# 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 $a\leq b$ .
6. If $m\neq 0$ , $a|b$  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 $\geq$  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 $a|\sum _{j=1}^{n}b_{j}x_{j}$  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 — 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.