Assume the bipartition of $G$ is $V=A\uplus B$. Prove that $|A|=|B|$

Question

Let $G$ be a bipartite graph whose every vertex has the same degree $d$. Assume that the bipartition of $G$ is $V=A\uplus B$. Prove that $|A|=|B|$.

I'm not great at figuring out how to start proofs, would appreciate some guidance.

EDIT: This is the first part of a 4-part question. The question is set up as follows:

An edge coloring of a graph $G$ with $n$ colors is a function $f:E \to \{1,...,n\}$. We say that an edge coloring is proper if any two edges which share a common endpoint are different colors. Just as with vertex colorings, we can define the edge chromatic number $\chi'(G)$ to be the smallest number of colors needed to properly color the edges of $G$. Now let $G$ be a bipartite graph whose every vertex has the same degree $d$. In this problem we will prove that $\chi'(G) = d$.

That is all the information given to solve the title question. I'm thinking maybe its something to do with pigeonhole principle but I have no clue really where to start.

Answer 1

The first step, before any logical steps are worked out, is some investigation. What goes wrong if $|A| \neq |B|$? Try some specific examples.

Let's say that we're building a graph where $|A| = 3$ and $|B| = 2$. Start by drawing these points, three in a column on the left, and two in a column on the right. Now, try drawing some edges between these columns of points. Can you ensure that each point has the same number of edges coming from it? Presumably it's impossible, but what seems to be going wrong when you try it?

What happens if you just force both the $B$ vertices to have the same degree. What happens to the $A$ vertices? You'll hopefully notice that you simply don't have enough edges to match that degree for each of the three $A$ vertices. Correspondingly, if you force the $A$ vertices to have the same degree, some of the $B$ vertices will have too many edges.

Try it with a couple of other examples too. What if $|A| = 6$ and $|B| = 4$? Do these observations continue to hold?

This investigation step never makes it into the final proof, but it's often the first step in making a proof. You need to understand why the result holds, before you can even think about presenting it as a proof. Brian M. Scott's hint in the comments is a good one, but this is how you start a proof.