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  • Brainly User
2016-03-28T06:32:21+05:30

Solution 



(i) (A ∩ B)' = A' ∪ B'



L.H.S. = (A ∩ B)'



A ∩ B = {3} 



(A ∩ B)' = {2, 4, 5, 6, 7, 8}     ……………….. (1) 



R.H.S. = A' ∪ B'



A’ = {5, 7, 8}



B’ = {2, 4, 6} 



A’∪B’ = {2, 4, 5, 6, 7, 8}     ……………….. (2) 



From (1) and (2), we conclude that;


(A ∩ B)' = (A' ∪ B') 



(ii) (A ∪ B)' = A' ∩ B' 



L.H.S. = (A ∪ B)'



A∪B = {2, 3, 4, 5, 6, 7} 



(A ∪ B)' = {8}     ……………….. (1) 



R.H.S. = A' ∩ B'



A' = {2, 4, 6, 8} 



B' = {5, 7, 8} 



A' ∩ B' = {8}     ……………….. (2) 



From (1) and (2), we conclude that;


(A ∪ B)' = A' ∩ B'



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2016-03-28T07:09:30+05:30

(i) (A ∩ B)' = A' ∪ B'

L.H.S. ⇒ (A ∩ B)'

⇒A ∩ B = {3} 

(A ∩ B)' = {2, 4, 5, 6, 7, 8}    -----(1) 

R.H.S. ⇒ A' ∪ B'

A’ = {5, 7, 8}

∴B’ = {2, 4, 6} 

A’∪B’ = {2, 4, 5, 6, 7, 8}    ---- (2) 

From (1) and (2), we get,

∴(A ∩ B)' = (A' ∪ B') 

(ii) (A ∪ B)' = A' ∩ B' 

L.H.S. ⇒ (A ∪ B)'

A∪B = {2, 3, 4, 5, 6, 7} 

⇒(A ∪ B)' = {8}    ----- (3) 

R.H.S. ⇒ A' ∩ B'

A' = {2, 4, 6, 8} 

B' = {5, 7, 8} 

⇒A' ∩ B' = {8}  ----(4) 

From (3) and (4), we get;

∴(A ∪ B)' = A' ∩ B'

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