Division Theorem/Positive Divisor/Uniqueness
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Theorem
For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:
- $\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
In the above equation:
- $a$ is the dividend
- $b$ is the divisor
- $q$ is the quotient
- $r$ is the principal remainder, or, more usually, just the remainder.
Proof 1
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
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:
\(\ds a\) | \(=\) | \(\ds q_1 b + r_1, 0 \le r_1 < b\) | ||||||||||||
\(\ds a\) | \(=\) | \(\ds q_2 b + r_2, 0 \le r_2 < b\) |
This gives:
- $0 = b \paren {q_1 - q_2} + \paren {r_1 - r_2}$
Aiming for a contradiction, suppose that $q_1 \ne q_2$.
Without loss of generality, suppose that $q_1 > q_2$.
Then:
\(\ds q_1 - q_2\) | \(\ge\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r_2 - r_1\) | \(=\) | \(\ds b \paren {q_1 - q_2}\) | |||||||||||
\(\ds \) | \(\ge\) | \(\ds b \times 1\) | as $b > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r_2\) | \(\ge\) | \(\ds r_1 + b\) | |||||||||||
\(\ds \) | \(\ge\) | \(\ds b\) |
This contradicts the assumption that $r_2 < b$.
A similar contradiction follows from the assumption that $q_1 < q_2$.
Therefore $q_1 = q_2$ and so $r_1 = r_2$.
Thus it follows that $q$ and $r$ are unique.
$\blacksquare$
Proof 2
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$.
Thus:
- $\dfrac a b = q + \dfrac r b$
and:
- $0 \le \dfrac r b \le \dfrac {b - 1} b < 1$
So:
- $q = \floor {\dfrac a b}$
and so:
- $r = a - b \floor {\dfrac a b}$
Thus, given $a$ and $b$, the numbers $q$ and $r$ are unique determined.
$\blacksquare$
Proof 3
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.
Suppose that:
- $a = b q_1 + r_1 = b q_2 + r_2$
where both $0 \le r_1 < b$ and $0 \le r_2 < b$.
Without loss of generality, suppose $r_1 \ge r_2$.
Then:
- $r_1 - r_2 = b \paren {q_2 - q_1}$
That is:
- $b \divides \paren {r_2 - r_1}$
where $\divides$ denotes divisibility.
But:
- $r_1 - r_2 < b$
while from Absolute Value of Integer is not less than Divisors: Corollary:
- $r_1 - r_2 \ge b$
unless from Integer Divides Zero $r_1 - r_2 = 0$.
So $r_1 = r_2$ and it follows directly that $q_1 = q_2$.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 6$: The division process in $I$: Theorem
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-1}$ Euclid's Division Lemma: Theorem $\text {2-1}$