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4. Integers
We now look at what we can say about division in Z. [Lecture example: how does school ‘long division’ work?]
4.1 Theorem (the division algorithm for N). Let a, b ∈ N with b 6 = 0. Then there exist unique natural numbers q and r such that a = qb + r and 0 6 r < b.
Proof. Assume a < b. Then a = 0b + a so q = 0 and r = a do the job.
Now assume a > b > 0. Consider
S = { m ∈ N | mb − a > 0 }.
Then S 6 = ∅ because N (as a subset of R) is not bounded above. By the Well-ordering Property there exists a least element, k say, in S. Note 0 ∈/ S and so k > 1. Let q = k − 1 and r = a − qb. Then q ∈ N r S (by leastness of k) so r > 0. Also q + 1 = k ∈ S so that (q + 1)b > a. Hence b > r.
Corollary (the division algorithm for Z). Let a, b ∈ Z with b 6 = 0. Then there exist unique integers q and r such that a = qb + r and 0 6 r < |b|.
Proof. [omitted from lecture]
(i) a > 0, b > 0. Apply the theorem. (ii) a > 0, b < 0. Then −b > 0 and by the theorem there exist q′, r′^ ∈ N with a = q′(−b) + r′^ and 0 6 r′^ < −b = |b|. Let q = −q′^ and r = r′. (iii) a < 0, b > 0. By the theorem, there exist q′′, r′′^ ∈ Z such that −a = q′′b + r′′^ and 0 6 r′′^ < b. If r′′^ = 0, take q = −q′′^ and r = 0. If r′′^ > 0, take r = b − r′′^ and note that 0 < r < b. Also let q = (−q′′^ − 1) and note that a = −q′′b − r′′^ = −q′′b + (r − b) = qb + r, as we require. (iv) a < 0, b < 0. Apply the result from case (iii) with b replaced by −b, in the same way as in case (ii).
Uniqueness of q and r in the Division Algorithm
Assume there is another pair of integers ˜q and ˜r such that a = ˜qb + ˜r with 0 6 ˜r < |b|. Then, from a = qb + r = ˜qb + ˜r, we deduce that (q − q˜)b = ˜r − r. So ˜r = r would force q = ˜q (note that property (ID) is used here).
So suppose for a contradiction that ˜r 6 = r. Suppose b > 0 (the case b < 0 works similarly). Without loss of generality, ˜r > r and hence q > ˜q. But then
b > ˜r > ˜r − r = (q − q˜)b > 0.
But this would give 0 < (q − q˜) < 1, with (q − q˜) ∈ N and this is impossible.
We next give an application of the Division Algorithm which will be important later on, specifically in our study of groups.
4.2 Theorem (subrings of Z).
(1) Let S be a non-empty subset of Z closed under addition and subtraction. Then there exists n ∈ N such that S = nZ; = {nk | k ∈ Z}.
(2) The subrings of Z are exactly the sets of the form nZ for n ∈ Z.
Proof. (1) Since S 6 = ∅ there exists some m ∈ S. Then 0 = m − m ∈ S.
If S = { 0 }, then S = 0Z. Now assume S 6 = { 0 }. Then there exists some m 6 = 0 in S. Also −m = 0 − m ∈ S. Therefore S contains a positive integer. By the Well-order Property of N there exists a least positive member of S; call it n. We claim S = nZ.
We can prove by induction that n ∈ S implies nZ ⊆ S (see Problem sheet 2, question 2). Now let a ∈ S. By the Division Algorithm there exist q, r ∈ Z such that a = qn + r, with 0 6 r < n. But then r = a − qn ∈ S. By leastness of n we must have r = 0. Therefore a ∈ nZ.
(2) We noted earlier that every set nZ is a subring of Z. (1) supplies the converse.
4.4 The highest common factor of two integers: definitions and uniqueness. Let a and b be two non-zero integers. Assume that c is a positive integer such that
(HCF1) c|a and c|b; (HCF2) for all integers d such that d|a and d|b, then d|c.
Then c is called the highest common factor (or the greatest common divisor) of a and b, and denoted hcf(a, b) (or gcd(a, b)). It is convenient, by convention, to extend the definition by taking, for any integer a, hcf(a, 0) = hcf(0, a) = a.
Notes: Observe that if c satisfies (HCF1) and (HCF2) above then so does (−1)c = −c. By requiring the hcf of a and b to be positive we ensure it is unique, provided it exists (Exercise: check uniqueness, using the Divisors Lemma from 4.3). So we really can talk about the hcf rather than an hcf of given integers, provided there is an hcf at all.
Observe that for any a, b ∈ Z r { 0 },
hcf(a, b) = hcf(a, |b|) = hcf(|a|, b) = hcf(|a|, |b|).
We can therefore restrict attention to positive integers.
But do hcf’s exist? How do we find them if they do? The following lemma is a key ingredient in proving the existence of hcf’s.
4.5 Invariance Lemma (for hcf’s of positive integers) Let a and b be positive integers. Write
a = qb + r with q, r ∈ N and 0 6 r < b.
Assume hcf(b, r) exists. Then hcf(a, b) exists and equals hcf(b, r).
Proof. Let c := hcf(b, r). We shall show that c satisfies conditions (HCF1) and (HCF2) in the definition of hcf(a, b). First note that c|b and c|r. Hence c|qb + 1r = a by Divisors Lemma, part (iii). Now assume that d|a and d|b. From property (iii) in the Divisors Lemma we get d|r = a − qb. Hence d|c by condition (HCF2) applied to the pair b and r.
