# Talking through Ben Orlin’s exponent problem

Saw this problem last week via a Joshua Bowman retweet:

I thought it would be a good problem to talk through with the boys this morning, and it turned out to be an even better discussion that I was expecting.

Talking through this problem with kids gives you the opportunity to discuss lots of specific ideas in arithmetic. Three that come to mind are:

(i) exponent rules,
(ii) different ways to represent numbers,
(iii) the differences between adding and multiplying.

There’s also a great opportunity to discuss problem solving, and that discussion happens in the 2nd video when my older son says “we don’t really know anything about [the number] $2^{1/3}$.”

Here’s the beginning of our discussion. The first idea that my older son has is writing both numbers in the form $a^{1/10}$.

Before we went down that path, though, I wanted to make sure that my younger son understood a little bit about the two different ways to write numbers as nth roots. So, the discussion here is half about exponent rules and half about our start at solving the problem:

In the last video we found that Orlin’s problem was equivalent to comparing 10 with $2^{10/3}$ but we sort of got stuck there. Getting stuck at this step was fascinating to me because [to me!] it was so obvious what the next step was. The ensuing discussion taught me a lot about how kids see numbers.

In the next part of the project we explored comparing 10 and $2^{10/3}$ a bit more. The first new idea in this part of the discussion was to remove a factor of 2. Now we were left comparing 5 and $2^{7/3}$. What now?

My older son has a neat observation – if we knew if $2^{1/3}$ was larger or smaller than 1.25 we’d be able to answer the question. This particular observation comes from understanding a little bit about the relationship between addition and multiplication.

How, though, do we compare $2^{1/3}$ to 1.25?

So, having solved the problem at the end of the last video, I wanted to show the boys the actual values on the calculator. Unfortunately, I decided to zoom in on the calculator and then not zoom out for the rest of the discussion. Hopefully the words in that discussion are enough to know what we were talking about 🙂 Sorry about the camera goof up – at least the camera mistake was in the shortest video!

Definitely a fun project. I think Orlin’s problem is a great one to talk through with kids learning about exponents. There are so many different ways that conversation can go, and so many opportunities along the way for kids to think about and discussion ideas about numbers.

# Linear vs. Non-Linear ideas in teaching math

Three tweets over the last two days have had me thinking about linear and non-linear ideas that arise when students are learning math.

In the order I saw them:

(1) A linear approach to understanding fractional exponents:

(2) A student mistaking a non-linear idea in trig for a linear one:

(3) A second example of mistaking a non-linear process (functions / logarithms) for a linear one:

Something in the example in the first tweet left me uneasy and I couldn’t quite put my finger on what it was. Seeing the next two tweets, though, helped clear the fog – at least a little.

The first example takes a non-linear process – a geometric series – and uses it to illustrate how to understand a linear process – adding exponents. BUT, it is maybe a little surprising, especially to students, that the linear / non-linear relationship works so nicely in this situation. As the next two examples from students show, a non-linear idea doesn’t always simplify so easily. I worry that the first example subtly plants the idea that everything is linear and could lead to the type of misconceptions illustrated in the 2nd two tweets.

Still trying to think through all of this, though.

# Graham’s number and Skewes’ Number

Saw another great video from Numberphile today:

One thing that got my attention in the video was the comparison to Graham’s Number. Some of their previous videos about Graham’s number had inspired a few projects with the boys. For instance:

The last 4 digits of Graham’s number

The fun (and sometimes frustrating) thing about talking about Graham’s number is that it is so large that it is nearly impossible to describe. In fact, the title of this Evelyn Lame piece on Graham’s number sums it up perfectly: Graham’s numbers is too big for me to tell you how bit it is!

Skewes’number doesn’t have this little problem – you can even write the number 🙂

Skewes’ number = $10^{10^{10^{34}}}$

After seeing the video I thought these two questions might be interesting to kids learning about powers (or logarithms):

Question 1: Which is larger, Skewes’ number or this tower of 5 powers of 3 -> $3^{3^{3^{3^3}}}$

Question 2: Which is larger, Skewes’ number or this tower of 6 powers of 3 -> $3^{3^{3^{3^{3^3}}}}$

The reason I thought it would be interesting to compare to power of 3 is because of how Graham’s number is constructed.

