[1.2a] Notes on 1.2 Continued
Theorem 2 Every positive integer can be written as a product of primes in only one way, except for the
order of the factors. [Proof later: pp167 and 177]
Example 4 Factors of 120: 1,3,5,15,2,6,10,30, 4,12,20,60,8,24,40,120
or 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120
Common divisors and common factors. Any pair of positive integers m and n have a greatest common
divisor, which is denoted gcd(m, n). gcd(18, 24) = 6 gcd(72, 108) = 36
min{a, b} is a if a#b or b if b<a.
Relatively prime means gcd(a,b) = 1
Common multiples: Any pair of positive integers m and n have a least common multiple, which is
denoted lcm(m, n). lcm(18, 24) = 72 lcm(72, 108) = 216.
max{a, b} is a if a $b or b if b > a.
Used for lowest common denominator.
Theorem 3 For positive integers m and n, mn = gcd(m, n)Alcm(m, n)
[See proof in text]
The Division Algorithm If m is a positive integer, then for each integer n there are unique integers q and
r such that n = mAq + r and 0 # r < m
[A constructive proof is given in 4.1]
Dividing by m gives:
1999=45A44 + 19
We see that q =
1999 = 30A66 + 19
and
1999 = 12A166 + 7
Theorem 4 Let m and n be positive integers
(a) Every common divisor of m and n is a divisor of gcd(m, n). [see page 11]
(b) Every common multiple of m and n is a multiple of lcm(m, n). [by Div. Alg.]
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Ross & Wright Discrete Mathematics, 5th edition
EXERCISES 1.2, page 14 (continued)
9. For positive integers m and n, gcd(m, n)Alcm(m, n) = mAn
a) If lcm(m, n) = mAn, then gcd(m, n) = 1; so we know that m and n must be relatively prime.
b) If lcm(m, n) = n, then gcd(m, n) = m; so we know that m|n.
c) If gcd(m, n) = m, then lcm(m, n) = n; so we know that m|n.
10. If p and q are different primes, then gcd(p, q) = 1 and lcm(p, q) = pAq.
11. “The positive integers m and n are relatively prime” means gcd(m, n) = 1.
a) Because 64 = 2A2A2A2A2A2 and 729 is not divisible by 2, gcd(64, 729) = 1; relatively prime.
b) Because 31 is prime, gcd(27, 31) = 1; relatively prime.
c) Because 45 = 3A3A5 and 56 = 2A2A2A7, gcd(45, 56) = 1; relatively prime.
d) Because 69 = 3A23 and 87 = 3A29, gcd(69, 87) = 3; so not relatively prime.
12. A pair of even integers cannot be relatively prime since they have at least 2 as a common divisor.
13. To be relatively prime means a pair of positive integers has a gcd = 1.
a) To be relatively prime to 2 means the integer is not even.
b) To be relatively prime to 3 means the integer is not a multiple of 3.
c) To be relatively prime to 4 = 2A2 means the integer is not even.
d) To be relatively prime to a fixed prime p means the integer is not a multiple of that prime p.
e) To be relatively prime to a fixed positive power of prime p means the integer is not a multiple of p.
14. Suppose m and n are integers that are multiples of d, say m = ad and n = bd.
a) Because lm = lad, we see that d|lm for every integer l.
b) Because m + n = ad + bd = (a + b)Ad, we see that d divides m + n.
Because m – n = ad – bd = (a – b)Ad, we see that d divides m – n.
c) Because 17m – 72n = 17ad – 72bd = (17a – 72b)Ad, we see that d divides m – n.
15. a) If gcd(m, n) = d, then n = bd and m = ad, then n – m = bd – ad = (b – a)d.
b) By part a), we see that gcd(m, m+2) must divide the difference (m+2) – m = 2; so only d = 1 or 2.
c) gcd(1000, 1002) = 2 because both numbers are even; gcd(1001, 1003) = 1 because they are odd.
d) gcd(1000, 1003) = 1 because the difference is 3, but 1000 is not a multiple of 3.
gcd(1000, 1005) = 5 because the difference is 5, and both are multiples of 5.
16. a) Because 30 = 2A3A5, the numbers relatively prime to 30 cannot be even, nor multiples of 3, nor
multiples of 5. That leaves: 1, 7, 11, 13, 17, 19, 23, 29.
b) Because 36 = 2A2A3A3, the numbers relatively prime to 36 cannot be even, nor multiples of 3. That
leaves: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35.
17. If n is a positive integer and if n has positive factors k and l [that is, n = kl such that 1 # k # l < n],
then a) 1 # k # l implies 1 # k # k2 # kl = n. So by taking square roots, we have k # /n.
b) Any example in which k = l, that is a perfect square, will give k = /n. Say n = 2A2 = 4.
c) If n $ 2 and n is not prime, then it has factors k and l that meet the conditions n = kl and 1 # k # l < n.
If p is any prime factor of k, then p must be # k, p must divide n, and by part a) p must also be # /n.
d) The argument in part c) shows that if n $2 has no prime factors # /n, then n must itself be prime.
18. Because 90 = 2A3A3A5, we have 12 factors in a 3-D diagram like Figure 1 on page 10 in the text;
namely:1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Because 8 = 2A2A2, we have only 4 factors in a 1-D diagram: 1, 2, 4, 8.
Because 24 = 2A2A2A3, we have 8 factors in a 2-D diagram: 1, 2, 3, 4, 6, 8, 12, 24.