In the process of moving from the pair (a, b) to the pair (b, r)
- the hcf is left unchanged (it is called the ‘invariant’);
- the second component (called the ‘variant’) changes from one positive integer to another whose magnitude is strictly smaller.
[:Lecture example : computing hcf(72, 15).]
4.6 Theorem (Euclid’s algorithm, from Euclid’s Elements, Book 7, Propositions 1 and 2).
(1) Let a and b be positive integers, then hcf(a, b) exists; (2) (the hcf formula) there exist integers m and n such that
hcf(a, b) = ma + nb.
Proof. Without loss of generality we may assume that 0 < b 6 a. In what follows, all quantities appearing are integers and at each step the division algorithm is invoked. We can write
a = q 0 b + r 0 where 0 < r 0 < b b = q 1 r 0 + r 1 where 0 < r 1 < r 0 r 0 = q 2 r 1 + r 2 where 0 < r 2 < r 1
......... rk− 3 = qk− 1 rk− 2 + rk− 1 where 0 < rk− 1 < rk− 2 rk− 2 = qkrk− 1 + rk where 0 = rk
where we stop as soon as we reach a zero remainder. This must happen in a finite number of steps, by the Going Down Lemma (see Section 3). The Invariance Lemma 4.5 tells us that
hcf(a, b) = hcf(b, r 0 ) =... = hcf(rk− 2 , rk− 1 ) = rk− 1.
To obtain the hcf formula we retrace our steps: write rk− 1 in terms of rk− 2 , then substitute the resulting formula for rk− 1 into the equation for rk− 3 and so on until a formula in a and b of the required form is obtained. More formally: argue by induction on k.
Exercise example. Find the highest common factor of a = 38793 and b = 89531 and express it as ma + nb. [Ans: hcf(a, b) = 13 = (−1705) · 38793 + 738 · 89531 (note m and n are not unique −1705 and 738 are not the only possibilities; can you find others?)]
Tabulating the sums: Form a table, one row at a time, as follows. The first row has entries -, r 0 = max{a, b}, m 0 = 1, n 0 = 0 and the second has entries -, r 1 = min{a, b}, m 1 = 0 and n 1 = 1. Then, so long as ri 6 = 0, the (i + 1)th row is filled in from left to right by first calculating qi+1 by the division algorithm and then ri+1, mi+1 and ni+1 are calculated as per Step 2. ;
Example. To find the hcf of 134 and 28.
Q R M N
We have 2 = (−5) · 134 + 24 · 28.
Exercise. Rework the exercise from 4.6 tabulating the calculations.
4.8 Coprime integers: definition and a useful fact. Two non-zero integers a and b are said to be coprime if hcf(a, b) = 1. In this case, by the hcf formula from 4.6, there exist integers m and n such that 1 = ma + nb. We claim the converse is true too. Assume that 1 = ma + nb for some integers m and n. Let d be a positive integer such that d|a and d|b. Then d|ma + nb = 1 by property (iii) in the Divisors Lemma. By property (ii) in the same lemma, d = 1. Hence hcf(a, b) = 1, as claimed.
4.9 Definition: prime numbers. Let n ∈ N with n 6 = 0, 1. Then n is called prime or a prime if n has exactly two positive divisors (itself and 1). According to our definition, 1 is NOT a prime!
4.10 Proposition (division by primes). Assume that a and b are integers and that p is a prime such that p|ab. Then p|a or p|b.
Proof. If p|a there is nothing to prove. On the other hand, if p does not divide a, then hcf(a, p) = 1 and we can find m, n ∈ Z such that 1 = ma + np. Now b = mab + bmn. By the Divisors Lemma (iii) we get p|b.
4.11 The Fundamental Theorem of Arithmetic. Every positive integer a > 1 can be written as a product of prime factors, and this factorisation is unique up to the order in which the factors are written. Equivalently, for every positive integer a > 1 there exist a unique positive integer k, unique primes p 1 ,... , pk and unique positive integers β 1 ,... , βk such that
a = pβ 1 1 · · · pβ k k.
We can now see the reason for disqualifying 1 from being a prime:
1 = 1 · 1 = · 1 · 1 · 1 ·...
so we don’t get a unique factorisation.
The proof of the Fundamental Theorem of Arithmetic is not examinable in Mods. You can find a proof in A guide to Abstract Algebra, C. Whitehead (Macmillan Mathematical Guides). The result will be revisited in Part A Algebra next year, and treated in a more general context.
Remark: There is an obvious extension of the theorem to the factorisation of an integer ∈ {/ 0 , ± 1 }, in terms of products of elements of the form ±p, where p is prime. Now we only have uniqueness ‘up to order of factors and ±s’.
4.12 A classic theorem. There are infinitely many primes. [web notes only]
Proof. We assume for a contradiction that there are only finitely many primes p 1 , p 2 ,... , pn. Let a = p 1 p 2 · · · pn + 1. Note that since a is bigger than every pi it cannot itself be a prime. As a is not a prime, it must have a divisor other than a and 1. In particular it must have a prime divisor. Hence there exists j ∈ { 1 ,... , n} such that pj |a. We can also see that pj |p 1 p 2 · · · pn. Therefore pj divides a − p 1 p 2 · · · pn = 1. We have a contradiction, since 1 is the only positive divisor of 1 and 1 is not prime.
