My teacher of real analysis gave this question to us:
"Being $p,\,m\in\mathbb{N}$ such that $p>m$ and $p$ is not a multiple of $m$. How can we show that there are $q,\,r\in\mathbb{N}$ with $r<m$ such that $p=m\cdot q+r$ and that the natural numbers $q$ and $r$ that satisfy this situation are unique?"
She said that we will consider $0\notin \mathbb{N}$. This is ALL the information given to us. Probably we have to solve this with the Induction's Principle but I can't see how this is made.
On
I will give you some hints.
Lemma. Given $m,n \in \Bbb{N}$ with $n > m$ there is $q \in \Bbb{N}$ such that $$qm \leq n < (q+1)m.$$
You can prove it using the Well-ordering Principle on $X = \{xm \mid xm > n; x \in \Bbb{N}\}$.
Euclidean Division. Given $n > m \in \Bbb{N}$ there are $r,q \in \Bbb{N}$ (and they are unique) such that $$n = qm + r$$ where $0 \leq r < m$.
Now using the lemma for get $q \in \Bbb{N}$ such that $$qm \leq n \leq (q+1)m.$$
Study the cases $qm = n$ and $qm < n$.
Note that this generalize your problem. More precisely, $``p$ is not multiple of $m"$ can be excluded, making the appropriate changes.
Uniqueness: If $mq+r=mq'+r'$ and $q'=q$, then immediately $r=r'$. So assume $q\ne q'$ and wlog $q'>q$. Then $r=m(q'-q)+r'>m$, contradiction.
Existence: There are solutions $(q,r)$ to $p=mq+r$ if we do not require $r<m$. For example, $q=1,r=p-m$ is such a solution because we are given that $p>m$. Among all such pairs, pick one that minimizes $r$. Show that $r<m$ for this pair.