- Verify the commutative property of union and intersection of sets for the following.
A = {l, m, n, o, p, q}
B = {m, n, o, r, s, t}
Solution:
A ∪ B = {l, m, n, o, p, q, r, s, t}
B ∪ A = {l, m, n, o, p, q, r, s, t}
Therefore A ∪ B = B ∪ A
Hence, union of sets is commutative.
A ∩ B = {m, n, o}
B ∩ A = {m, n, o}
Therefore A ∩ B = B ∩ A
Hence, intersection of sets is commutative.
- Given P = {a, b, c, d, e}, Q={a, e, i, o, u} and R = {a, c, e, g}, verify associative property of union and intersection of sets.
Solution:
Consider the three sets P,Q and R
P = {a, b, c, d, e} , Q = {a, e, i, o, u} and R = {a, c, e, g}
PUQ = {a, b, c, d, e, i, o, u}
(PUQ)UR = {a, b, c, d, e, i, o, u} U {a, c, e, g}
(PUQ)UR = {a, b, c, d, e, g, i, o, u}………………….(1)
Now, we find the union of Q and R and later its union with set P
PU(QUR) = {a, b, c, d, e} U {a, c, e, i, o, g, u}
PU(QUR) = {a, b, c, d, e, i, o, u, g}…………………………(2)
Therefore, by comparing (1) and (2) we can observe that they are same.
(PUQ)UR = PU(QUR)
Therefore union of sets is associative
Associative property of intersections of sets:
P={a, b, c, d, e} ; Q = {a, e, i, o, u} ; R = {a, c, e, g}
P∩Q = {a, b, c, d, e} ∩ {a, e, i, o, u}
P∩Q = {a, e}
(P∩Q) ∩ R = {a, e} ∩{a, c, e, g} ={a, e}……………..(3)
Now, let us find the intersection of P with (Q∩R) (Q∩R)= {a, e, i, o, u} ∩ {a, c, e, g}
(Q∩R) = {a, e}
P∩ (Q ∩ R) = {a, b, c, d, e} ∩ {a, e} ={a, e}…………………..(4)
By comparing (3) and (4) we can observe that they are same.
Therefore (P∩Q) ∩ R = P∩ (Q ∩ R)
Therefore intersection of sets is associative.
- If A={-3, -1, 0, 4, 6, 8, 10}, B = {-1, – 2, 3, 4, 5, 6} and C = {-6, -4, -2, 2, 4, 6}
Show that A∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
Solution:
Consider, L.H.S = 𝐴 ∪ (𝐵 ∩ 𝐶)
(𝐵 ∩ 𝐶) = {−1, −2, 3, 4, 5, 6} ∩ {−6, −4, −2, 2, 4, 6}
= {-2, 4, 6}
Then, 𝐴 ∪ (𝐵 ∩ 𝐶) = {−3, −1, 0, 4, 6, 8, 10} ∪ {−2, 4, 6}
𝐴 ∪ (𝐵 ∩ 𝐶) = {−3, −2, −1, 0, 4, 6, 8, 10}…………………..(1)
Consider RHS = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
(A∪ 𝐵) = {−3, −1, 0, 4, 6, 8, 10} ∪ {−1, −2, 3, 4, 5, 6} = {-3, -2, -1, 0, 3, 4, 5, 6, 8, 10}
(𝐴 ∪C)={-3,- 1, 0, 4, 6, 8, 10} ∪ {−6, −4, −2, 2, 4, 6} ={-6, -4, -3, -2, -1, 0, 2, 4, 6, 8, 10}
(𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)={−3, −2, −1, 0, 3, 4, 5, 6, 8, 10} ∩{- 6,-4, -3, -2, -1, 0, 2, 4, 6, 8, 10}
(𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)={−3, −2, −1, 0, 4, 6, 8, 10}…………………….(2)
∴ from (1) and (2) A∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
- If U = {4, 8, 12, 16, 20, 24, 28}, A={8, 16, 24}, B={4, 16, 20, 28}. Verify that
(i) (A∪B)’=A’∩ 𝐵 ′
(𝑖𝑖)(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
Solution:
(i) (𝐴 ∪ 𝐵) = {8, 16, 24} ∪ {4, 16, 20, 28}
= {4, 8, 16, 20, 24, 28}
(A∪B)’= U-{A∪B}
={4, 8, 12, 16, 20, 24, 28} – {4, 8, 16, 20, 24, 28} = {12}
A’ = U – A = {4, 8, 12, 16, 20, 24, 28} – {8, 16, 24}
= {4, 12, 20, 28}
B’= U – B = {4, 8, 12, 16, 20, 24, 28} – {4, 16, 20, 28}
= {8, 12, 24}
A’∩ B’={4, 12, 20, 28} ∩ {8, 12, 24}
= {12}
