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?
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?? 🙂