Mathematics 220 Homework 4 - Solutions. Solution: We must prove the two statements: (1) if A = B, then A B = A B, and (2) if A B = A B, then A = B.

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1 1. (4.46) Let A and B be sets. Prove that A B = A B if and only if A = B. Solution: We must prove the two statements: (1) if A = B, then A B = A B, and (2) if A B = A B, then A = B. Proof of (1): Suppose that A = B. Then A B = B B = B. Also, A B = B B = B. Hence, A B = B = A B and so (1) is proved. Proof of (2): Suppose that A B = A B. We will show that A = B by first proving that A B and second that B A. First, let x A. Then x A B, and since A B = A B this means that we also have x A B. Hence x B, and so A B. Next, let x B. Then x A B, and since A B = A B this means that we also have x A B. Hence x A, and so B A. Since A B and B A, we have proved that A = B. both been proved, A B = A B if and only if A = B. Thus, since (1) and (2) have 2. (4.54) Prove that A B = A B for every two sets A and B (Theorem 4.22(4b)). Solution: We first show A B A B. Let x A B. Then x A B. Thus, x A or x B. Without loss of generality, assume that x A. Then it follows that x A and so x A B. Next, we show that A B A B. Let x A B. Then x A or x B. Without loss of generality, assume that x A. Then it follows that x A and so x A B. Hence, x A B. Thus, A B = A B for every two sets A and B. 3. (4.56) Let A, B, and C be sets. Prove that (A B) (A C) = A (B C). Solution: We first show that (A B) (A C) A (B C). Let x (A B) (A C). Then x A B or x A C. Without loss of generality, assume that x A B. Then x A and x B. Hence, x A and x B C. Therefore, x A (B C). Next, we show that A (B C) (A B) (A C). Let x A (B C). Then Page 1 of 6

2 x A and x B C. Hence, x A, and x B or x C. Without loss of generality, assume that x B. Then, we have x A and x B and so x A B. Hence, x (A B) (A C). Thus, (A B) (A C) = A (B C). 4. (4.64) For sets A and B, find a necessary and sufficient condition for (A B) (B A) =. Verify that this condition is necessary and sufficient. Solution: The necessary and sufficient condition is that A and B are disjoint. That is, A B =. To verify this, we prove the statement For sets A and B, (A B) (B A) = if and only if A B =. We must prove two implications: 1. If A B =, then (A B) (B A) =. 2. If (A B) (B A) =, then A B =. Proof of 1. We prove the contrapositive: if (A B) (B A), then A B Assume that (A B) (B A), and let x (A B) (B A). Then x A B and x B A. Since x A B, there exists c A and d B such that x = (c, d). Also, however, since x B A we also must have c B and d A. Hence, c A B and so A B. Proof of 2. We prove the contrapositive: if A B, then (A B) (B A). Assume that A B and let x A B. Then x A and x B, and hence (x, x) A B and also (x, x) B A. Therefore, (x, x) (A B) (B A), and so (A B) (B A). Since both implications are proved, the statement is true. 5. (4.60) For A = {x, y}, determine A P(A) Solution: Here we have P(A) = {, {x}, {y}, {x, y}}, so A P(A) = {(x, ), (x, {x}), (x, {y}), (x, {x, y}), (y, ), (y, {x}), (y, {y}), (y, {x, y})} 6. (4.62) Let A and B be sets. Prove that A B = if and only if A = or B =. Page 2 of 6

