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The division algorithm is a theorem in mathematics which precisely expresses the outcome of the usual process of division of integers. The name is something of a misnomer, as it is a theorem, not an algorithm, i.e. a well-defined procedure for achieving a specific task — although the division algorithm can be used to find the greatest common divisor of two integers.

It should be noted that the term "division algorithm" in the study of algebra is commonly applied to the more general variant of this theorem, shown to hold in integral domains which are principal ideal domains.

Contents 1 Statement of theorem 2 Examples 3 Proof 3.1 Existence 3.2 Uniqueness 4 Generalizations 5 External links

Specifically, the division algorithm states that given two integers a and d, with |d| ≠ 0

There exist unique integers q and r such that a = qd + r and 0 ≤ r < | d |, where | d | denotes the absolute value of d.

The integer

• q is called the quotient
• r is called the remainder
• d is called the divisor
• a is called the dividend

• If a = 7 and d = 3, then q = 2 and r = 1, since 7 = (2)(3) + 1.
• If a = 7 and d = −3, then q = −2 and r = 1, since 7 = (−2)(−3) + 1.
• If a = −7 and d = 3, then q = −3 and r = 2, since −7 = (−3)(3) + 2.
• If a = −7 and d = −3, then q = 3 and r = 2, since −7 = (3)(−3) + 2.

The proof consists of two parts — first, the proof of the existence of q and r, and secondly, the proof of the uniqueness of q and r.

Consider the set

We claim that S contains at least one nonnegative integer. There are two cases to consider.

• If d < 0, then −d > 0, and by the Archimedean property, there is a nonnegative integer n such that (−d)n ≥ −a, i.e. a − dn ≥ 0.
• If d > 0, then again by the Archimedean property, there is a nonnegative integer n such that dn ≥ −a, i.e. a − d(−n) = a + dn ≥ 0.

In either case, we have shown that S contains a nonnegative integer. This means we can apply the well-ordering principle, and deduce that S contains a least nonnegative integer r. If we now let q = (a − r)/d, then q and r are integers and a = qd + r.

It only remains to show that 0 ≤ r < |d|. The first inequality holds because of the choice of r as a nonnegative integer. To show the last (strict) inequality, suppose that r = |d|. Since d ≠ 0, r > 0, and again d > 0 or d < 0.

• If d > 0, then r = d. Let q' = q + 1; then q' is an integer and q'd = (q + 1)d = qd + d = qd + r = a, i.e. a − q'd = 0.
• If d < 0, then r = −d. Let q' = q − 1; then q' is an integer and q'd = (q − 1)d = qd − d = qd + r = a, i.e. a − q'd = 0.

In either case, we have shown that r > 0 was not really the least nonnegative integer in S, after all. This is a contradiction, and so we must have r < |d|. This completes the proof of the existence of q and r.

Suppose with such that and Without loss of generality we may assume that .

Subtracting the two equations yields: .

If then and , and so . Similarly, if then and , and so . Combining these gives .

The original equation implies that divides ; therefore either or . Because we just established that , the first possibility cannot hold by trichotomy. Thus, .

Substituting this into the original two equations quickly yields (Recall: d is not 0) and thus proving uniqueness.

There is nothing particularly special about the set of remainders {0, 1, ..., |d| − 1}. We could use any set of |d| integers, such that every integer is congruent to one of the integers in the set. This particular set of remainders is very convenient, but it is not the only choice. See also coset and equivalence relation.

• Informal discussion of the division algorithm and well-ordering principle