Question

PROVE THE FOLLOWING COMMUTATIVE LAW: A+B=B+A A.B=B.A DE MORGAN’S LAW: (A+B)’=A’.B’ (A.B)’=A’+B’​

Answer 1

Answer:

The proving is mentioned below :-

Step-by-step explanation:

Let,

U = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }

A =.{ 1, 2, 3, 4, 5.}

B = { 2, 4, 6, 8 10 }

We have,

For A + B = B + A

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

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

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

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

Hence, A + B = B + A : proved !

For A n B = B n A

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

= { 2, 4 }

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

= { 2, 4 }

Hence, A n B = B n A : proved

For ( A + B )' = A' n B'

( A + B )' = U - ( A + B )

= { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } - { 1, 2, 3, 4, 5, 6, 8, 10}

= { 7, 9 }

A' = U - A

= { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } - { 1, 2, 3, 4, 5 }

= { 6, 7, 8, 9, 10 }

B' = U - B

= { 1, 2, 3, 4, 5, 6, 6, 8, 9, 10 } - { 2, 4, 6, 8, 10 }

= { 1, 3, 5, 7, 9 }

A' n B' = { 6, 7, 8, 9, 10 } n { 1, 3, 5, 7, 9 }

= { 7, 9 }

Hence, ( A + B )' = A' n B'

Mark my answer as brainliest if it helped you. Also things like A.B and B.A doesn't exist here since those all are ordered pairs forms and here we don't need ordered pair forms.