Question

state and prove De- morgans theorem for rational index

Answer 1

Heya.....

Here's your answer....

There are two theorems -

De Morgan's First Theorem:-

Statement - The complement of a logical sum equals the logical product of the complements.

Logic equation - A+B¯=A¯.B¯

De Morgan's Second Theorem:-

Statement - The complement of a logical product equals the logical sum of the complements.

Logic equation - A.B¯=A¯+B¯

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Answer 2

Answer:

the complement of the union of two sets is the same as the intersection of their complements; and

the complement of the intersection of two sets is the same as the union of their complements.

or

not (A or B) = not A and not B; and

not (A and B) = not A or not B

Step-by-step explanation:

The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan’s laws.

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