I'm asked to prove or disprove whether there are any isomorphisms between $\Bbb{R}$ and $\Bbb{R^2}$, as well as between $\Bbb{Z}$ and $\Bbb{Z^2}$.
It's all very confusing since I understand injective and surjective functions, but I don't know how to find an isomorphism between these two, and I don't know how to prove that all functions are not isomorphisms.
My intuition tells me that $\Bbb{R}$ and $\Bbb{R^2}$ are probably not isomorphic (if they were, $\Bbb{R}$ and $\Bbb{R^{1000}}$ would also be isomorphic, which sounds wrong). I also know that $\Bbb{Z}$ and $\Bbb{Q}$ are not isomorphic, and since $\Bbb{Q}$ is kind of like $\Bbb{Z^2}$, maybe I can do something with that?
Not everything that sounds wrong is wrong.
;-)Specifically, $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as vector spaces over $\mathbb{Q}$, in particular as additive groups.
This is because both have the same dimension over $\mathbb{Q}$. And yes, $\mathbb{R}$ and $\mathbb{R}^{1000}$ are isomorphic as additive groups as well.
To the contrary, $\mathbb{Z}$ and $\mathbb{Z}^2$ are not isomorphic as additive groups, because the former is cyclic and the latter isn't. You should be able to prove this fact, just by considering the subgroup of $\mathbb{Z}^2$ generated by a single element.
Note that “isomorphism” is a relative notion; for instance $\mathbb{R}$ and $\mathbb{R}^2$ are certainly not isomorphic as vector spaces over $\mathbb{R}$, because in this case their dimensions differ. But it's not a contradiction: we have added structure and we can expect something changes.
Also $\mathbb{R}$ and $\mathbb{R}^2$ are not “isomorphic as topological spaces” (the term commonly used is homeomorphic), but this is quite deeper.
The existence of an isomorphism of additive groups between $\mathbb{R}$ and $\mathbb{R}^2$ relies on the axiom of choice (for the general concept of basis of non finitely generated vector spaces), so it's even impossible to “write down explicitly”. It is certainly not a continuous map with respect to the usual topologies on them.