Last modified on 24 July 2009, at 15:30

Famous Theorems of Mathematics/Number Theory/Basic Results (Divisibility)

Definition of divisibilityEdit

An integer b is divisible by an integer a, not zero, if there exists an integer x such that b = ax and we write  a|b . In case b is not divisible by a, we write  a \nmid b. For example  3|6 and  4 \nmid 6.

If  a | b and 0 < a < b then a is called a proper divisor of b. The notation  a^k || b is used to indicate that  a^k | b but  a^{k+1} \nmid b. For example  3||6.

Basic propertiesEdit

  1.  a|b implies  a|bc for any integer c.
  2.  a|b and  b|c imply  a|c.
  3.  a|b and  a|c imply  a|(bx + cy) for any integers x and y.
  4.  a|b and  b|a imply  a=\pm b.
  5.  a|b, a > 0, b > 0, imply a \le b.
  6. If  m \ne 0,  a|b implies and is implied by  ma|mb.

Proof:

  1. Clearly if  a | b then \exists x such that b = ax. Then bc = a(xc) = ay and so  a | bc.
  2.  a | b and  b | c imply that m,n exist such that b = am and c = bn. Then clearly c = bn = (am)n = ap and so  a | c.
  3.  a | b and  a | c imply existence of m,n such that b = am and c = an so that bx + cy = amx + any = az and so  a | bx + cy.
  4.  a | b and  b | a imply that m,n exist such that b = am and a = bn. Then a = bn = amn and so mn = 1. The only choice for m and n is to be simultaneously 1 or -1. In either case we have the result.
  5. b = ax for some x > 0 as otherwise a and b would have opposite signs. So b = (a + a + ...) x times  \ge a.
  6.  a | b implies b = ax for some x which imples mb = amx i.e.  ma | mb. Conversely  ma | mb implies mb = max from which b = ax and so  a | b follows.

Remarks:

  • Properties 2 and 3 can be extended by the principle of mathematical induction to any finite set. That is,  a_1 | a_2,  a_2 | a_3,  a_3 | a_4 \cdots a_{k-1} | a_k implies  a_1 | a_k; and  a | b_1,  a | b_2 \cdots a | b_n implies that  a | \sum_{j=1}^n b_jx _j for any integers x_j.
  • Property 4 uses the fact that the only units in the set of integers are 1 and -1. (A unit in a ring is an element which possesses a multiplicative inverse.)

The division algorithmEdit

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.

Note: The integer

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

Proof: 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.

Let us first consider existence.

Consider the set

S = \left\{a - nd : n \in \mathbb{Z}\right\}

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. adn ≥ 0.
  • If d > 0, then again by the Archimedean property, there is a nonnegative integer n such that dn ≥ −a, i.e. ad(−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 = (ar)/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 rd implies a-qdd. This implies that a-qd-d ≥0, further implying that a-(q+1)d ≥ 0. Therefore, a-(q+1)d is in S and, since a-(q+1)d=r-d with d>0 we know a-(q+1)d<r, contradicting the assumption that r was the least nonnegative element of S.
  • If d<0 then r ≥ -d implying that a-qd ≥ -d. This implies that a-qd+d ≥0, further implying that a-(q-1)d ≥ 0. Therefore, a-(q-1)d is in S and, since a-(q-1)d=r+d with d<0 we know a-(q-1)d<r, contradicting the assumption that r was the least nonnegative element of S.

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.

Now let us consider uniqueness.

Suppose there exists q, q' , r, r' with 0 ≤ r, r' < |d| such that a = dq + r and a = dq' + r' . Without loss of generality we may assume that qq' .

Subtracting the two equations yields: d(q' - q) = (r - r' ).

If d > 0 then r'r and r < dd+r' , and so (r-r' ) < d. Similarly, if d < 0 then rr' and r' < -d ≤ -d+r, and so -(r- r' ) < -d. Combining these yields |r- r' | < |d|.

The original equation implies that |d| divides |r- r' |; therefore either |d| ≤ |r- 'r' | (if |r- r' | > 0 so that |d| is also > 0 and property 5 of Basic Properties above holds), or |r- r' |=0. Because we just established that |r-r' | < |d|, we may conclude that the first possibility cannot hold. Thus, r=r' .

Substituting this into the original two equations quickly yields dq = dq' and, since we assumed d is not 0, it must be the case that q = q' proving uniqueness.

Remarks:

  • The name division algorithm is something of a misnomer, as it is a theorem, not an algorithm, i.e. a well-defined procedure for achieving a specific task.
  • 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.