3 Solution: Proof: First, by the definition of Cartesian product, if A = or B = then A B =. So, we need only to prove the converse statement, which is if A B = then A = or B =. To prove this statement, we prove the contrapositive: if A and B, then A B. So, assume that A and B. Since A, there exists an element a A. Since B, there exists an element b B. Thus, (a, b) A B. Hence, A B. 7. (4.66) Result 4.23 states that if A, B, C, and D are sets such that A C and B D, then A B C D. (a) Show that the converse of Result 4.23 is false. Solution: The converse is if A B C D, then A C and B D. We can show a counterexample to this to show it is false. Take, for instance, A =, B = {1}, C = {2} and D = {3}. Then and so A B C D, but B D. A B =, C D = {(2, 3)} (b) Under what added hypothesis is the converse true? Prove your assertion. Solution: The additional hypothesis required for the converse to be true is that A and B must both be non-empty. We can then prove the statement: If A and B are non-empty and A B C D, then A C and B D. Proof: Let C and D be sets and let A and B be non-empty sets such that A B C D. First, we prove that A C. Let a A. Then, since B, there exists b B. Hence, (a, b) A B. Because A B C D, it follows that (a, b) C D and hence a C. Thus, A C. Next, we prove that B D. Let b B. Then, since A, there exists a A. Hence, (a, b) A B. Because A B C D, it follows that (a, b) C D and hence b D. Thus, B D. 8. (4.70) Let A and B be sets. Show, in general, that A B A B. Solution: We need only show that a counterexample exists. For instance, consider the sets A = {1}, B = {2} with the universal set U = {1, 2}. Note that the corresponding universal set for A B is U U = {(1, 1), (1, 2), (2, 1), (2, 2)}. Then we have A B = {(1, 2)}, so A B = {(1, 1), (2, 1), (2, 2)} Page 3 of 6

4 and Here we clearly have A B A B. A = {2}, B = {1}, so A B = {(2, 1)} 9. (4.4) Let x, y Z. Prove that if 3 x and 3 y, then 3 (x 2 y 2 ). Solution: Assume that 3 x and 3 y. Then x = 3p + 1 or x = 3p + 2 for some integer p and y = 3q +1 or y = 3q +2 for some integer q. We then consider the following four cases: Case 1: x = 3p + 1 and y = 3q + 1. Then x 2 y 2 = (3p + 1) 2 (3q + 1) 2 = 3(3p 2 + 2p 3q 2 2q) Since 3p 2 + 2p 3q 2 2q is an integer, 3 (x 2 y 2 ). Case 2: x = 3p + 2 and y = 3q + 2. Then x 2 y 2 = (3p + 2) 2 (3q + 2) 2 = 3(3p 2 + 4p 3q 2 4q) Since 3p 2 + 4p 3q 2 4q is an integer, 3 (x 2 y 2 ). Case 3: x = 3p + 1 and y = 3q + 2. Then x 2 y 2 = (3p + 1) 2 (3q + 2) 2 = 3(3p 2 + 2p 3q 2 4q 1) Since 3p 2 + 2p 3q 2 4q 1 is an integer, 3 (x 2 y 2 ). Case 4: x = 3p + 2 and y = 3q + 1. Then x 2 y 2 = (3p + 2) 2 (3q + 1) 2 = 3(3p 2 + 4p 3q 2 2q + 1) Since 3p 2 + 4p 3q 2 2q + 1 is an integer, 3 (x 2 y 2 ). Since we have shown it is true in all possible cases, if 3 x and 3 y, then 3 (x 2 y 2 ). 10. (4.10) Let n Z. Prove that 2 (n 4 3) if and only if 4 (n 2 + 3). Solution: We first prove the implication If 4 (n 2 + 3), then 2 (n 4 3). Assume that 4 (n 2 + 3). Then n = 4m for some integer m. Hence n 2 = 4m 3 and so n 4 3 = (4m 3) 2 3 = 16m 2 24m + 6 = 2(8m 2 12m + 3). Since 8m 2 12m + 3 is an integer, 2 (n 4 3). Page 4 of 6

