hw1sol

# hw1sol - € Problem 1 a 6 ∈ B 6 ∉ A b ∈ C b ∉ B{2...

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Unformatted text preview: € Problem 1 a) 6 ∈ B , 6 ∉ A , b ∈ C , b ∉ B , {2, 4, 6} ⊄ A , {2, 4, 6} ⊆ B , {4, 6, 8,10} ⊄ C b) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9,11,12} A ∩ B = {4, 8} A − B = {1, 3, 5, 7, 9,11,12} € € € € € A + B = {1, 2, 3, 5, 6, 7, 9,11,12} € € A ∩ B ∩C 2 = {∅,{4},{8},{4, 8}} € ( A ∩ B) × {a, b} = {( 4, a), ( 4, b), (8, a), (8, b)} € A ∩ ( B ∪ C ) = {4, 5, 7, 8} € c) A − ( B ∪ C ) = {1, 3, 9,11,12} € ( A − B) − C = {1, 3, 9,11,12} € ( A + B) + C = {a, b, c,1, 2, 3, 4, 8, 9,11,12} € A + ( B + C ) = {a, b, c,1, 2, 3, 4, 8, 9,11,12} € A ∪ B = 11, A ∩ B = 2 , A − B = 7 , A + B = 9 , 2 A ∩ B ∩C = 4 , € d) ( A ∩ B) × {a, b} = 4 , A ∩ ( B ∪ C ) = 4 , A − ( B ∪ C ) = 5 , ( A − B) − C = 5 , € ( A + B) + C = 11, A + ( B + C ) = 11 € € € € € € Problem 2 € € € € € a) A ∩ B ∩ C € € € € b) C − B, C ∩ B, C − ( B ∩ C ) C) ( A ∩ B) − C, ( A − C ) ∩ B, ( B − C ) ∩ A Problem 3 a) P(4) True P(12) False P(2010) False P(‐302) True ∃ x, P(x) True ∀ integer x, P(x); False ( ∀ positive integer n) ( ∃ an interger x>n) P(x); True b) P(shark) False € P(rat) True € P(dolphin) True € € P(diamond) False ( ∀ living creature x in the Washington Zoo)(P(x)) False ( ∀ sea s in the world)( ∃ some living creture x inside x) P(x) False ! depends on the assumption of what is a sea. If a lake as Dead Sea is considered as landlocked sea, then proposition is false. € € € Problem 4 a) There is at least one current GW student in Spring 2010, who is an undergraduate CS student. b) There is some current student who does not take all the courses offered at GW in spring 2010. c) There is some current undergraduate CS student who does not take all the courses offered at GW in spring 2010. d) There is some current undergraduate CS student who does not take any course offered at GW in spring 2010. e) Every current GW student takes at least one course offered at GW in spring 2010. f) There is some current student who does not take any courses offered at GW in spring 2010. Problem 5 a) p ∨ (q ∧ r) ( p ∨ q) ∧ ( p ∧ r) p q r T T T T T T T F T F T F T T T € T F F € T F F T T F F F T F F F F F T F T F F F T T b) When n=5, 3n + 2 = 35 + 2 = 243 + 2 = 245 = 5 × 7 × 7 , which is not a prime number. c) When x=5, − x 2 + 4 x + 2 = −5 2 + 4 × 5 + 2 = −25 + 20 + 2 = −3 < 0 . So with x=5, a positive integer, − x 2 + 4 x + 2 < 0 . € Problem 6 € € a) A ∪ B = A ∪ C ⇒ B = C False Disprove by counterexample: A = {1, 2}, B = {1}, C = {2} A ∪ B = A ∪ C = {1, 2}, B ≠ C € b) A ∩ B = C ∩ B ⇒ A = C False Disprove by counterexample: € A = {1, 2}, B = ∅, C = {2} € € A ∩ B = C ∩ B = ∅, A ≠ C c) € € € A − B = C − B ⇒ A = C False Disprove by counterexample: A = {1, 2}, B = {1, 2, 3}, C = {3} A − B = C − B = ∅, A ≠ C d) A + B = A + C ⇒ B = C True Proof: In order to show B = C , need to show 1) B ⊆ C , 2) C ⊆ B 1) To show B ⊆ C ∀x ∈ B , there are two cases: x ∈ A, x ∉ A x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∉ ( A + B) ⇒ x ∉ ( A + C ) ⇒ x ∉ (( A − C ) ∪ (C − A)) € € € ⇒ ( x ∉ ( A − C )) ∧ ( x ∉ (C − A)) € x ∈ A, x ∉ ( A − C ) € x ∈ A ∩ C ∴ € ( A ∩ C) ⊆ C ∴ x ∈ C x ∉ A x ∈ ( A + B) ⇒ x ∈ ( A + C ) ⇒ x ∈ (( A − C ) ∪ (C − A)) ⇒ ( x ∈ ( A − C )) ∨ ( x ∈ (C − A)) x ∉ A ⇒ x ∉ ( A − C ) ∴ x ∈ (C − A) € (C − A) ⊆ C ∴ x ∈ C ∀x ∈ B, ( x ∈ A) ∨ ( x ∉ A) ⇒ x ∈ C ∴ B ⊆ C 2) C ⊆ B can be shown the same as above, by exchanging B and C. € With both 1) and 2), A + B = A + C ⇒ B = C . € € € ...
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