Let \(A = \{-1, \sqrt{2}, \pi\}\) and
\(B = \{ n \in \mathbb{R} : n^2 \in \mathbb{N}\}\).
What is \(\mid A \cap B \mid\)?
The answer is 2.
Note that \(B\) is the set of real numbers whose squares are natural
numbers. Since the squares of \(-1\) and \(\sqrt{2}\) are \(1\) and \(2\),
respectively, \(B\) contains both \(-1\) and \(\sqrt{2}\).
But \(B\) does not contain \(\pi\) because \(\pi^2\) is not a natural
number. Thus, \(-1\) and \(\sqrt{2}\) are the only elements in both
\(A\) and \(B\).