Ex 2 Solutions

7/31/2019 Ex 2 Solutions

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CM121A, Abstract Algebra Solution Sheet 2

1. Suppose that a, b and c are positive integers.

(a) Prove that ifa|b and b|c, then (b/a)|(c/a).Solution: Ifa|b and b|c, then m = b/a and n = c/b are integers, andc/a = (c/b)(b/a) = nm is divisible by m = b/a.

(b) Prove that if c is a common divisor of a and b, then gcd(a/c, b/c) =(gcd(a, b))/c.Solution: Let g = gcd(a, b). Since c|a and c|b, we know that c|g(Theorem 2.2.2), so g/c is an integer. We have to prove that g/c = h,where h = gcd(a/c, b/c).We know (from part (a)) that g/c divides a/c, and g/c divides b/c.Being a common divisor of a/c and b/c, we know that g/c h.On the other hand, we also know (again by Theorem 2.2.2) thatg = ax + by for some x, y Z. Therefore g/c = (a/c)x + (b/c)y

is divisible by h (since h|(a/c) and h|(b/c)). Also note that g/c ispositive (since both c and g are), so h g/c. Therefore h = g/c.

2. Suppose that a, b, c and d are integers, with a = 0. Suppose that a and bare relatively prime.

(a) Prove that ifd|a, then b and d are relatively prime.Solution: Let g = gcd(b, d). Then g|b and g|d, and since d|a, weknow that g is a positive common divisor of a and b. Since a and bare relatively prime, we must have g = 1.(Alternatively, apply Corollary 2.3.2: we know that ax + by = 1 forsome x, y Z, and a = dk for some k Z, so d(kx) + by = 1, and itfollows that b and d are relatively prime.)

(b) Prove that gcd(a, c) = gcd(a,bc).

Solution: Let g = gcd(a, c) and h = gcd(a,bc). Since g|a and g|c,it follows that g|bc, so g is a common divisor of a and bc. Thereforeg h.On the other hand, we know h|a and h|bc. Since a and b are relativelyprime and h|a, it follows from part (a) that h and b are relativelyprime. So it follows from Corollary 2.3.4 that h|c. Now we know his a positive common divisor of a and c, so h g. Therefore g = h.

(c) Prove that a2 and b2 are relatively prime.Solution: Applying part (b) in the case b = c, we see that if a andb are relatively prime, then so are a and b2. Now apply this againwith our b2 as a and our a as b to conclude that b2 and a2 arerelatively prime.

3. Suppose that m and n are positive integers and g = gcd(m, n).

(a) Prove that if k is an integer divisible by both m and n, then k is di-visible by mn/g. (Hint: Use the fact that m/g and n/g are relativelyprime to show that k/g is divisible by (m/g)(n/g).)Solution: Since g|m and m|k, we have m = ga and k = mc forsome a, c Z. Therefore k/g = gac/g = ac is an integer divisible by

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m/g = a. Similarly, writing n = gb and k = nd for some c, d Z,we see that k/g = bd is an integer divisible by n/g = b. Since m/gand n/g are relatively prime (Cor. 2.3.3) and both divide k/g, their

product (m/g)(n/g) divides k/g (Cor. 2.3.5). Multiplying throughby g, we conclude that mn/g divides k.

(b) Show that mn/g is the least positive common multiple of m and n.Solution: Since m/g and n/g are integers, mn/g is a multiple ofboth n and n. We saw in part (a) that every common multiple ofm and n is divisble by mn/g, so if k is a positive multiple of m andn, then k mn/g. Therefore mn/g is the least positive commonmultiple of m and n.

4. Find all solutions with x, y Z to each of the following linear Diophantineequations (or show it has no solutions):

(a) 42x + 15y = 7;Solution: Using the Euclidean Algorithm 42 = 215+12, 15 = 12+3,

and 12 = 4 3, so g = gcd(42, 15) = 3. Since 7 is not divisible by 3,there are no solutions.

(b) 154x + 345y = 18;Solution: Using the Euclidean Algorithm 345 = 2 154 + 37, 154 =437+6 and 37 = 66+1, so gcd(154, 345) = 1 and there are solutions.To get a solution, first write 1 as an integer linear combination of154 and 345 by unwinding the equations gotten from the Euclideanalgorithm:

1 = 37 6 6 = 37 6(154 4 37)= 25 37 6 154 = 25(345 2 154) 6 154 = 25 345 56 154.

