how is the abs val of excess of p(n|odd # of parts) OVER p(n|even #of parts) = p(n|distinct odd parts?)

Question

A question in The Elementary Theory Of Partitions asks the reader to show that the absolute value of excess of the number of partitions $n$ with an odd number of parts over the number of those with an even number of parts equals the number of partitions of $n$ into distinct odd parts.

I think that i have seen proof that $|p_{even}(n) - p_{odd}(n)| = p_{do}(n)$ but I do not believe that is what the question is asking, I think it wants: $$\left|\frac{p_{odd}(n)}{p_{even}(n)}\right|=p_{do}(n)$$ Is this what the question asks or am I misunderstanding? If it is true how is it true?

Answer 1

I think they want the equation with the difference, not the quotient.

The “excess of $A$ over $B$” means $A-B$. In other words, the amount by which $A$ exceeds $B$.