We can see this by noting that A | My Assignment Tutor

1. (i) True(v) True(ii) False(vi) False(iii)False(vii) False(iv) False(viii) True 2. We note that D = f0; 1; 2g and E = f1; 2g. Hence A = D but A 6= E and D 6= E. In fact, no other pairs ofthe sets are equal. We can see this by noting that A, D and E are finite while B and C are infinite, and that-1 2 B while -1 2= C.3. A [ B = (0; 1] A B = ; B C = B A [ C = C A C = A4. (⊆): Let x 2 f0; 1g. For this, we use a proof by cases. Case 1: x = 0 ) x = 1-12 =2 . Hence, 9k 2 Z s.t. x = 1-(-2 1)k , thus, x 2 nn 2 Z : 9k 2 Z s.t. n = 1-(-1)21-(-2 1)k o.Case 2: x = 1 ) x = 1-(-1)2 =1-(-1)12 . Hence, 9k 2 Z s.t. x = 1-(-2 1)k, thus, x 2 nn 2 Z : 9k 2 Z s.t. n = 1-(-2 1)k o.Therefore, f0; 1g ⊆ nn 2 Z : 9k 2 Z s.t. n = 1-(-2 1)k o.(⊇): Let y 2 nn 2 Z : 9k 2 Z s.t. n = 1-(-2 1)k o. Now k can be odd or even, so again we consider cases.Case 1: Let k be odd. Then y = 1-(-2 1)k = 1-(2-1) = 2 2 = 1.Case 2: Let k be even. Then y = 1-(-2 1)k = 1-2 1 = 02 = 0.Hence, y 2 f0; 1g, thus, nn 2 Z : 9k 2 Z s.t. n = 1-(-2 1)k o ⊆ f0; 1g.Therefore, f0; 1g = nn 2 Z : 9k 2 Z s.t. n = 1-(-2 1)k o.5. A = B, P = B – f2g [ x : x is odd ^ x is not prime = x : x is compositeg [ f1g,P – A = f2g, B – P = B – f2g, A – B = A.A and B are disjoint as A B = ;; P 6⊆ A, since 2 2 P but 2 2= A.6. (i) (C U) [ C = (C [ C) (U [ C) = U U = U(ii) (A U) [ A = (A [ 😉 [ A = A [ A = A(iii) (C [ 😉 [ C = (C [ 😉 C = C C = ;(iv) (A B) A = A B A = (A A) B = ; B = ;7. (i) The statement is true. We haveA – B = A B Definition of set difference = B A= B – ACommutativityDefinition of set difference Hence A – B = B – A.(ii) The statement is false; here is a counterexample. Let U = N, A = f1g, B = ; and C = f1g. ThenA; B; C 2 P(U) andA – (B – C) = f1g – (; – f1g) = f1g – ; = f1g ;(A – B) – C = (f1g – 😉 – f1g = f1g – f1g = ; :As 1 2 A – (B – C) and 1 62 (A – B) – C we have that A – (B – C) 6= (A – B) – C.8. There are 35 permutations of 3 things taken 5 at a time. But 35 = 0 since k > n. There are 53 =5!(5-3)! = 60 permutations of 5 things taken 3 at a time.9. There are 25 = 32 binary words of length 5. There is one word, 00000, that does not contain a 1 and one word,11111, that does not contain a 0. Therefore, there are 32 – 2 = 30 words that contain at least one 0 and one 1.10. There are 3 ways to assign a; for each of those there are 3 ways to assign b, etc. In total, there are 3·3·3·3 = 81different functions.1

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