How can I prove this with induction?

Question

Being $p,\,m\in\mathbb{N}$ such that $p>m\ge1$ and $p$ is not a multiple of $m$. How can I show that there are $q,\,r\in\mathbb{N}$ with $r<m$ such that $p=m\cdot q+r$? We take $0\in\mathbb{N}$.

Answer 1

Induct on $p\ge m+1$. For the base case $p=m+1$ take $q=1,\,r=1$. If $p=mq+r$, either $r=m-1$ so write $p+1=(m+1)q+0$, or else write $p+1=mq+(r+1)$.