Ex 1 Solutions

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CM121A, Abstract Algebra Solution Sheet 1

1. Consider the following subsets ofZ:

A = { multiples of 3 }, B = { a Z |6 < a 6 }, C = {3, 2, 4, 9, 31 }.

(a) Find B (A C).Solution: A C = {3, 9}, so

B (A C) = {5,4,3,2,1, 0, 1, 2, 3, 4, 5, 6, 9 }.

(b) Find (B A) C.Solution: The only element of C which is not in B A is 31, so(B A) C = {3, 2, 4, 9}.

(c) Find (B \ A) C (where X\ Y denotes the set of elements of X

which are not in Y).Solution: B \ A = {5,4,2,1, 1, 2, 4, 5}, so (B \ A) C ={2, 4}.

2. Let A, B and C be subsets of a set X. Prove that

(A B) C = (A C) (B C).

Solution: Suppose that x (A B) C. Then x A B, andx C. Since x A B, we have that x A or x B. Supposefirst that x A. Since also x C, it follows that x A C, andtherefore x (A C) (B C). Simlarly, if x B, then x B C,so x (A C) (B C). We have now shown that

(A B) C (A C) (B C).

Now suppose that x (AC)(BC). Then x AC, or x BC.Suppose first that x A C. Then x A A B and x C, sox (A B) C. Similarly if x B C, then x B A B andx C, so x (A B) C. We have now shown that

(A B) C (A C) (B C).

Combining the two inclusions shows the two sets are equal.

3. Decide which of the following assertions are true for all x, y , z R. Iffalse, give a counterexample.

(a) Ifx2 y2, then x y.Solution: False, take x = 0, y = 1.

(b) 0 x y x2 y2.Solution: True.

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(c) x z y z if and only if x y.Solution: True.

(d) x 2 x2 2x.Solution: False, take x = 1.

4. Prove by induction that the sum of the first n odd positive squares is1

3(4n3 n), i.e., that

1 + 9 + 25 + + (2n 1)2 =1

3(4n3 n).

Solution: Let P(n) be the equation we need to prove. Then P(1) isobviously true since 1 = 1

3(4 13 1). Suppose now that n 1 and

P(n) is true, and we will show that P(n +1) is true. Adding (2n + 1)2

(the next odd square) to both sides of P(n) gives

1 + 9 + 25 + + (2n + 1)2 = 13

(4n3 n) + (2n + 1)2

= 43

n3 13

n + 4n2 + 4n + 1= 4

3n3 + 4n2 + 11

3n + 1.

We need to show that this agrees with the right-hand-side of P(n +1),which is

1

3(4(n + 1)3 (n + 1)) = 1

3(4(n3 + 3n2 + 3n + 1) (n + 1))

= 43

n3 + 4n2 + 113

n + 1.

Since P(1) is true, and P(n) P(n + 1) for all n 1, it follows that

P(n) is true for all n N

.5. Suppose that a,b,c,d Z and n N. Prove the following:

(a) Ifa|b and b|c, then a|c.Solution: If a|b then b = sa for some s Z (by definition ofdivisibility), and if b|c then c = tb for some t Z (again bydefinition). Therefore c = tb = t(sa) = (ts)a, and since ts Z,we conclude that c is divisible by a.

(b) Ifa|n, then a n.Solution: First note that ifa 0, then a n (since we assumedn > 0). So we may assume that a > 0. If a|n, then n = sa for

some s Z

. Since a and n are positive, so is s. Therefore s 1.Since a is positive, it follows that as a 1, so n a.

(c) Ifn|a and n|b, then n|(ac + bd).Solution: If n|a and n|b, then a = sn and b = tn for somes, t Z. Therefore ac + bd = (sn)c + (tn)b = (sc + tb)n, so ac + bdis divisible by n.

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6. Use the Euclidean algorithm to find gcd(1066, 2009) and express it inthe form 1066m + 2009n with m, n Z.Solution: The Euclidean algorithm gives:

2009 = 1 1066 + 943,1066 = 1 943 + 123,

943 = 7 123 + 82,123 = 1 82 + 41,

82 = 2 41.

Therefore gcd(1066, 2009) = 41.Writing 41 in terms of the preceding two remainders and iterativelysubstituting gives:

41 = 123 82 = 123 (943 7 123) = 8 123 943

= 8(1066 943) 943 = 8 1066 9 943= 8 1066 9(2009 1066) = 17 1066 9 2009,

so let m = 17 and n = 9.

7. Suppose that a1, a2, . . . , ak are integers, not all of which are 0.

(a) Write down the definition of gcd(a1, a2, . . . , ak).Solution: g = gcd(a1, a2, . . . , ak) (the greatest common divisorof a1, a2, . . . , ak) if:1) g Z and g|ai for i = 1, . . . , k, and2) d Z and d|ai for i = 1, . . . , k d g.Remark: Variants on this with the same meaning are fine, forexample:gcd(a1, a2, . . . , ak) is the greatest integer g such that g|a1, g|a2,. . . , g|ak.

(b) Prove that gcd(a1, a2, . . . , ak) is the least positive integer of theform n1a1 + n2a2 + + nkak with n1, n2, . . . , nk Z.Solution: Let

S = { n1a1 + n2a2 + + nkak | n1, n2 . . . , nk Z }.

Then S contains a positive integer, for example choose i so thatai = 0 and then |ai| is a positive element ofS. Let m be the leastpositive element of S.We show that m|ai for each i = 1, . . . , k. By the Division Algo-rithm, ai = mq+ r for some q, r Z, and 0 r < m. We willprove by contradiction that r = 0. Suppose that r > 0. Sincem = n1a1 + n2a2 + + nkak for some n1, n2, . . . , nk Z, we

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have

r = ai qm

= ai q(n1a1 + n2a2 + + nkak)= (qn1)a1 + (qn2)a2 + + (1 qni)ai + + (qnk)ak.

Therefore r S. Recall that 0 r < m. If r > 0, then sincer < m, we get a contradiction to m being the least element of S.We therefore conclude that r = 0 and ai = mq is divisible by m.We have now shown that m is a common divisor of a1, a2, , ak.Now suppose that d is any common divisor ofa1, a2, , ak. Thenfor each i = 1, 2, . . . , k, we have ai = dci for some ci Z. There-fore

m = n1a1 + n2a2 + + nkak = d(n1c1 + n2c2 + + nkck)

is divisible by d. Since d|m and m > 0, it follows that d m.We have now shown that m = gcd(a1, a2, . . . , ak).

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Ex 1 Solutions