5 Next, we prove the implication If 2 (n 4 3), then 4 (n 2 + 3). Assume that 2 (n 4 3). Then n 4 3 = 2m for some integer m. Thus, n 4 = 2m + 3 = 2(m + 1) + 1. Since m + 1 Z, this shows that n 4 is odd. Now, by theorem 3.12 (in fact, by this theorem stated in terms of contrapositives), n 2 is odd. By applying theorem 3.12 again, we have that n is odd. Hence n = 2k + 1 for some integer k, and so n = (2k + 1) = 4k 2 + 4k + 4 = 4(k 2 + k + 1) and since k 2 + k + 1 is an integer, 4 (n 2 + 3), as required. Since both implications have been proved, 2 (n 4 3) if and only if 4 (n 2 + 3). 11. (4.16) Let a, b Z. Prove that if a 2 + 2b 2 0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. Solution: We prove the contrapositive: If one of a or b is congruent to 0 modulo 3 and the other is not congruent to 0 modulo 3, then a 2 + 2b 2 0 (mod 3). We consider 2 cases: Case 1: a 0 (mod 3) and b 0 (mod 3) Since a 0 (mod 3), by Result 4.11 (useful congruence result (3) from class), a 2 0 (mod 3). Since b 0 (mod 3), either b 1 (mod 3) or b 2 (mod 3). We now have 2 subcases: 1. If b 1 (mod 3), then by Result 4.11 b 2 1 (mod 3), and by Result 4.9 (useful congruence result (1) from class) 2b 2 2 (mod 3). Hence, by Result 4.10 (useful congruence result (2) from class) a 2 + 2b 2 2 (mod 3) and so a 2 + 2b 2 0 (mod 3), as required. 2. If b 2 (mod 3), then by Result 4.11 b (mod 3), and by Result 4.9, 2b 2 2 (mod 3). Hence, by Result 4.10, a 2 + 2b 2 2 (mod 3) and so a 2 + 2b 2 0 (mod 3), as required. Case 2: b 0 (mod 3) and a 0 (mod 3) Since b 0 (mod 3), by Result 4.11, b 2 0 (mod 3) and then by Result 4.10 we have 2b 2 0 (mod 3). Since a 0 (mod 3), either a 1 (mod 3) or a 2 (mod 3). We now have 2 subcases: 1. If a 1 (mod 3), then by Result 4.11 a 2 1 (mod 3). Hence, by Result 4.10, a 2 + 2b 2 2 (mod 3) and so a 2 + 2b 2 0 (mod 3), as required. 2. If a 2 (mod 3), then by Result 4.11 a (mod 3). Hence, by Result 4.10, a 2 + 2b 2 2 (mod 3) and so a 2 + 2b 2 0 (mod 3), as required. In all cases, a 2 + 2b 2 0 (mod 3), thus proving the contrapositive and so the original statement is true. Page 5 of 6

6 12. (4.24) Let x and y be even integers. Prove that x 2 y 2 (mod 16) if and only if either (1) x 0 (mod 4) and y 0 (mod 4) or (2) x 2 (mod 4) and y 2 (mod 4). Solution: Proof: Let x and y be even integers. We must prove the two implications: (I) if either (1) x 0 (mod 4) and y 0 (mod 4) or (2) x 2 (mod 4) and y 2 (mod 4), then x 2 y 2 (mod 16) and (II) if x 2 y 2 (mod 16), then either (1) x 0 (mod 4) and y 0 (mod 4) or (2) x 2 (mod 4) and y 2 (mod 4), then x 2 y 2 (mod 16) To prove (I), we consider the two cases: Case 1: Suppose that x 0 (mod 4) and y 0 (mod 4). Then there exist integers p and q such that x = 4p and y = 4q. Hence x 2 y 2 = (4p) 2 (4q) 2 = 16p 2 16q 2 = 16(p 2 q 2 ) and since p, q Z, this shows that x 2 y 2 (mod 16). Case 2: Suppose that x 2 (mod 4) and y 2 (mod 4). Then there exist integers p and q such that x = 4p + 2 and y = 4q + 2. Hence x 2 y 2 = (4p + 2) 2 (4q + 2) 2 = 16p p q 2 16q 4 = 16(p 2 + p q 2 q) and since p, q Z, this shows that x 2 y 2 (mod 16). To prove (II), we prove the contrapositive: if neither (1) nor (2) hold, then x 2 y 2 (mod 16). Assume that neither (1) nor (2) is true. Then, since x and y are even, there are only 2 possible cases: Case 1: Suppose x 0 (mod 4) and y 2 (mod 4). Then, x = 4p and y = 4q + 2 for some p, q Z. Thus, x 2 y 2 = (4p) 2 (4q +2) 2 = 16p 2 16q 2 16q 4 = 16(p 2 q 2 q 1)+12 and so x 2 y 2 (mod 16). Case 2: Suppose x 2 (mod 4) and y 0 (mod 4). Then, x = 4p + 2 and y = 4q for some p, q Z. Thus, x 2 y 2 = (4p + 2) 2 (4q) 2 = 16p p q 2 = 16(p 2 + p q 2 ) + 4 and so x 2 y 2 (mod 16). Since we have proved both (I) and (II), the biconditional is proved. Page 6 of 6