9 points I. Consider the following subsets of Z. A = {2n + E Z} B = {3nin E C={3n+2InEZ} Find each of the following sets, and express it in set-builder notation. a. A— B b. BnC c. C n a 9 points 2....

Q1:
A = {2n+ 1 | n ∈ Z}
B = {3n | n ∈ Z}
C = {3n+ 2 | n ∈ Z}
Hence,
A = {· · · − 9,−7,−5,−3,−3, 1, 3, 5, 7, 9, 11, 13, 15, . . . }
B = {· · · − 9,−6,−3, 0, 3, 6, 9, 12, 15, . . . }
C = {· · · − 7,−4,−1, 2, 5, 8, 11, . . . }
(a)
A−B = {· · · − 7,−5,−1, 1, 5, 7, 11, 13, . . . }
Thus A-B contains all odd numbers except the ones which are divisible by 3.
In set-builder notation,
A−B = {x ∈ Z | x = 2m+ 1,m ∈ Z, x 6= 3n, n ∈ Z}
(b)
B ∩ C = {φ}
because set B contains all odd integers while set C contains all integers 2 more than the
odd integers. hence both sets cannot have any number in common
(c)
B = {· · · − 5,−4,−2,−1, 1, 2, 4, 5, 7, 8, 10, 11 . . . }
C = {· · · − 7,−4,−1, 2, 5, 8, 11, . . . }
C ∩B = {· · · − 7,−4,−1, 2, 5, 8, 11, . . . } = C
Hence, C ∩B = {3n+ 2 | n ∈ Z}
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Q2:
To prove: a) A ∩B = φ
b) A−B = A
c) A∆B = A ∪B
are equivalent
We already know that, A−B = A ∩B′
Using that,
A−B = A ∩B′
= A ∩ (U −B)
= (A ∩ U)− (A ∩B)
= A− (A ∩B)
From a), we know that A ∩B = φ
Putting that in above equation,
A−B = A− φ = A
1
Thus proved that a) and b) are equivalent, as a) helps to prove b). - [1]
To prove a) and c) are equivalent,
A∆B = (A ∪B)− (A ∩B)
From a), A ∩B = φ
Hence, A∆B = A ∪B − φ
Hence, A∆B = A ∪B
Thus a) helped prove c). hence both a) and c) are equivalent. -[2]
Statements [1] and [2] show that a), b) and c) are all equivalent.
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Q3:
To find out if A∆(B ∩ C) = (A∆B) ∩ (A∆C)
Figure 1: Venn Diagram for question 3
In this venn diagram, all independent regions are shown by numbers from 1 to 7.
To calculate right hand side of equation,
A∆B contains regions 2,4,3,5.
A∆C contains regions 2,6,1,5.
Hence, (A∆B) ∩ (A∆C) contains regions 2 and 5. -[1]
To calculate left hand side of equation,
(B ∩ C) contains regions 5 and 7.
A contains regions 2,4,6,7.
Hence, A∆(B ∩ C) contains regions 2,4,6,5. -[2]
Thus we can see that statements [1] and [2] do not contain same regions.
Hence we prove that the left hand side and right hand side of the equations are not equiva-
lent.
Hence the statement is False and the above venn...