More questions like these can be found in Richard Evan Schwartz’s book “Really Big Numbers.” A few projects that we did from that book are here (with the 3rd one being similar to the two questions above):

A few porjects for kids from Ricahrd Evan Schwartz’s “Really Big Numbers”

Oh, and by luck my older son came home from school as I was finishing this post, so I tried out question 1 with him:

# Talking a little math and physics with MIT’s dark matter boxes

I was visiting MIT last week and was lucky to receive two “dark matter boxes” from the MIT Physics department:

Tonight we finally had some time to sit down and talk about them. It was a fun conversation about both math and physics.

We started by just talking about what the boys saw written on the boxes and about what some of the terms met. At the end of this part of the conversation my older son asked a really great question!

So, at the end of the last video my older son wondered why the box said that the universe was made up of 23% dark matter if there was only one particle of dark matter in the box. Basically 0% of the box is dark matter – how do you get to 23%?

The boys have some pretty cool ideas as to how to resolve this mystery 🙂

The last thing that we talked about was how to understand the exponent $10^{-27}$. It turns out that there are (at least) two places in the universe where a number close to $10^{27}$ appears. First, the mass of the Earth is approximately $6x10^{27}$ grams. Second, the diameter of the observable universe is about $10^{27}$ meters. We use those ideas to try to wrap our heads around $10^{27}$.

So, a really fun conversation touching on a bunch of fun topics. I’m happy that my son wondered about how one particle per box led to 23% of the universe being dark matter – that was a really fun surprise. Happy, too, to have found a few ways to talk about $10^{27}$ – you don’t see that number too often!

Thanks to MIT’s Physics department for the boxes – look for my younger son’s dark matter store coming soon 🙂

# Larry Guth’s “No Rectangles” problem

[note: sorry for the rushed write up – I wanted to get this written up before 9:00 this morning because of a work project that’s going to have me tied up for most of the rest of the day]

Yesterday I was lucky to attend a lecture that Larry Guth gave at MIT:

Larry Guth’s “No Rectangles” problem

It was sort of doubly lucky because I’d already set up a meeting in Cambridge and a little bit of juggling with some times allowed me to have the meeting and attend the lecture. The problem that Guth discussed has the wonderful property of being simultaneously accessible to kids and interesting to research mathematicians. His talk began with a discussion about a simple 3×3 grid and ended with a discussion of intersecting lines in the field Z mod p! Today I used some of the basic ideas in Guth’s talk as the basis for a fun little project with the boys.

I really loved talking through Guth’s problem with them. As often happens in our projects, we ended up going in a slightly different direction than I’d anticipated – but the discussion we had as a result was also much better than I’d anticipated!

We started with a quick introduction to Guth’s “no rectangles” problem and the boys tried to work through the 3×3 case:

The first part of the project ended with the boys finding a way to place 6 snap cubes on a 3×3 square without forming a rectangle. Now we try to see if 6 is the maximum number or is there a way to place 7?

They approached this question in a way that surprised me a little. The first thing they wanted to do was count the total number of rectangles in the grid (which has one subtle point). This approach turned out to be a pretty interesting way to look at the original problem. It is also a pretty interesting counting problem for kids all by itself:

So, having counted 9 rectangles in the 3×3 grid, the boys then tried to see if they could eliminate all 9 of these rectangles by taking away just two of the snap cubes. The conversation here was super fun – lots of great opportunities for the kids to talk about patterns and explore different ideas. For example, my younger son noticed that you are forced to take away one of the cubes on a corner of the 3×3 grid.

In the 2nd half of this video I show them a different way to see that you can’t have 7 cubes on the grid without forming a rectangle. This approach – which is the way Guth explained it in his talk – involves the pigeonhole principle.

For the next part of the project I thought it would be fun for the kids to try to figure out the total number of different ways to place snap cubes onto an NxN grid. Thinking through this problem helps you understand that trying to work through the “no rectangles” problem by brute force with a computer is going to take a long, long time.

