**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)**

## 1 thought on “Sets – Exercise 1.1 – Class X”

Comments are closed.