Division Theorem/Proof 3

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For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:

$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$



Consider the arithmetic sequence:

$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$

which extends in both directions.

Then by the Well-Ordering Principle, there must exist a smallest non-negative element, denoted by $r$.

So $r = a - q b$ for some $q \in \Z$.

$r$ must be in the interval $\hointr 0 b$ because otherwise $r - b$ would be smaller than $r$ and a non-negative element in the sequence.



Suppose we have another pair $q_0$ and $r_0$ such that $a = b q_0 + r_0$, with $0 \le r_0 < b$.


$b q + r = b q_0 + r_0$

Factoring we see that:

$r - r_0 = b \paren {q_0 - q}$

and so:

$b \divides \paren {r - r_0}$

Since $0 \le r < b$ and $0 \le r_0 < b$, we have that:

$-b < r - r_0 < b$


$r - r_0 = 0 \implies r = r_0$

So now:

$r - r_0 = 0 = b \paren {q_0 - q}$

which implies that:

$q = q_0$

Therefore the solution is unique.