## To prove that |B| < |A|, we need to show that the size of B is strictly less than the size of A.

Since A is a finite set, we can assign a size or cardinality to it, denoted as |A|. Similarly, we can assign a size or cardinality to B, denoted as |B|.

Given that B is a proper subset of A, this means that there are elements in A that are not in B. We can denote the set of elements in A that are not in B as A - B.

The hint suggests considering the union of disjoint sets B and A - B. The union of two sets, denoted as A âˆª B, is the set that contains all the elements that are either in A, in B, or in both.

In this case, the union of sets B and A - B can be written as B âˆª (A - B).

Since B and A - B are disjoint sets, meaning they have no elements in common, the union of these sets will simply be the combination of all the elements from B and all the elements from A - B.

Therefore, B âˆª (A - B) will be equal to A. This can be understood intuitively, as B is a proper subset of A, so the union of B and the elements not in B (A - B) will essentially cover all the elements in A.

Now, let's consider the size or cardinality of B âˆª (A - B). Since B âˆª (A - B) is equal to A, the size of A, denoted as |A|, is equal to the size of B âˆª (A - B), denoted as |B âˆª (A - B)|.

We can express the size of B âˆª (A - B) as the sum of the sizes of B and A - B, denoted as |B| + |A - B|.

Since A - B represents the elements in A that are not in B, the size of A - B will be greater than or equal to 0 (it could be 0 if B is equal to A). Therefore, |A - B| â‰¥ 0.

So, |B| + |A - B| â‰¥ |B|.

Taking into account that |A| = |B âˆª (A - B)|, we have:

|A| = |B âˆª (A - B)| = |B| + |A - B| â‰¥ |B|

Since |A| is the size of A and |B| is the size of B, we have shown that |B| < |A|.

Hence, we have proven that if A is a finite set and B is a proper subset of A, then |B| < |A|.