UNION & INTERSECTION OF SET WITH PROBLEM & SOLUTION
Example-1
If the Universal set U = { x / x < 12, x ϵ W } and Z = { x / x is Odd } then verify n(Z’) = n(U) – n(Z)
Ans.) U = { x / x < 12, x ϵ W }
=> U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }, Z = { 1, 3, 5, 7, 9, 11 }, Z’ = { 2, 4, 6, 8, 10 }
So, n(U) = 12, n(Z) = 6, n(Z’) = the set of members of U which does not belongs to Z = 6
So, as per the given condition, n(Z’) = n(U) – n(Z)
=> n(Z’) = 12 – 6 = 6 (Ans.)
Example – 2
If A = { x /x is a letter of the word SWITZERLAND } and B = { E, N, G, L, A, D }, verify that n( A U B ) = n(A) + n(B) – (A⋂ B )
Ans.) n( A U B ) = n(A) + n(B) – (A⋂ B )
A = { x /x is a letter of the word SWITZERLAND }
=> A = { S, W, I, T, Z, E, R, L, A, N, D }, So n(A) = 11
Where B = { E, N, G, L, A, D }, So n(B) = 6,
Now, A U B = { S, W, I, T, Z, E, R, L, A, N, D } U { E, N, G, L, A, D }
= { S, W, I, T, Z, E, R, L, A, D, N, G }, so n(A U B) = 12
And, A ⋂ B = { S, W, I, T, Z, E, R, L, A, N, D } ⋂ { E, N, G, L, A, D }
= { E, N, L, A, D }, So n(A ⋂ B) = 5
Now, as per given condition to prove n(A U B) = n(A) + n(B) – (A⋂B )
= 11 + 6 - 5
= 12 = n( A U B ) (Proven)
Example – 3
If A = { x /x is a letter of the word SWITZERLAND } and B = { E, N, G, L, A, D }, verify that n (A ⋂ B) = n(A) + n(B) – (A U B )
b) n( A ⋂ B ) = n(A) + n(B) – (A U B )
A = { x /x is a letter of the word SWITZERLAND }
=> A = { S, W, I, T, Z, E, R, L, A, N, D }, So n(A) = 11
Where B = { E, N, G, L, A, D }, So n(B) = 6,
Now, A U B = { S, W, I, T, Z, E, R, L, A, N, D } U { E, N, G, L, A, D }
= { S, W, I, T, Z, E, R, L, A, D, N, G }, so n(A U B) = 12
And, A ⋂ B = { S, W, I, T, Z, E, R, L, A, N, D } ⋂ { E, N, G, L, A, D }
= { E, N, L, A, D }, So n(A ⋂ B) = 5
Now, as per given condition to prove n(A ⋂ B) = n(A) + n(B) – (A U B )
= 11 + 6 – 12 = 5 = n( A ⋂ B ) (Proven)
Example - 4
If U = { -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 } , X = { -4, -2, 0, 2, 4 } and Y = { -4, -3, -2, -1, 0, 1, 2, 3, 4 } then verify (X U Y )’ = X’ ⋂ Y’
Ans.) here U = { -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 } , X = { -4, -2, 0, 2, 4 } and Y = { -4, -3, -2, -1, 0, 1, 2, 3, 4 }
So, X U Y = { -4, -2, 0, 2, 4 } U { -4, -3, -2, -1, 0, 1, 2, 3, 4 } = { -4, -3, -2, -1, 0, 1, 2, 3, 4 }
Then, (X U Y)’ = the set of members of U that do not belong to -
X ⋃ Y = { -6, -5, 5, 6 }
Now, X’ = the set of members of U that do not belong to -
X = { -6, -5, -3, -1, 0, 1, 3, 5, 6 }
So, Y’ = the set of members of U that do not belong to Y = { -6, -5, 6, 5 }
To prove or given condition is that, (X U Y )’ = X’ ⋂ Y’
= { -6, -5, -3, -1, 0, 1, 3, 5, 6 } ⋂ { -6, -5, 6, 5 }
= { -6, -5, 6, 5 } = (X U Y )’ (Proven)
Example- 5
If U = { -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 } , X = { -4, -2, 0, 2, 4 } and Y = { -5, -3, -1, 0, 1, 3, 5 } then verify (X ⋂ Y)’ = X’ U Y’
Ans.) Here, U = { -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 } , X = { -4, -2, 0, 2, 4 } and Y = { -5, -3, -1, 0, 1, 3, 5 }
So, X ⋂ Y = { -4, -2, 0, 2, 4 } ⋂ { -5, -3, -1, 0, 1, 3, 5 } = { 0 }
Then, (X ⋂ Y)’ = the set of members of U that do not belong to -
X ⋂ Y = { -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6 }
Now, X’ = the set of members of U that do not belong to -
X = { -6, -5, -3, -1, 1, 3, 5, 6 }
So, Y’ = the set of members of U that do not belong to Y = { -6, -4, -2, 2, 4, 6 }
To prove or given condition is that, (X ⋂ Y)’ = X’ U Y’
= { -6, -5, -3, -1, 1, 3, 5, 6 } U { -6, -4, -2, 2, 4, 6 }
= { -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6 }
= (X ⋂ Y)’ (Proven)