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

Definition of divisibility
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 properties

 * 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 algorithm
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 &mdash; 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 &minus;d > 0, and by the Archimedean property, there is a nonnegative integer n such that (&minus;d)n ≥ &minus;a, i.e. a &minus; dn ≥ 0.
 * If d > 0, then again by the Archimedean property, there is a nonnegative integer n such that dn ≥ &minus;a, i.e. a &minus; d(&minus;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 &minus; 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 implies 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 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 q ≤ q' .

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

If d > 0 then r'  ≤ r and r < d ≤ d+r' , and so (r-r' ) < d. Similarly, if d < 0 then r ≤ r'        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| &minus; 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.