Is $\mathbb{Q}^2$ connected? 
Is $(\mathbb Q \times \mathbb Q)$ connected?

I am assuming it isn't because $\mathbb Q$ is disconnected. There is no interval that doesn't contain infinitely many rationals and irrationals.
But how do I show $\mathbb Q^2$ isn't connected? Is there a simple counterexample I can use to show that it isn't? What would the counterexample look like?
 A: Suppose $\mathbb{Q}\times \mathbb{Q}$ was connected, then by calling $\pi:\mathbb{Q}\times \mathbb{Q} \to \mathbb{Q}$ one the projections, you would get that $\mathbb{Q}$ is connected, since it would be a continuous image of a connected space.
A: Hint:  ${\mathbb Q} \times {\mathbb Q} = (A \times {\mathbb Q}) \cup (B \times {\mathbb Q})$ if ...
A: Let $X$ be a disconnected topological space. Then $X = A \cup B$, where $A$ and $B$ are open subsets of $X$ and $A \cap B = \varnothing$. But then $X \times X = (X \times A) \cup (X \times B)$. Because $A \cap B = \varnothing$, $(X \times A) \cap (X \times B) = \varnothing$, and these sets are open by definition of the product topology.
A: What topology is $\mathbb{Q}^2$ equipped with? I assume you are considering the subspace topology inherited from $\mathbb{R}^2$.
To prove $\mathbb{Q}^2$ is disconnected, we can find  two disjoint open subset which constitute the whole space $\mathbb{Q}^2$. Consider the two open subset of $\mathbb{Q}^2$: $\{(x,y) \in \mathbb{Q}^2\: x < \pi\}$ and $\{(x,y) \in \mathbb{Q}^2\: x > \pi\}$. (Recall the definition of a subspace topology). They form a disjoint union of $\mathbb{Q}^2$ and are both open.
So $\mathbb{Q}^2$ is disconnected by definition.
