Division Theorem/Proof 1

Theorem
For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \left|{b}\right|$.


 * $\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \left|{b}\right|$

In the above equation:
 * $q$ is called the quotient
 * $r$ is called the principal remainder, or, more usually, just the remainder.

Non-Negative Integers
First we need to prove $\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$.

That is, we prove the theorem for non-negative $a$ and positive $b$.

Proof of Existence
Let us define the set $S$ as:


 * $S = \left\{{x \in \Z: x = a - z b, z \in \Z, x \ge 0}\right\}$

$S \ne \varnothing$ because, by putting $z = 0$, we find that $a \in S$.

Now $S$ is bounded below by $0$ and therefore has a least element, which we will call $r$.

Thus:
 * $\exists q \in \Z: a - q b = r$

and so:
 * $a = q b + r$

So we have proved the existence of $q$ and $r$ such that $a = q b + r$.

Now we need to show that $0 \le r < b$.

We already know that $0 \le r$ as $r \in S$ and is therefore bounded below by $0$.

Suppose $b \le r$. As $b > 0$, we see that $r < r + b$.

Thus $b \le r < r + b \implies 0 \le r - b < r$.

So $r - b = \left({a - q b}\right) - b = a - b \left({q + 1}\right)$.

So $r - b \in S$ as it is of the correct form.

But $r - b < r$ contradicts the choice of $r$ as the least element of $S$.

Hence $r < b$ as required.

So we have now established the existence of $q$ and $r$ satisfying $a = q b + r, 0 \le r < b$.

Proof of Uniqueness
Now we need to prove that $q$ and $r$ are unique.

Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$.

That is:

This gives $0 = b \left({q_1 - q_2}\right) + \left({r_1 - r_2}\right)$.

If $q_1 \ne q_2$, let $q_1 > q_2 \implies q_1 - q_2 \ge 1$.

Since $b > 0$, we get $r_2 - r_1 = b \left({q_1 - q_2}\right) \ge b \times 1 = b$.

So $r_2 \ge r_1 + b \ge b$ which contradicts the assumption that $r_2 < b$.

Similarly for if $q_1 < q_2$.

Therefore $q_1 = q_2$ and so $r_1 = r_2$, and so $q$ and $r$ are unique after all.

Thus we have proved the Division Theorem for $a \ge 0, b > 0$.

Negative $a$
Now we need to prove the Theorem for $a < 0$.

We know that:
 * $\exists \tilde q, \tilde r \in \Z: \left|{a}\right| = \tilde q b + \tilde r, 0 \le \tilde r < b$

Since $\left \vert {a} \right \vert = -a$, this gives:

If $\tilde r = 0$, then $q = -\tilde q, r = \tilde r = 0$, which gives $a = q b + r, 0 \le r < b$ as required.

Otherwise we have $0 < \tilde r < b \implies 0 < b - \tilde r < b$, which suggests we rearrange the expression for $a$ above:

Now if we take $q = \left({-1 - \tilde q}\right)$ and $r = \left({b - \tilde r}\right)$, we have the required result.

Negative $b$
Now the proof is extended to take on negative values of $b$.

Let $b < 0$.

Consider $\left|{b}\right| = -b > 0$.

By the above, we have the existence of $\tilde q, \tilde r \in \Z$ such that $a = \tilde q \left|{b}\right| + \tilde r, 0 \le \tilde r < \left|{b}\right|$.

Since $\left|{b}\right| = -b$, we have:


 * $a = \tilde q \left({-b}\right) + \left({\tilde r}\right) = \left({-\tilde q}\right) b + \tilde r$

We define $q = -\tilde q, r = \tilde r$ and we have proved the existence of integers that satisfy the requirements.

The proof that they are unique is the same as that for the proof for positive $b$, but with $\left|{b}\right|$ replacing $b$.