Question

A={x:x is a prime factor of 210}
B={x:x ≤10 and x € N, show that (AUB)-(ANB)=(A-B)U(B-A).​

Answer 1

A U B -   A ∩ B =  (A - B) U (B - A)  Verified

Step-by-step explanation:

A={x:x is a prime factor of 210}

A = { 2 , 3 , 5 , 7 }

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

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

A ∩ B = {  2 , 3 , 5 , 7 }

A U B -   A ∩ B =  {  1 , 4 , 6 , 8 , 9 , 10 }

A - B  =  { }

B - A  = {  1 , 4 , 6 , 8 , 9 , 10 }

(A - B) U (B - A) = {  1 , 4 , 6 , 8 , 9 , 10 }

A U B -   A ∩ B =  (A - B) U (B - A)

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Answer 2

Answer:

Given,

A={x:x is a prime factor of 210}

B={x:x ≤10 and x ∈ N }

Roster form of these sets are,

A={2, 3, 5, 7}

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

A∪B = all elements of A and B

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

= B

A∩B = common elements of A and B

= {2, 3, 5, 7}

= A

Thus,

(AUB)-(A∩B) = elements of AUB which are not in A∩B

= {1, 4, 6, 8, 9, 10}

Also, A - B = {}

B - A = {1, 4, 6, 8, 9, 10}

So,

(A-B)U(B-A) = {1, 4, 6, 8, 9, 10}

Hence, (AUB)-(A∩B)=(A-B)U(B-A).​