∴ (A ∪ B)’= A’∩ 𝐵′
(ii) (𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
(𝐴 ∩ 𝐵) = {8, 16, 24} ∩ {4, 16, 20, 28} = {16}
A’= U – A = {4, 8, 12, 16, 20, 24, 28} – {8, 16, 24}
= {4, 12, 20, 28}
B’= U – B = {4, 8, 12, 16, 20, 24, 28} – {4, 16, 20, 28}
= {8, 12, 24}
(𝐴 ∩ 𝐵)’= U – (A ∩ B) ={4, 8, 12, 16, 20, 24, 28} – {16}
= {4, 8, 12, 20, 24, 28}………………..(1)
A’ ∪ B’ = {4, 12, 20, 28} ∪ {8, 12, 24} = {4, 8, 12, 20, 24, 28}……………(2)
From (1) and (2),
(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
- If A = {1, 2, 3} and B = {2, 3, 4, 5} are the subsets of U = {1, 2, 3, 4, 5, 6, 7, 8} verify
(i) (A∪B)’=A’∩ 𝐵′
(𝑖𝑖)(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
Solution:
(i) (𝐴 ∪ 𝐵) = {1, 2, 3} ∪ {2, 3, 4, 5} = {1, 2, 3, 4, 5}
(A ∪ B)’= U – {A ∪ B} ={1, 2, 3, 4, 5, 6, 7, 8} – {1, 2, 3, 4, 5} = {6, 7, 8}……………………(1)
A’= U – A ={1, 2, 3, 4, 5, 6, 7, 8} – {1, 2, 3} = {4, 5, 6, 7, 8}
B’= U – B = {1, 2, 3, 4, 5, 6, 7, 8} – {2, 3, 4, 5} = {1, 6, 7, 8}
A’∩B’= {4, 5, 6, 7, 8} ∩ {1, 6, 7, 8} = {6, 7, 8}…………….(2)
From (1) and (2)
∴(A ∪ B)’= A’∩ 𝐵 ′ ={6,7,8}
(ii) (𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
(𝐴 ∩ 𝐵) = {1, 2, 3} ∩ {2, 3, 4, 5} = {2, 3}
(𝐴 ∩ 𝐵)’=U – (A∩B) ={1, 2, 3, 4, 5, 6, 7, 8} – {2, 3} ={1, 4, 5, 6, 7, 8}………..(3)
Consider A’∪B’
A’ = U – A = {1, 2, 3, 4, 5, 6, 7, 8} – {1, 2, 3} = {4, 5, 6, 7, 8}
B’ = U – B = {1, 2, 3, 4, 5, 6, 7, 8} – {2, 3, 4, 5} = {1, 6, 7, 8}
A’ ∪ B’ = {4, 5, 6, 7, 8} ∪ {1, 6, 7, 8} = {1, 4, 5, 6, 7, 8}………….(4)
From (3) and (4)
(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
- If A = {2, 3, 5, 7, 11, 13}, B = {5, 7, 9, 11, 15} are the subsets of U = {2, 3, 5, 7, 9, 11, 13, 15}, Verify De Morgan’s laws. De Morgan’s laws: (i) (A ∪ B)’= A’∩ 𝐵 ′ and (𝑖𝑖)(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
Solution:
(i) (𝐴 ∪ 𝐵) = {2, 3, 5, 7, 11, 13} ∪ {5, 7, 9, 11, 15} = {2, 3, 5, 7, 9, 11, 13, 15}
(A ∪ B)’ = U – {A ∪ B} = {2, 3, 5, 7, 9, 11, 13, 15} – {2, 3, 5, 7, 9, 11, 13, 15} = { } = ∅ ………………….(1)
Consider, A’∩ 𝐵 ′
A’= U – A = {2, 3, 5, 7, 9, 11, 13, 15} – {2, 3, 5, 7, 11, 13} = {9, 5}
B’= U – B = {2, 3, 5, 7, 9, 11, 13, 15} – {5, 7, 9, 11, 15} = {2, 3, 13}
A’∩ B’ = {9, 5} ∩ {2, 3, 13} = { } = ∅ ……………..(2)
From (1) and (2),
(A ∪ B)’ = A’ ∩ 𝐵 ′ = ∅
(ii) (𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
(𝐴 ∩ 𝐵) = {2, 3, 5, 7, 11, 13} ∩ {5, 7, 9, 11, 15} = {5, 7, 11}
(𝐴 ∩ 𝐵)’ = U – (A ∩ B) = {2, 3, 5, 7, 9, 11, 13, 15} – {5, 7, 11} = {2, 3, 9, 13, 15}……………………(3)
Consider A’∪B’
A’= U – A = {2, 3, 5, 7, 9, 11, 13, 15} – {2, 3, 5, 7, 11, 13} = {9, 15}
B’= U – B = {2, 3, 5, 7, 9, 11, 13, 15} – {5, 7, 9, 11, 15} = {2, 3, 13}
A’ ∪ B’ = {9, 15} ∪ {2, 3, 13} = {2, 3, 9, 13, 15}…………….(4)
From (2) and (4)
(𝐴 ∩ 𝐵) ′ = 𝐴 ′ ∪ 𝐵′
Thus De morgan’s laws verified.
- Draw Venn diagrams to illustrate the following:
i) AUB
ii) (AUB)’
iii) A’∩B
iv) A∩B’
v) A\B
vi) A∩(B\C)
vii) AU(B∩C)
viii) C∩(BUA)
ix) C∩(B\A)
x) A\(B∩C)
xi) (A\B)U(A\C)
xii)(AUB) \ (AUC)
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