# A problem that’s always bugged me a little

Note – just to be clear from the start – this post is only half serious, and I understand that I’m playing a little fast and loose with a few things – especially the definition of a limit – but it is true that the concept I’ve outlined below has always bugged me.

I started reading (actually listening to) Eugenia Cheng’s How to Bake Pi today. It is a fun book so far, and certainly made working out at the Harvard stadium this morning much more enjoyable than usual.

One of her statements that got me thinking was the idea that nearly everyone hits a wall in mathematical abstraction at some point. I may not have Cheng’s exact words exactly right, but the idea was pretty interesting to me. It made me wonder about some fairly abstract ideas that I’d never completely reconciled in my mind.

What came to mind is strange property of the harmonic series (1 + 1/2 + 1/3 + 1/4 + 1/5 + . . .). Nearly everyone I’ve shared this idea with thinks I’ve completely lost my mind, so with that warning (and with the extra note at the start!), I guess, here we go 🙂

A standard proof that the harmonic series diverges goes something like this:

Step 0: 1 is bigger than 1/2
Step 1: 1/2 + 1/3 is bigger than 1/2
Step 2: 1/4 + 1/5 + 1/6 + 1/7 is bigger than 1/2
Step 3: 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 is bigger than 1/2

The reason that the sum in step 3 is larger than 1/2 is that all 8 terms are larger than 1/16, so the sum is greater than 8/16 which is 1/2.

Step n: $1/2^n + 1 / (2^n + 1) + \ldots + 1/(2^{n + 1} - 1)$ is greater than 1/2

The reason that this sum is greater than 1/2 is more or less the same reason given for step 3. Each of the $2^n$ terms is larger than $1/(2^{n+1})$ so the sum is larger than $2^n / 2^{n+1}$ which is 1/2.

At each step m, you have a sum with $2^m$ terms which is greater than 1/2, and that tells you that the finite sum $1 + 1/2 + 1/3 + \ldots 1/2^m$ is greater than $m/2$ and so as you let m approach infinity, the sum also goes to infinity.

Totally seperately, when you learn about set theory you learn about some of Cantor’s theorems about power sets. The “power set” of a set is the set of all subsets of the set, including the null set and the original set itself. The notation that I learned for the power set of a set $S$ was $2^S$. That notation, I assume, comes from the fact that for a finite set $S$ with $n$ elements, the number of elements of the power set of $S$ is $2^n$.

Cantor proved that the cardinality of the power set of a set $S$ is always greater than the cardinality of the set itself. In particular the cardinality of the set of positive integers – the “smallest” infinity and usually denoted by $\aleph_0$, is larger than $\aleph_o$. In fact, the cardinality of the power set of the set of positive integers is equal to the cardinality of the real numbers. If we call the cardinality of the real numbers $R$, what I’ve written above can be summarized by saying:

$R = 2^{\aleph_0} > \aleph_0$

Loosely speaking, that equations tells you that there are “more” real numbers than there are integers.

Ok, so back to the standard proof that the harmonic series diverges. The way that I’ve constructed the proof above, at every step m we show that the sum of $2^{m+1}$ terms starting at $1/2^m$ is greater than 1/2.

So, as $m$ goes to infinity, the number of terms we add in each additional step – namely $2^{m+1}$ terms – also gets *really* large. In fact, so large that it is actually a different infinity! Right?? That means that the “last” sum that is greater than 1/2 in the proof above – namely the step that “finally” gets you to infinity requires adding up more terms than there are integers!

A little crazy, I know, but at least I felt less crazy about being puzzled by this idea after learning that $1 + 2 + 3 + \ldots = -1/12$ 🙂

# Basic stats: A math club homework problem and a problem from the 1989 AIME

My older son had this interesting basic statistics problem as part of his math club homework:

For a list of 8 positive integers, the mean, median, unique mode and range are 8. What is the greatest integer that could be in this set?

My older son and I talked through this problem yesterday as part of our short, morning math talks: MathyMath15. It is such a nice problem, though, that I wanted to turn into a project.

Part of the reason was the similarity to this problem from the 1989 AIME:

Problem 11 from the 1989 AIME

Here’s the slightly simplified version of that problem I used in our project today:

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let \$D\$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of \$D\$?

I really like both of these problem for getting kids to think about different ways numbers can be combined to produce similar (basic) statistical results.

The 20-ish minute conversation we has about these two problems is below. I let my younger son tackle the first problem mostly on his own since my older son had already worked through it yesterday. There are several fun “aha” moments in the conversation:

Part 1: Introducing the first problem (including defining some of the words) and my younger son’s initial thoughts. His first thought uses all 8’s:

Part 2: My younger son tries a few more ideas based on noticing that the sum of the numbers needs to stay at 64. He finds a solution that uses 12 as the largest number.

Part 3: Here the three of us work together to find the largest integer that can be in the set:

Part 4: Now we move on to the problem from the 1989 AIME. There are more numbers to keep track of, and the arithmetic is a little more difficult, but the ideas we need to solve this problem are pretty much exactly the same as the first one. Both kids have several really nice ideas about how to tackle this problem over the next two videos:

Part 5: While skipping over the arithmetic, we walk down the path to the solution of the problem.

So, a fun project connecting some ideas from one math club problem to an old AIME problem. AIME problems obviously aren’t usually going to be great problems for kids, but I thought this would be an interesting exception. It is pretty neat to me that more or less the same ideas solve both problems 🙂