Suppose $a = qb + r$, for $a,q,b,r \in \mathbb{Z}$ and $0 \le |r|<|b|$, $|b| > 0$, then $q$ is called the quotient and $r$ the remainder. Is the following proof of the existence of $q$ and $r$ valid?
Let $P(a,b)\ $ be the statement that there exists a quotient and a remainder for this $a$ and $b$. Now first I will prove $P(0,b)$: take $q=0, r=a$.
Now I will prove $P(a+1,b)$. Suppose $P(a,b)$. Then there are integers $q$ and $r$ satisfying $a=qb+r$. So $a+1=qb+(r+1)$. There are two cases: $(1)\ $ $|r+1|<|b|$, and $(2)\ $ $|r+1|\ge|b|$.
$(1)\ $ If $|r+1|<|b|$ then we are done: since $0 \le |r+1| < |b|$, we found a quotient $q$ and a remainder $r+1$.
$(2)\ $ Now the other part: $|r+1|\ge|b|$. Since $|r|<|b|$, we have $|r+1|=|b|$, which implies $ a=qb+r \Rightarrow a+1=qb+(r+1) \Rightarrow a+1=qb \pm b \Rightarrow a+1 = (q \pm 1) b. $ So we found quotient a $q \pm 1$ and a remainder $0$. So $P(a+1,b)$ holds. We can prove $P(a-1,b)$ in a similar way. And thus, for all $a \in \mathbb{Z}$, the statement $P(a,b)$ holds.