Question

2.
If u = {2,3,4,5), A = {2,4} and B = {3,4), then verify the following laws.
i) A-B = AB
ii) (ANB)' = A' UB! iii) (AUB)' = A' NB
3.
4.
If n(A) = 10, n(B) = 12 and n(An B) = 6, then find n(AUB).
If A {2,3,4,5), B= {1, 2, 3) and C = {5,6), then verify
( AUBUC)+P(A) B)+n( B0C)+ n(COA) = n(A)+n(B)+ n(C)+P(A0BC)​

Answer 1

Answer:

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Step-by-step explanation:

Examples on De Morgan’s law:

1. If U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}.

Proof of De Morgan's law: (X ∩ Y)' = X' U Y'.

Solution:

We know, U = {j, k, l, m, n}

X = {j, k, m}

Y = {k, m, n}

(X ∩ Y) = {j, k, m} ∩ {k, m, n}

= {k, m}

Therefore, (X ∩ Y)' = {j, l, n} ……………….. (i)

Again, X = {j, k, m} so, X' = {l, n}

and Y = {k, m, n} so, Y' = {j, l}

X' ∪ Y' = {l, n} ∪ {j, l}

Therefore, X' ∪ Y' = {j, l, n} ……………….. (ii)

Combining (i)and (ii) we get;

(X ∩ Y)' = X' U Y'. Proved

2. Let U = {1, 2, 3, 4, 5, 6, 7, 8}, P = {4, 5, 6} and Q = {5, 6, 8}.

Show that (P ∪ Q)' = P' ∩ Q'.

Solution:

We know, U = {1, 2, 3, 4, 5, 6, 7, 8}

P = {4, 5, 6}

Q = {5, 6, 8}

P ∪ Q = {4, 5, 6} ∪ {5, 6, 8}

= {4, 5, 6, 8}

Therefore, (P ∪ Q)' = {1, 2, 3, 7} ……………….. (i)

Now P = {4, 5, 6} so, P' = {1, 2, 3, 7, 8}

and Q = {5, 6, 8} so, Q' = {1, 2, 3, 4, 7}

P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7}

Therefore, P' ∩ Q' = {1, 2, 3, 7} ……………….. (ii)

Combining (i)and (ii) we get;

(P ∪ Q)' = P' ∩ Q'.