# CLASS-7UNION & INTERSECTION OF SET

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) – (AB )

=  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)