Irrationals of the form a+b√c

I made the claim in a post last week that the set of irrationals of the form a+b\sqrt{c} 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 a+b\sqrt{c}, with a,b \neq 0,c>0\in\mathbb{Q}, is countable.

Proof. Consider the set of irrationals of the form a+b\sqrt{c}, with a,b \neq 0,c >0\in\mathbb{Q}. More formally, define

\mathbb{I}=\left\{a+b\sqrt{c}\in\mathbb{R}: a,b \neq 0,c>0\in\mathbb{Q}\right\}

And also require that c is ‘square free’–that is, we require that neither the numerator or denominator of c contain factors that are perfect squares. So a+b\sqrt{c} is in ‘simplest’ form. We aim to show that \mathbb{I} is countable.

Now, consider the function

\mathbb{I} \overset{f}{\rightarrow} \mathbb{Q}^3

defined

f(a+b\sqrt{c})=(a,b,c)

This function is one-to-one since we require c to be in simplest form–that is, the image of any number a+b \sqrt{c} under f is unique. So f is an injection from \mathbb{I} into\mathbb{Q}^3.

We know that \mathbb{Q} is countable. Since a finite Cartesian product of countable sets is countable, \mathbb{Q}^3 must also be countable. And we have constructed a function f which is an injection from \mathbb{I} into \mathbb{Q}^3. So the cardinality of \mathbb{I} must be no greater than the cardinality of \mathbb{Q}^3. Thus \mathbb{I} must also be countable, as desired.

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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s