Answer
See explanation
Work Step by Step
Recall that we have already shown that:
1. The set of rational numbers, \(\mathbb{Q}\), is countable.
2. The union of two countable sets is countable.
Also, it is well known (via Cantor's diagonal argument, for example) that the set of real numbers, \(\mathbb{R}\), is uncountable.
Now, notice that every real number is either rational or irrational. That is,
\[
\mathbb{R} = \mathbb{Q} \cup \bigl(\mathbb{R}\setminus\mathbb{Q}\bigr),
\] where \(\mathbb{R}\setminus\mathbb{Q}\) is the set of all irrational numbers.
Suppose, for the sake of contradiction, that the set of irrational numbers is countable. Then both \(\mathbb{Q}\) (by known result) and \(\mathbb{R}\setminus\mathbb{Q}\) (by assumption) are countable. But then their union, being the disjoint union of two countable sets, would also be countable. That is,
\[
\mathbb{R} = \mathbb{Q} \cup (\mathbb{R}\setminus\mathbb{Q})
\] would be countable.
However, we know that \(\mathbb{R}\) is uncountable. This contradiction implies that our assumption must be false.
Therefore, the set of all irrational numbers is uncountable.
