Division Theorem/Positive Divisor/Existence/Proof 2
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Theorem
For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
- $\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
Proof
Let:
- $q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$
where $\floor {\, \cdot \,}$ denotes the floor function.
Thus $q \in \Z$ and $t \in \hointr 0 1$.
So:
- $\dfrac a b = q + t$
and so:
- $(1): \quad a = q b + r$
where $r = t d$.
Since $a, q, b \in \Z$, it follows from $(1)$ that:
- $r = a - q b$
and so $r \in \Z$ also.
Since $0 \le t < 1$ and $b > 0$, it follows that:
- $0 \le t b < b$
that is:
- $0 \le r < b$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 11.1$: The division algorithm
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$