Question

Let ξ = {x | x ε N, x is a factor of 144}, A = {x | x ε N, x is a factor of 24}, B = {x | x ε N, x is a factor of 36}, C = {x | x ε N, x is a factor of 48}. Find:
(i) A’
(ii) B’
(iii) C’
(iv) A ⋃ B
(v) B ⋃ C
(vi) A ⋃ C
(vii) A ⋂ B’
(viii) B ⋂ C’
(ix) C – A
(x) A – (B ⋂ C)


(Will mark Brainliest if the answer helps)

Answer 1

Answer:

= {x: x ∈ N, x is a factor of 12}

The above set A is written in the set – builder form. The property of x here is that it is a natural number and is a factor of 12. So, the natural factors of 12 are the elements that contain in set A.

The factors of 12 are: 1, 2, 3, 4, 6 and 12

So, set A can be written as: n (A) = {1, 2, 3, 4, 6, 12}.

Now, writing the elements of set B.

B = {x: x ∈ N, x is a factor of 18}

The set – builder form of the above set B is similar to A, the only difference is that here x is a factor of 18. So, we are going to write the factors of 18.

The factors of 18 are: 1, 2, 3, 6, 9 and 18.

So, set B can be written as: n (B) = {1, 2, 3, 6, 9, 18}.

As you can see form the set A and set B, the common elements among both the sets are 1, 2, 3 and 6.

Hence, A∩B = {1, 2, 3, 6}.

Note: While writing the factors for 12, don’t forget to write 1 and 12. Similarly for writing the factors for 18 also, don’t forget to write 1 and 18. Usually, we forget to consider 1 and the number itself as a factor of a numb