Abstract Algebra Theory and Applications - Computer Science ...

Hints and SolutionsChapter 0. Preliminaries1. (a) {2}. (b) {5}.2. (a) {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)}.(d) ∅.6. If x ∈ A∪(B ∩C), then either x ∈ A or x ∈ B ∩C ⇒ x ∈ A∪B and A∪C ⇒x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C).Conversely, x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ x ∈ A ∪ B and A ∪ C ⇒ x ∈A or x is in both B and C ⇒ x ∈ A ∪ (B ∩ C) ⇒ (A ∪ B) ∩ (A ∪ C) ⊂A ∪ (B ∩ C). Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).10. (A ∩ B) ∪ (A \ B) ∪ (B \ A) = (A ∩ B) ∪ (A ∩ B ′ ) ∪ (B ∩ A ′ ) = [A ∩ (B ∪B ′ )] ∪ (B ∩ A ′ ) = A ∪ (B ∩ A ′ ) = (A ∪ B) ∩ (A ∪ A ′ ) = A ∪ B.14. A \ (B ∪ C) = A ∩ (B ∪ C) ′ = (A ∩ A) ∩ (B ′ ∩ C ′ ) = (A ∩ B ′ ) ∩ (A ∩ C ′ ) =(A \ B) ∩ (A \ C).17. (a) Not a map. f(2/3) is undefined.(c) Not a map. f(1/2) = 3/4 and f(2/4) = 3/8.18. (a) One-to-one but not onto. f(R) = {x ∈ R : x > 0}.(c) Neither one-to-one nor onto.20. (a) f(n) = n + 1.22. (a) Let x, y ∈ A. Then g(f(x)) = (g ◦ f)(x) = (g ◦ f)(y) = g(f(y)) ⇒ f(x) =f(y) ⇒ x = y, so g ◦ f is one-to-one.(b) Let c ∈ C, then c = (g ◦f)(x) = g(f(x)) for some x ∈ A. Since f(x) ∈ B,g is onto.23. f −1 (x) = (x + 1)/(x − 1).24. (a) Let y ∈ f(A 1 ∪ A 2 ) ⇒ there exists an x ∈ A 1 ∪ A 2 such that f(x) = y ⇒y ∈ f(A 1 ) or f(A 2 ) ⇒ y ∈ f(A 1 ) ∪ f(A 2 ) ⇒ f(A 1 ∪ A 2 ) ⊂ f(A 1 ) ∪ f(A 2 ).391