Sheldon Axler’s Precalculus: 2nd ed, 0.3.78

— an Exercise

Abstract

I have embarked on a grand adventure to learn math well, which I may write more on in the future. This piece gives my solution to exercise 0.3.78 in Sheldon Axler’s Precalculus (2nd ed), and the process I went through to solve it. This piece partly functions as a test of the MathJax-Latex plugin for writing pretty maths.

Exercise text

0.3.78: Give an example of a set of real numbers such that the average of any two numbers in the set is in the set, but the set is not an interval.

Def: An interval is a set of real numbers that contains all numbers between any two numbers in the set. (And may or may not contain the two endpoints.))

Solution process

The challenge stems from how an initial set of numbers will produce a set of averaged numbers that must be in the set, but then those averaged numbers produce new averaged numbers again, and so forth. In other words, we have a kind of recursive set that generates new elements from the elements in the set.

e.g. \(S=\{a, b, …\}\cup\{\frac{x+y}{2}|x,y_\in S\}\), where a, b, … are initial “populating constants”.

On the evening I first attempted the exercise, I initially “ran” a few checks (that is, mentally looked at the early cases of constructing S) with various starting constants to see if I could find any pattern to the construed numbers that meant they wouldn’t populate everywhere along the interval between the endpoint constants. But the average of two numbers is the midpoint of the two numbers, and by successive averaging, we can construe numbers halfway to anywhere on the interval, which means that the numbers in the set will spread out along the entire interval with infinite precision.

At least, that was the general case. Perhaps there was some case where the populating constants created other behaviour? Not really. The midpoint observation meant that no matter what populating constants we start with, they’ll still spread out in the same way. A trivial solution for a set where the average of the numbers in the set are in the set is any set of one number, but then we can’t average any two numbers in the set, since there aren’t any two numbers…

So, after having spent ~20 minutes or so fruitlessly trying to spot a better angle of attack, I decided that I had had enough math that day.

Solution found:

The next morning while walking to the university, I started thinking on the exercise again… And after a few minutes of thinking of midpoints and averages and endpoints, I realized that any such S whose initial populating constants are rational numbers (including integers) must simply “produce” new integers through the averaging process. Then, when I arrived, I hurriedly wrote down a shoddy note about this, attended class etc. and then wrote an elementary little proof.

Let \(a,b_\in\mathbb{Q}\). Then \(a=\frac{p_1}{q_1}\) and \(b=\frac{p_2}{q_2}\) where \(p,q_\in\mathbb{Z}\) and \(q\ne0.\) But then \(\frac{a+b}{2}=\frac{p_1q_2+p_2q_1}{2q_1q_2}=\frac{p\prime}{q\prime}\). Since summing integers gives an integer, and multiplying integers gives an integer, \(p\prime\) and \(q\prime\) must be integers. Therefore, the average of a and b is also a rational number.

But that means \(S=\{a, b, …\}\cup\{\frac{x+y}{2}|x,y_\in S\}\) where a, b, … are initial “populating constants” and \(a,b,…_\in\mathbb{Q}\) is the set of all rational numbers between the endpoints given by the smallest and largest populating constants. Since an interval is defined as the set of all real numbers between the endpoints of the set, that means S is not an interval.

For a concrete example, construing \(S\) from \({1, 2}\) will lead to a set of rational numbers that doesn’t include \(\sqrt{2}\) despite 1<\(\sqrt{2}\)<2.

Conclusion

I am pretty satisfied by how the MathJax-Latex plugin output looks, and while I’ve cheated in a few places (like the previous paragraph), I found writing this to be decent fun. While the exercise above is obviously not groundbreaking stuff, I found it a fun teaser. More importantly, the solution process serves as an example of what Barbara Oakley calls the “diffuse mode” of thinking. I wasn’t able to find the solution the first evening I fiddled with the exercise, but it came to me easily while walking the next day. Though, if I hadn’t spent those ~20 minutes on the first day, it probably wouldn’t have come to me the next day either! As a final note, I intend to post solutions to other interesting problems as they pop up in my mathematical adventure.