Multiplication function with sets

Question

Prove there is a unique function $$* :\mathbb N \times \mathbb N\to\mathbb N$$ such that:

1. $m * 0 = 0 $ for all $m \in \mathbb N$

2. $m * (n+1) = m * n + m$ for all $m,n \in \mathbb N$

Answer 1

Suppose there are two different functions that satisfy these rules, $*$ and $\otimes$. If they are different, there must be some $(a,b)\in\mathbb{N}\times\mathbb{N}$ such that $a*b\neq a\otimes b$. Call such a pair $(a,b)$ a discrepancy.

Of all the discrepancies, choose one $(a,b)$ where $b$ is minimal. We have: $$a*b\neq a\otimes b$$

We now consider two cases. Hidden since it's homework, but they both lead to contradiction.

If $b=0$, then $a*b=0=a\otimes b$, a contradiction. If $b\neq 0$, then $a*b=a*(b-1)+a$ and $a\otimes b=a\otimes(b-1) +a$. But $(a,b)$ was chosen as a discrepancy so that $b$ is minimal. Hence $(a,b-1)$ is not a discrepancy. Hence $a*(b-1)=a\otimes (b-1)$. This is a contradiction as well.