Please prove that A-B = A ∩ B’ with words Prove that A-B is equal to A intersection B complement.
The Correct Answer and Explanation is:
To prove that:A−B=A∩B′A – B = A \cap B’A−B=A∩B′
we will show that both sets contain exactly the same elements. This is done by proving two inclusions:
1. Show that A−B⊆A∩B′A – B \subseteq A \cap B’A−B⊆A∩B′:
Let x∈A−Bx \in A – Bx∈A−B.
By the definition of set difference, this means:
- x∈Ax \in Ax∈A, and
- x∉Bx \notin Bx∈/B
But if x∉Bx \notin Bx∈/B, then by definition of complement:
- x∈B′x \in B’x∈B′
So:
- x∈Ax \in Ax∈A and x∈B′x \in B’x∈B′
Hence:
- x∈A∩B′x \in A \cap B’x∈A∩B′
This proves that:A−B⊆A∩B′A – B \subseteq A \cap B’A−B⊆A∩B′
2. Show that A∩B′⊆A−BA \cap B’ \subseteq A – BA∩B′⊆A−B:
Let x∈A∩B′x \in A \cap B’x∈A∩B′.
By the definition of intersection, this means:
- x∈Ax \in Ax∈A, and
- x∈B′x \in B’x∈B′
Since x∈B′x \in B’x∈B′, then:
- x∉Bx \notin Bx∈/B
Thus:
- x∈Ax \in Ax∈A, and x∉Bx \notin Bx∈/B
By the definition of set difference, this means:
- x∈A−Bx \in A – Bx∈A−B
So:A∩B′⊆A−BA \cap B’ \subseteq A – BA∩B′⊆A−B
Conclusion:
We have shown both:
- A−B⊆A∩B′A – B \subseteq A \cap B’A−B⊆A∩B′, and
- A∩B′⊆A−BA \cap B’ \subseteq A – BA∩B′⊆A−B
Therefore:A−B=A∩B′A – B = A \cap B’A−B=A∩B′
Explanation (Like in Books):
The set difference A−BA – BA−B is the set of all elements that are in set AAA but not in set BBB. On the other hand, the complement of B, denoted by B′B’B′, includes all elements not in BBB (relative to the universal set). When we take the intersection A∩B′A \cap B’A∩B′, we are selecting all elements that are both in AAA and not in BBB, which is exactly the same condition as set difference A−BA – BA−B. Therefore, both expressions describe the same set, and we conclude:A−B=A∩B′A – B = A \cap B’A−B=A∩B′