Question

lf U= {1, 2, 3, 4, 5, 6, 7, 8, 9}, A= {2, 4, 6, 8} and B= {2, 3, 5, 7}. Verify that $(A \cup B)^{'} = A^{'} \cap B^{'}$

Answer 1

Here given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
first of all, find $A\cup B$,
$A\cup B$ means the set of all distinct elements or objects that are in both A or B.
so, $A\cup B$ = {2, 4, 6, 8} U {2, 3, 5 , 7}
= {2, 3, 4, 5, 6, 7, 8}

now, $(A\cup B)'$ = U - (A U B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5, 6, 7, 8}
= {1, 9}

now, A' = U - A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8 }
= {1, 3, 5, 7, 9}

B' = U - B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7}
= {1, 4, 6, 8, 9}

now, $A'\cap B'$ means the set of all distinct elements or objects that are in both A' and B'.
so, $A'\cap B'$ = {1, 3, 5, 7, 9} - {1, 4, 6, 8, 9} = {1, 9}

here it is clear that
$(A\cup B)'=A'\cap B'=\{1,9\}$
hence, varified.

Answer 2

Solution :

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

A = { 2,4,6,8 }

B = { 2,3,5,7 }

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Complement of set A is denoted by A'.

A' = U - A

= { x:x€U and x doesn't belongs to A }

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i ) AUB

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

= { 2,3,4,5,6,7 ,8 }----( 1 )

ii ) ( AUB)'

= U - ( AUB )

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

= { 1,9 } ----( 2 )

iii ) A' = U - A

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

= { 1,3,5,7,9 } ----( 3 )

iv ) B' = U - B

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

= { 1,4,6,8,9 } ----( 4 )

Now ,

A' cap B'

= {1,3,5,7,9} cap {1,4,6,8,9}

= { 1, 9 } ---( 5 )

Therefore ,

( A U B )' = A' cap B'

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