Is $\mathbb{Q}$ isomorphic to $\mathbb{Z^2}$? Most of us are aware of the fact that $\mathbb{C}$ is isomorphic to $\mathbb{R^2}$, as we can define $\mathbb{C}$ as follows :
$$\mathbb{C} := \left\{z : z=x+iy \ \ \ \text{where} \ \ \langle x,y \rangle \in \mathbb{R^2}\right\}$$
Now this implies
$$\mathbb{C} \cong \mathbb{R^2} $$
But, by this same argument, can't we make the same argument for $\mathbb{Q}$, the set of rational numbers and  $\mathbb{Z^2}$? We can define $\mathbb{Q}$ as follows:
$$\mathbb{Q} := \left\{\frac{m}{n} :  \langle m,n \rangle \in \mathbb{Z^2} \ \ \land \ n \neq 0 \right\}$$
Which would imply
$$\mathbb{Q} \cong \mathbb{Z^2} $$
And we could go further and explain that $\{\mathbb{Q}\} = \mathbb{Z}r\mathbb{Z} - 0$, where $r$ is a relation on $\mathbb{Z}$.
So is $\mathbb{Q}$ isomorphic to $\mathbb{Z^2}$?
 A: You have to be more specific about what you mean by "isomorphic". It depends on what category you are working in. For example, as sets $\mathbb{Z}^n \cong \mathbb{Q}$ for any natural number $n \geq 1$, since both sets are countable. On the other hand, it is certainly false that $\mathbb{Z}^2 \cong \mathbb{Q}$ as groups. 
Perhaps more shockingly, if you've taken linear algebra, all finite dimensional vector spaces are isomorphic as sets! On the other hand, they are obviously not isomorphic as vector spaces. So obviously bijection is the wrong notion of isomorphism if you're studying linear algebra,
Here's the general idea. Depending on what you are interested in studying, you define a category of objects and morphisms in that category. Usually (but not always) the objects will be presented by sets with some structure and the morphisms will be functions that preserve the structure. So if you are studying groups, your objects will be groups and the morphisms will be group homomorphisms. If you are studying topological spaces the objects will be spaces and the morphisms will be continuous maps. Anyway, now that you've decided on a category to study, the isomorphisms will be the invertible morphisms in that category. 
If you want to learn more about this structural perspective on mathematics, check out Lawvere and Schanuel's book Conceptual Mathematics: A First Introduction to Categories. It's very accessible for beginners and teaches you some very important tools and ideas for doing mathematics.
A: It's important to specify in what sense two objects are isomorphic or not. In your example, $\mathbb{R}^2$ and $\mathbb{C}$ are isomorphic as abelian groups or $\mathbb{R}$-vector spaces, but not as rings (if the multiplication on $\mathbb{R}^2$ is done componentwise).
As it happens, $\mathbb{Q}$ and $\mathbb{Z}^2$ are not isomorphic groups, since for instance $\mathbb{Q}$ is divisible but $\mathbb{Z}^2$ isn't.
A: No. An equivalence relation is missing to do the "isomorphism" (or representation, visualization) correctly.
$Q$ is defined by axioms, and this way of expressing $Q$ as $Z ^ 2$ and the relation $R$ is one of the multiple ways of "visualizing" the axioms of $Q$.
The equivalence relation $R$ that is missing is that two elements $(a, b)$ and $(c, d)$ are realigned, if and only if, $a * d = b * c$.
Then $Z ^ 2$, under the equivalence ralation mentioned above, is isomorphic to $Q$. (Defining correctly the two operations of Q, addition and multiplication, on all the pairs $(a, b)$).
