Homework 10 Solutions

Homework 10 Solutions - Homework 10 Section 8.1 1.(M) 48a)...

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Homework 10 Section 8.1 1.(M) 48a) Since R and S are both having all pairs like (x,x) where x is an element of set A the union will also have it. Thus R S is also reflexive. 48b) R S will be reflexive as both R and S have pairs like (x,x). 48c) R S will be irreflexive as it does not contain any of the (x,x) type pairs. 48d) Since both R and S have all (x,x) kind of pairs R - S would have none. So R - S would be irreflexive. 48e) Since both R and S have all (x,x) kind of pairs the S R would have them too and hence would be reflexive. 2.(E) 54a) R 2 is R R . So R 2 = { (1,2), (1,3), (1,4), (1,1), (1,5), (2,1), (2,4), (2,5), (2,2), (3,1), (3,2), (3,3), (3,5), (3,4), (4,3), (4,4), (4,1), (4,2), (5,1), (5,2), (5,3), (5,4), (5,5) } 54b) R 3 = { (1,2), (1,3), (1,4), (1,1), (1,5), (2,1), (2,3), (2,4), (2,5), (2,2), (3,1), (3,2), (3,3), (3,5), (3,4), (4,3), (4,4), (4,1), (4,2), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) } 3.(C) 56) We prove this by induction. For n = 1 this is trivial as R is given as symmetric. Assume the result to be true for
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Homework 10 Solutions - Homework 10 Section 8.1 1.(M) 48a)...

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