Proof d-regular graph has an equal number of vertices in its bipartition

Question

Let $G$ be a $d$-regular graph. Suppose that $G$ is bipartite with bipartition $(A,B)$. Prove that if $d>0$ then $|A| = |B|$. Also why is this statement false when $d=0.$

I'm not sure how to show this. It makes sense in my head but that isn't enough.

Answer 1

Each vertex has $d$ edges and no two vertices in $A$ (or $B$) are connected. So in total, $A$ has $|A|d$ edges coming out of it and $B$ has $|B|d$ edges going into it. Can those be different?

Answer 2

I wanted to vote up the Karolis's answer and add a comment for $d=0$ but I don't have enough reputation, so I write an answer.

If $d=0$ then your graph has no edges. Any partitioning into disjoint sets $A$ and $B$ would render a legal bipartition, since there are no edges among the vertices in either of the sets. In particular different number of vertices in each of those sets would also be a legal bipartition.

For the case $d>0$: $d$-regular means that each vertex has degree $d$. FTSOC assume (wlog) $|A| > |B|$. Since $G$ is $d$-regular $A$ has to be $d$-regular. Hence we have $|A|d$ incident edges in $A$. Since $|B| < |A|$ we have in average $\frac{|A|d}{|B|}>d$ edges per vertex in $B$. Thus $B$ is not $d$-regular and hence neither is $G$.