# College Math Teaching

## October 11, 2016

### The bias we have toward the rational numbers

Filed under: analysis, Measure Theory — Tags: , , — collegemathteaching @ 5:39 pm

A brilliant scientist (full tenure at the University of Chicago) has a website called “Why Evolution is True”. He wrote an article titled “why is pi irrational” and seemed to be under the impression that being “irrational” was somehow special or unusual.

That is an easy impression to have; after all, almost every example we use rationals or sometimes special irrationals (e. g. multiples of $pi$, $e^1$, square roots, etc.

We even condition our students to think that way. Time and time again, I’ve seen questions such as “if $f(.9) = .94, f(.95) = .9790, f(1.01) = 1.043$ then it is reasonable to conclude that $f(1) =$. It is as if we want students to think that functions take integers to integers.

The reality is that the set of rationals has measure zero on the real line, so if one were to randomly select a number from the real line and the selection was truly random, the probability of the number being rational would be zero!

So, it would be far, far stranger had “pi” turned out to be rational. But that just sounds so strange.

So, why do the rationals have measure zero? I dealt with that in a more rigorous way elsewhere (and it is basic analysis) but I’ll give a simplified proof.

The set of rationals are countable so one can label all of them as $q(n), n \in \{0, 1, 2, ... \}$ Now consider the following covering of the rational numbers: $U_n = (q(n) - \frac{1}{2^{n+1}}, q(n) + \frac{1}{2^{n+1}})$. The length of each open interval is $\frac{1}{2^n}$. Of course there will be overlapping intervals but that isn’t important. What is important is that if one sums the lengths one gets $\sum^{\infty}_{n = 0} \frac{1}{2^n} = \frac{1}{1-\frac{1}{2}} = 2$. So the rationals can be covered by a collection of open sets whose total length is less than or equal to 2.

But there is nothing special about 2; one can then find new coverings: $U_n = (q(n) - \frac{\epsilon}{2^{n+1}}, q(n) + \frac{\epsilon}{2^{n+1}})$ and the total length is now less than or equal to $2 \epsilon$ where $\epsilon$ is any real number. Since there is no positive lower bound as to how small $\epsilon$ can be, the set of rationals can be said to have measure zero.