I made the claim in a post last week that the set of irrationals of the form is countable. I said that pretty quickly, without justification. I’ve never proved statements like this before, but here I’m going to try.

**Theorem**. The set of irrationals of the form , with , is countable.

**Proof**. Consider the set of irrationals of the form , with . More formally, define

And also require that is ‘square free’–that is, we require that neither the numerator or denominator of contain factors that are perfect squares. So is in ‘simplest’ form. We aim to show that is countable.

Now, consider the function

defined

This function is one-to-one since we require to be in simplest form–that is, the image of any number under is unique. So is an injection from into.

We know that is countable. Since a finite Cartesian product of countable sets is countable, must also be countable. And we have constructed a function which is an injection from into . So the cardinality of must be no greater than the cardinality of . Thus must also be countable, as desired.

I think I did that right. Any suggestions, math readers?

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I don’t think that your care to express yourself in simplest form is even required. If , and are rational (with no other restrictions), then the number of real numbers of the form is the number of ordered triples of rational numbers, which is countable.

Good point, Will. What in the world are you doing surfing my blog posts from seven years ago?? 🙂