Question

(a) If A = {1, 3, 4, 8, 9, 12}, B = {1, 4, 9} and C = {2, 4, 8, 10}

Find (i) A ∪ (B ∩ C) (ii) A ∩ (B ∪ C) (iii) (A ∪ B) ∩ (A ∪ C) (iv) (A ∩ B) ∪ (A ∩ C)

(b) If A = (2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6} and C = (1, 3, 5, 7, 9, 11, 13}

Verify (i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) (iv) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

(c) If X = {x : x is a prime number less than 12}

Y = {x : x is an even number less than 12}

Z = {x : x is an odd number less than 12}

Show that (i) union of sets of distributive over intersection of sets.

(ii) intersection of sets is distributive over union of sets.

Answer 1

(a) A = {1, 3, 4, 8, 9, 12}, B = {1, 4, 9} and C = {2, 4, 8, 10}

(i) B ∩ C = {1, 4, 9} ∩ {2, 4, 8, 10} = {4}

∴ A ∪ (B ∩ C) = {1, 3, 4, 8, 9, 12} ∪ {4} = {1, 3, 4, 8, 9, 12}

(ii) B ∪ C = {1, 4, 9} ∪ {2, 4, 8, 10} = {1, 2, 4, 8, 9, 10}

∴ A ∩ (B ∪ C) = {1, 3, 4, 8, 9, 12} ∩ {1, 2, 4, 8, 9, 10} = {1, 4, 8, 9}

(iii) A ∪ B = {1, 3, 4, 8, 9, 12} ∪ {1, 4, 9} = {1, 3, 4, 8, 9, 12}

A ∪ C = {1, 3, 4, 8, 9, 12} ∪ {2, 4, 8, 10} = {1, 2, 3, 4, 8, 9, 10, 12}

∴ (A ∪ B) ∩ (A ∪ C) = {1, 3, 4, 8, 9, 12} ∩ {1, 2, 3, 4, 8, 9, 10, 12} = {1, 3, 4, 8, 9, 12}

(iv) A ∩ B = {1, 3, 4, 8, 9, 12} ∩ {1, 4, 9} = {1, 4, 9}

A ∩ C = {1, 3, 4, 8, 9, 12} ∩ {2, 4, 8, 10} = {4, 8}

∴ (A ∩ B) ∪ (A ∩ C) = {1, 4, 9} ∪ {4, 8} = {1, 4, 8, 9}

(b) A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6} and C = {1, 3, 5, 7, 9, 11, 13}

(i) B ∩ C = {1, 2, 3, 4, 5, 6} ∩ {1, 3, 5, 7, 9, 11, 13} = {1, 3, 5}

∴ A ∪ (B ∩ C) = {2, 4, 6, 8, 10} ∪ {1, 3, 5} = {1, 2, 3, 4, 5, 6, 8, 10}

A ∪ B = {2, 4, 6, 8, 10} ∪ {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6, 8, 10}

A ∪ C = {2, 4, 6, 8, 10} ∪ {1, 3, 5, 7, 9, 11, 13} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}

∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}

= {1, 2, 3, 4, 5, 6, 8, 10}

Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) B ∪ C = {1, 2, 3, 4, 5, 6} ∪ {1, 3, 5, 7, 9, 11, 13} = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}

∴ A ∩ (B ∪ C) = {2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6, 7, 9, 11, 13} = {2, 4, 6}

A ∩ B = {2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6} = {2, 4, 6}

A ∩ C = {2, 4, 6, 8, 10} ∩ {1, 3, 5, 7, 9, 11, 13} =

∴ (A ∩ B) ∪ (A ∩ C) = {2, 4, 6} ∪ = {2, 4, 6}

Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ B)

(iii) (A ∩ B) ∪ C = {2, 4, 6} ∪ {1, 3, 5, 7, 9, 11, 13} = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}

A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}

B ∪ C = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}

∴ (A ∪ C) ∩ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13} ∩ {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}

= {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}

Hence, (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

(iv) A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}

∴ (A ∪ B) ∩ C = {1, 2, 3, 4, 5, 6, 8, 10} ∩ {1, 3, 5, 7, 9, 11, 13} = {1, 3, 5}

A ∩ C =

B ∩ C = {1, 3, 5}

∴ (A ∩ C) ∪ (B ∩ C) = ∪ {1, 3, 5} = {1, 3, 5}

Hence, (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

(c) X = {x: x is a prime number less than 12} = {2, 3, 5, 7, 11}

Y = {x: x is an even number less than 12} = {2, 4, 6, 8, 10}

Z = {x: x is an odd number less than 12} = {1, 3, 5, 7, 9, 11}

To show:

(i) Union of sets is distributive over intersection of sets, i.e., X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)

(ii) Intersection of sets is distributive over union of sets, i.e., X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)

Proof:

(i) Y ∩ Z = {2, 4, 6, 8, 10} ∩ {1, 3, 5, 7, 9, 11} =

∴ X ∪ (Y ∩ Z) = {2, 3, 5, 7, 11} ∪ = {2, 3, 5, 7, 11}

X ∪ Y = {2, 3, 5, 7, 11} ∪ {2, 4, 6, 8, 10} = {2, 3, 4, 5, 6, 7, 8, 10, 11}

X ∪ Z = {2, 3, 5, 7, 11} ∪ {1, 3, 5, 7, 9, 11} = {1, 2, 3, 5, 7, 9, 11}

∴(X ∪ Y) ∩ (X ∪ Z) = {2, 3, 4, 5, 6, 7, 8, 10, 11} ∩ {1, 2, 3, 5, 7, 9, 11} = {2, 3, 5, 7, 11}

Hence, X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)

(ii) Y ∪ Z = {2, 4, 6, 8, 10} ∪ {1, 3, 5, 7, 9, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

∴ X ∩ (Y ∪ Z) = {2, 3, 5, 7, 11} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}∴ = {2, 3, 5, 7, 11}

X ∩ Y = {2, 3, 5, 7, 11} ∩ {2, 4, 6, 8, 10} = {2}

X ∩ Z = {2, 3, 5, 7, 11} ∩ {1, 3, 5, 7, 9, 11} = {3, 5, 7, 11}

∴ (X ∩ Y) ∪ (X ∩ Z) = {2} ∪ {3, 5, 7, 11} = {2, 3, 5, 7, 11}

Hence, X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)