Claim: If $n$ belongs to $\mathbb{N}$, and $p$ and $q$ are natural numbers with maximum $n$, then $p=q$.

Let $S$ be the subset of the natural numbers for which the claim is true. $1$ belongs to $S$, since if $p$ and $q$ belong to $N$ and their maximum is $1$, then $p=q=1$. Now assume $k$ belongs to $S$, and that the maximum of $p$ and $q$ is $k+1$. Then the maximum of $p-1$, $q-1$ is $k$. But $k$ is in $S$, so $p-1=q-1$, thus $p=q$ and $k+1$ is in $S$, so the assertion is true for all $n$ in $N$.