Division Theorem

This page has been identified as a candidate for refactoring of advanced complexity. In particular: The links in the top section to dividend, divisor and quotient do not link to the correct definitions of these concepts in this context. The context is that of division in a field (for dividend and quotient), and divisor is specifically for the case where $r = 0$. So appropriately written pages need to be crafted. Until this has been finished, please leave {{Refactor}} in the code. New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code.

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 $0 \le r < \size b$:

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

In the above equation:

$a$ is the dividend

$b$ is the divisor

$q$ is the quotient

$r$ is the remainder.

Half Remainder Version

For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $-\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$:

$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, -\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$

Proof 1

From Division Theorem: Positive Divisor:

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

That is, the result holds for positive $b$.

$\Box$

It remains to show that the result also holds for negative values of $b$.

Let $b < 0$.

Consider:

$\size b = -b > 0$

where $\size b$ denotes the absolute value of $b$: by definition $\size b > 0$.

From Division Theorem: Positive Divisor, we have the existence of $\tilde q, \tilde r \in \Z$ such that:

$a = \tilde q \size b + \tilde r, 0 \le \tilde r < \size b$

Since $\size b = -b$:

$a = \tilde q \paren {-b} + \paren {\tilde r} = \paren {-\tilde q} b + \tilde r$

Taking:

$q = -\tilde q, r = \tilde r$

the existence has been proved of integers $q$ and $r$ that satisfy the requirements.

The proof that they are unique is the same as that for the proof for positive $b$, but with $\size b$ replacing $b$.

$\blacksquare$

Proof 2

Consider the set of integer multiples $x \size b$ of $\size b$ less than or equal to $a$:

$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$

We have that:

$-\size a \size b \le -\size a \le a$

and so $M \ne \O$.

From Set of Integers Bounded Above by Integer has Greatest Element, $M$ has a greatest element $h \size b$.

Then $h \size b \le a$ and so:

$a = h \size b + r$

where $r \ge 0$.

On the other hand:

$\paren {h + 1} \size b = h \size b + \size b > h \size b$

So:

$\paren {h + 1} \size b > a$

and:

$h \size b + \size b > h \size b + r$

Thus:

$r \le b$

Setting:

$q = h$ when $b > 0$

$q = -h$ when $b < 0$

it follows that:

$h \size b = q b$

and so:

$a = q b + r$

as required.

$\blacksquare$

Informal Proof

Existence

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.

$\Box$

Uniqueness

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

Then:

$b q + r = b q_0 + r_0$

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:

$-b < r - r_0 < b$

Hence:

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

So now:

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

which implies that:

$q = q_0$

Therefore the solution is unique.

$\blacksquare$

Also known as

Otherwise known as the Quotient Theorem, or (more specifically) the Quotient-Remainder Theorem (as there are several other "quotient theorems" around).

Some sources call this the division algorithm but it is preferable not to offer up a possible source of confusion between this and the Euclidean Algorithm to which it is closely related.

It is also known by some as Euclid's Division Lemma, and by others as the Euclidean Division Property.

Also see

• Definition:Integer Division

Sources

• 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): Appendix: Elementary set and number theory
• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.3$: Some properties of subrings and ideals
• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.28$
• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): remainder: 1.