Answer
The argument is Invalid due to the converse error
Work Step by Step
If there are as many rational numbers as there are irrational numbers, then the set of all irrational numbers is infinite $(p →q)$.
The set of all irrational numbers is infinite $(q)$
∴ There are as many rational numbers as there are irrational
numbers.$(p)$
The argument form is:
p →q
q
∴ p
Thus this is the typical converse error
