Solutions Manual for Abstract Algebra Structures and Applications, 1st Edition Stephen Lovet

Solutions Manual for Abstract Algebra Structures and Applications, 1e Stephen Lovet




Solutions Manual for Abstract Algebra Structures and Applications, 1e Stephen Lovet

,1 | Set Theory
1.1 – Sets and Functions
Exercise: 1 Section 1.1
Question: Let U = {n ∈ N | n ≤ 10} and consider the subsets A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 4, 5}, and
C = {1, 2, 5, 7, 8}. Calculate the following operations.
a) A ∩ B
b) (B ∪ C) — A
c) (A ∩ B) ∩ (A ∩ C) ∩ (B ∩ C)
d) ((A — B) — C) ∩ (A — (B — C))
Solution: We apply the definitions of set operations:
a) A ∩ B = {1, 3, 5}
b) (B ∪ C) — A = {1, 2, 3, 4, 5, 7, 8} — {1, 3, 5, 7, 9} = {2, 4, 8}
c) A ∩ B∩A ∩ C∩B ∩ C = {1, 3, 5}∩{1, 5, 7}∩{1, 2, 5} = {2, 4, 6, 7, 8, 9, 10}∩{2, 3, 4, 6, 8, 9, 10}∩{4, 6, 8, 9, 10}
d) ((A — B) — C) ∩ (A — (B — C)) = ({7, 9} — C) ∩ (A — {3, 4}) = {9} ∩ {1, 5, 7, 9} = {9}

Exercise: 2 Section 1.1
Question: Let U = a, { b, c, d, e, f, g }and consider the subsets A = { a, b, d} , B = {b, c, e} , and C = {c, d, f }.
Calculate the following operations.
a) C ∩ (A ∪ B)
b) (A ∪ C) — B
c) (A ∪ B ∪ C) — (A ∩ B ∩ C)
d) (A — B) ∪ (B — C)
Solution: We apply the definitions of set operations:
a) C ∩ (A ∪ B) = C ∩ {a, b, c, d, e} = {c, d}
b) (A ∪ C) — B = {a, b, c, d, f } — B = {a, d, f }
c) (A ∪ B ∪ C) — (A ∩ B ∩ C) = {a, b, c, d, e, f } — ∅ = {a, b, c, d, e, f }
d) (A — B) ∪ (B — C) = {a, d} ∪ {b, e} = {a, b, d, e}

Exercise: 3 Section 1.1
Question: As subsets of the reals, describe the differences between the sets {3, 5}, [3, 5] and (3, 5).
Solution: The set {3, 5 }contains the integers 3 and 5. The closed interval [3, 5] contains all real numbers
between 3 and 5 including 3 and 5, while the open interval (3, 5) contains all real numbers between 3 and 5 not
including 3 and 5.

Exercise: 4 Section 1.1
Question: Prove that the following are true for all sets A and B.
a) A ∩ B ⊆ A.
b) A ⊆ A ∪ B.
Solution: We use the definitions of subsets and the intersection and union of sets.
a) Let x ∈ A ∩ B. Then x ∈ A and x ∈ B =⇒ x ∈ A, so A ∩ B ⊆ A.
b) Let x ∈ A. We know that A ∪ B = {y | y ∈ A or y ∈ B}, so x ∈ A =⇒ x ∈ A ∪ B. Hence A ⊆ A ∪ B.

Exercise: 5 Section 1.1
Question: Let A and B be subsets of a set S.
a) Prove that A ⊆ B if and only if P(A) ⊆ P(B)
b) Prove that P(A ∩ B) = P(A) ∩ P(B).
c) Show that P(A ∪ B) = P(A) ∪ P(B) if and only if A ⊆ B or B ⊆ A.
Solution:

