ means there exists an injection from A to B. Using this and the prior knowledge that $$| \mathbb{N} |= | \mathbb{Z} |$$ I want to prove a few basics such as;
$$|\mathbb{N}|=|\mathbb{N \times N}|$$
Now my question about this is, is the following valid and in line with the above thereoms? 
We could take $f: \mathbb{N} \to \mathbb{N \times N}$ by $f(n)=(n,0)$
and $g: \mathbb{N \times N} \to \mathbb{N}$ by $g(n,m)=2^{n}3^{m}$ which are both injective, so by cantor bernstien there exists a bijection and thus they are of equal cardinality.
Next, using some of that I want to prove
$$| \mathbb{Z} | = | \mathbb{Z \times Z}|$$
Let $h: \mathbb{N} \to \mathbb{N \times N}$ be a bijection ( which from the previous paragraph we know exists)
My professor then hinted to me that a bijection $f: \mathbb{N} \to \mathbb{Z}$ induces a bijection $g=(f,f):  \mathbb{N \times N} \to \mathbb{Z \times Z}$. I am not sure I understand this part. Can anyone help to explain what that means, and how it is allowed? And from that I think we can form a function composition that allows us to use that composition of bijections is bijective to finish the proof.
And lastly,
I want to consider a proof of $$| \mathbb{N} | = | \mathbb{Q} |$$
Would it be correct to say if I can find some injective $f: \mathbb{Q} \to \mathbb{Z \times Z}$ and some injective $g: \mathbb{ Z \times Z} \to \mathbb{Q}$ and I could conclude a bijective $h:\mathbb{Q} \to \mathbb{Z \times Z}$  exists and hence from the previous paragraph I can directly conclude the proof? Is there a method I could use that incorporates the definition of rationals via equivalence classes of integers?
What would be the best injective functions for the cases,
$f(\frac{m}{n})=(m,n)$ $n \ne 0$
and $g(m,n)=\frac{m}{n}$ ?
Anyways, my apologies for the long post, but I wanted to make sure I got all my questions/ideas out here so I can work to better understand. Thank you all for any help and insight.
 A: Your reasoning up to that last part is very good.
Let me remark that $g=(f,f)$ is the function defined as $g(n,m)=(f(n),f(m))$. While it's not composition as you know it, it is a composition. It's not difficult, however, to check manually that $g$ defined like that is a bijection. Of course it follows from the fact that $f$ is a bijection.
Finally, as for $\Bbb Q$ and $\Bbb{Z\times Z}$. Note that $\frac12$ and $\frac24$ and $\frac48$ are all the same rational number. So you need to say something about what are the integers you've chosen to represent the fraction. Note that it is possible that both $m$ and $n$ are negative, or that only one is negative (either one).
So you need to refine that definition. In the case of $g$, the same problem applies, as $g(2,4)=g(1,2)$ so it is not injective. Here I would actually suggest you to construct an injection from $\Bbb{N\times N\to Q}$, then use the facts you have so far (assuming you have an injection from $\Bbb Q$ to $\Bbb{Z\times Z}$, of course) and have:
$$\Bbb{|N\times N|\leq|Q|\leq|Z\times Z|\leq|N\times N|}.$$