Muultiplying through by 18 gives

154 (1008) + 345 450 = 18,

so one solution is given by x0 = 1008, y0 = 450. Therefore byTheorem 2.4.5, the general solution is x = 1008 + 345k, y = 450 154k for k Z. (Or switch to a smaller x0, y0, for example takingk = 3 gives x = 27, y = 12 as a solution, so another expression forthe general solution is therefore x = 27 + 345k, y = 12 154k fork Z.)

(c) 1066x + 2009y = 164.Solution: According to the solution of Exercise 1.6, gcd(1066, 2009) =41 and 41 = 17 1066 9 2009. Since 164 = 4 41, we can mul-

tiply through by 4 to get the solution x0

= 68, y0

= 36. Since2009/41 = 49 and 1066/41 = 26, the general solution is

x = 68 + 49k, y = 36 26k, for k Z.

Or taking k = 1 to get smaller values gives

x = 19 + 49k y = 10 26k for k Z.

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5. Youre at a shop in the airport about to leave New York. You want tospend the last $15 of your US money on chocolate by buying a combinationof Mercury bars, which cost 39 cents apiece, and Andromeda bars, which

cost 47 cents apiece. How many of each do you need to buy so you wonthave any change?Solution: Let x be the number of Mercury bars and y the number ofAndromeda bars. We have to solve the equation

39x + 47y = 1500,

where x and y are non-negative integers.We work through the Euclidean algorithm for a = 39, b = 47:

47 = 1 39 + 839 = 4 8 + 7

8 = 1 7 + 17 = 7 1 + 0.

The last non-zero remainder is 1, so g = gcd(39, 47) = 1. Therefore theequation has integer solutions. Unwinding the above equations gives:

1 = 8 7 = 8 (39 4 8) = 5 8 39= 5 (47 39) 39 = 5 47 6 39.

Multiplying through by 1500 gives

9000 39 + 7500 47 = 1500,

so an integer solution is

x0 = 9000, y0 = 7500.

Since g = 1, Theorem 2.4.5 gives all integer solutions as:

x = 9000 + 47k, y = 7500 39k for k Z.

The values of k giving non-negative solutions must satisfy

9000 + 47k 0, 7500 39k 0,

This is equivalent to k 9000/47 191.5, and k 7500/39 192.3.Therefore the only possible value of k is 192, giving

x = 24, y = 12.

So you need to buy 24 Mercury bars and 12 Andromeda bars.

6. Suppose that p, q and r are distinct primes. What is the greatest commondivisor of p3q2r and p2r2?Solution: According to Cor. 2.5.6, the common positive divisors are theintegers of the form paqbrc, where a,b,c are non-negative integers satis-fying a min{3, 2} = 2, b min{2, 0} = 0, c min{1, 2} = 1. Thegreatest common divisor is therefore p2r.

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7. Suppose that a and b are positive integers with prime factorizations

a = pr11

pr22 prk

k, b = ps1

1ps22 psk

k

where p1, p2, . . . , pk are distinct primes and r1, r2, . . . , rk, s1, s2, . . . , sk arenon-negative integers.

(a) Prove that the least positive common multiple ofa and b is

pt11

pt22 ptk

k

where ti = max{ri, si} for i = 1, 2 . . . , k.Solution: Suppose that c is a positive integer. Write its primefactorization as

c = pu11

pu22 puk

kqv11

qv22 qv

,

where q1, q2, . . . , q are primes distinct from p1, p2, . . . , pk, andu1, u2, . . . , uk, v1, v2, . . . , v are non-negative integers. According toCor. 2.5.6, c is a common multiple of a and b if and only ui ri andui si. Therefore the least common multiple of a and c is gotten bytaking each ui = max{ri, si} and each vi = 0.

(b) Use this to give another solution to Exercise 3b.Solution: Write

m = pr11

pr22 prk

k, n = ps1

1ps22 psk

k

where p1, p2, . . . , pk are distinct primes and r1, r2, . . . , rk, s1, s2, . . . , skare non-negative integers. By part (a), we have that the least positivecommon multiple of m and n is

h = pt11

pt22 ptk

k

where ti = max{ri, si}. According to Cor. 2.5.6, we have

g = gcd(m, n) = pu11

pu22 puk

k

where ui = min{ri, si}. Since max{ri, si} + min{ri, si} = ri + si, wesee that gh = mn, and h = mn/g.

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Ex 2 Solutions