1

,2 CHAPTER 1. SET THEORY

a) (=⇒): Suppose A ⊆ B. Then, ∀a ∈ A, a ∈ B. Since P(B) contains all the possible subsets of B, all the possible
subsets of A must be in P(B) because A ⊆ B. Therefore, P(A) ⊆ P(B).
(⇐=): Suppose P(A) ⊆ P(B). Then ∀{a} ∈ P(A), {a} ∈ P(B). Therefore, there must exist a subset C of
P(B) that contains every {a} from P(A). The subset C leads to the conclusion that every a ∈ A
must also be in B. Therefore, A ⊆ B.
b) By definition, P(A ∩ B) = {{t1, t2, ..., tn} | ti ∈ A, ti ∈ B}. This implies {ti} ∈ P(A) and {ti} ∈ P(B).
Therefore, by definition of intersection, P(A ∩ B) = P(A) ∩ P(B).
c) (=⇒): Suppose there are two sets A and B such that neither A ⊆ B nor B ⊆ A. Let a ∈ A — B and b ∈
B — A. Then the set {a, b} is in P(A ∪ B) but not in P(A) or in P(B). Therefore by the
contrapositive, P(A ∪ B) = P(A) ∪ P(B) if A ⊆ B or B ⊆ A.
(⇐=): Suppose A ⊆ B. Then, A ∪ B = B so P(A ∪ B) = P(B). Now suppose B ⊆ A. Then A ∪ B = A so
P(A ∪ B) = P(A). Either way, P(A ∪ B) = P(A) ∪ P(B).


Exercise: 6 Section 1.1
Question: Give the list description of P({1, 2, 3, 4}).
Solution: Using the definition of a power set,

P({1, 2, 3, 4}) ={∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4},
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.




Exercise: 7 Section 1.1
Question: Give the list description of {{a1, a2, . . . , ak} ∈ P({1, 2, 3, 4, 5}) a1 + a2 + · · · + ak = 8}.
Solution: We need to find all the subsets of {1, 2, 3, 4, 5} whose elements add to a total of 8. Recall that no
subset has repeated elements so {4, 4, } does not make sense. The set is

{{1, 2, 5}, {1, 3, 4}, {3, 5}} .




Exercise: 8 Section 1.1
Question: Let A, B, and C be subsets of a set S.
a) Prove that (A — B) — C = (A — C) — (B — C).
b) Find and prove a similar formula for A — (B — C).

Solution:




a)
In the first Venn diagram, the lighter shade represents (A— B), and the darker shade, which overlaps some
of (A — B), represents (A — B) —C. In the second Venn diagram, the lighter shade represents (A— C),
while the darker shade represents (A — C)— (B —C). We observe from the diagrams that the darker regions
are equal.




Solutions Manual for Abstract Algebra Structures and Applications, 1e Stephen Lovet

, 1.1. SETS AND FUNCTIONS 3




b)
In the Venn diagram above, the lighter shade represents B —C, and the darker shade represents A—(B —C). In
the second diagram, the lighter region represents A — B, and the darker region represents A — C, which overlaps
some of A — B. Thus, (A — B) ∪ (A — C) = A — (B — C).

Exercise: 9 Section 1.1
Question: Let A, B, and C be subsets of a set S.
a) Prove that AΔB = ∅ if and only if A = B.
b) Prove that A ∩ (BΔC) = (A ∩ B)Δ(A ∩ C).
Solution: Let A, B, and C be subsets of a set S.
a) Suppose that AΔB = ∅. Then by definition of the symmetric difference

(A — B) ∪ (B — A) = ∅.

If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce that
A — B = ∅ and B — A = ∅. Now for and two sets U and T , the identity U — T = ∅ is equivalent to U ⊆ T . Hence
we deduce that A ⊆ B and B ⊆ A. Consequently, A = B.
The argument of the opposite direction is identical. Suppose that A = B. Then A ⊆ B and B ⊆ A. Thus
A — B = ∅ and B — A = ∅. We deduce that AΔB = (A — B) ∪ (B — A) = ∅.
b) There are a variety of ways to prove the identity∩ A Δ(B C) = (A∩ B)Δ (A ∩C). We could use a well
designed Venn diagram. We could also use a membership table which lists all possibilities of an element
whether it is in or not in one of the given three sets. Here is a membership table for both side of the equality.
√
In this table, we put an in a column to indicate membership and nothing to indicate non-membership.
√
Hence if there is a in the A and C column and nothing in the B column, that refers to the situations of
an element in A, not in B and in C.

(A ∩ B) (A ∩ C) (A ∩ B)Δ(A ∩ C)




Since the A ∩ (B ΔC) and column and the (A ∩ B) Δ(A ∩ C) of this membership table are the same, then
the sets are equal.

Exercise: 10 Section 1.1
Question: Let S be a set and let {Ai}i∈I be a collection of subsets of S. Prove the following.
[ \
a) Ai = Ai.
i∈I i∈I
\ [
b) Ai = Ai.
i∈I i∈I

Solution: Let S be a set and let {Ai}i∈I be a collection of subsets of S.