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Question

Question: If we have the sets as \(A=\left\\{ x:x\text{ is a multiple of 3} \right\\}\) and \(B=\left\\{ x:x\t...

If we have the sets as A=\left\\{ x:x\text{ is a multiple of 3} \right\\} and B=\left\\{ x:x\text{ is a multiple of 5} \right\\}, then ABA-B is equal to
(a) AB\overline{A}\cap B
(b) ABA\cap \overline{B}
(c) AB\overline{A}\cap \overline{B}
(d) AB\overline{A\cap B}

Explanation

Solution

We start solving the problem by finding the sets A\overline{A}, B\overline{B} and ABA\cap B by using the fact that compliment remove includes all elements in the universe other than the set we are taking the complement and intersection includes elements of both the sets. We then find the elements in the set ABA-B and then compare it with the intersections of different combinations of A, B, A\overline{A}, B\overline{B} to get the required answer.

Complete step-by-step solution
According to the problem, we are given that A=\left\\{ x:x\text{ is a multiple of 3} \right\\}, B=\left\\{ x:x\text{ is a multiple of 5} \right\\} and we need to find which of the given options is equal to ABA-B.
Let us find A\overline{A}, B\overline{B} and ABA\cap B.
We know that A\overline{A} is defined as the elements in the universe other than the elements in set A. So, we get \overline{A}=\left\\{ x:x\text{ is not a multiple of 3} \right\\}.
Similarly, we get \overline{B}=\left\\{ x:x\text{ is not a multiple of 5} \right\\}.
We know that the ABA\cap B is defined as the set of all elements that satisfy conditions of both sets A and B. So, we get A\cap B=\left\\{ x:x\text{ is multiple of both 3 and 5} \right\\}.
We know that ABA-B is defined as the set with elements of A by neglecting the elements of ABA\cap B.
So, we get A-B=\left\\{ x:x\text{ is a multiple of 3 but not 5} \right\\}.
We can see that the A\cap \overline{B}=\left\\{ x:x\text{ is multiple of 3 but not 5} \right\\}.
So, we have AB=ABA-B=A\cap \overline{B}.
∴ The correct option for the given problem is (b).

Note: We can also solve this problem by drawing Venn diagrams which gives a good view of different sets we just discussed in the problem. We should not be confused by the intersection of sets with the union of sets while solving this problem. We should make sure about the definitions of complement, intersection, the union of sets before solving this problem. Similarly, we can expect problems to find what BAB-A is equal to.