Just like the rectangle counting problem the boys came up with in the earlier part of the project, this problem is a nice counting problem for kids all by itself.

Now that we’ve found that the total number of ways to put the snap cubes onto an NxN square is $2^{n^2}$, we try to understand the size of some of these large powers of 2. We start by finding an approximation for $2^{25}$ – the number of ways to put snap cubes on a 5×5 grid – and then try to see if a computer could handle a number this large.

What about a 10×10 grid, though? This problem ends up bringing to the surface a little problem understanding the exponential notation, but that’s one of the great things about this project – you have a nice opportunity to talk about / review exponents while talking about a really interesting problem:

For the last part of the project we try to understand how long $10^{21}$ seconds is. It is actually so long that there’s probably no way for kids to have an intuitive understand of that many seconds. However, we can try to convert it to other units of time, and years seemed like a pretty natural choice. We found that if you can check 1 billion squares per second, searching through all of the possible ways to put snap cubes on a 10×10 grid would take about 300 trillion years!!

Since that’s more time than we’d probably want to spend on that problem, you need to look for a different approach. One really cool thing about Guth’s problem is that there’s a surprising (to me!) connection to geometry and number theory hiding beneath the surface. Guth ended his talk by explaining how to solve the problem using those ideas. That approach, by the way, is truly marvelous but there wasn’t enough room in the margin of my white board to explain it . . . . 🙂

I love being able to share projects like this one with the boys. There aren’t too many problems that are both interesting to research mathematicians and accessible to kids. The kids were able to understand the problem right way and had lots of interesting ideas about the 3×3 case. Exploring the 4×4 case would probably be lots of fun for kids, too. It always amazes me how much fun you can have and how far math conversations with kids can go when the problems keep them engaged. It really was a great bit of luck to have had the opportunity to attend Guth’s lecture yesterday.

# An interesting math example coming from the world of finance

The Best and Worst Investments of 2014

What caught my eye was the description of the best performing bond fund (about 2/3 of the way down the page):

“Vanguard’s index fund invests in U.S. government bonds that don’t mature for 20 to 30 years. They did well in 2014, reflecting expectations that inflation will remain low for quite a while. If you’d invested \$10,000 in VEDTX on Jan. 1, it would be worth \$14,506 today.”

Here’s the math that students interested finance (and, of course, anyone else) might find interesting:

According to Bloomberg data, the 30 year US Government bond had a yield of 3.969% on December 31, 2013, and a yield of 2.752% on December 31, 2014. So, throughout 2014 the yields of long-dated US government bonds – the exact type of bonds that the Vanguard fund invests in – were quite low. In that sort of yield environment how could a fund that invests in US government bonds have had a return of 45%?

Let’s start with a simple example. Say that on December 31, 2013 someone had offered to pay you a 3.969% per year return on your initial investment which would result in you receiving \$1,000 in 30 years. Just to be clear, you’ll receive no interest payments on this investment – the only time cash changes hands is on day 1 when you hand over the initial investment, and at year 30 when you are handed back \$1,000. How much would you have to give that person today to make this investment (ignoring taxes and all sorts of other potential complications – just to keep things easy)?

One way to get to the answer to this question is to take the final \$1,000 payment and divide it by the 1.03969% return that you are promised each year for 30 years. That math tells you the initial investment would be \$1,000 / (1.03969)^30 = \$311.09.

Now, one year later these long-dated yields have gone down to 2.752%. How much is the \$1,000 you receive in 29 years worth at that rate now? We can do the same math: \$1,000 / (1.02752)^29 = 455.07. Wow – what a difference!

In just one year the so-called “present value” of the future payment has gone from \$311 to \$455 – a gain of 46.3%! That gain is due to combination of interest rates falling by 1.25% and the \$1,000 payment being far off in the future so that the change in interest rates compounds for many years.

I’m cheating a little in the above calculation by using the wrong rates. For a single payment in the future I should be using so-called “zero coupon” rates, but I’m just trying to illustrate where these high returns that the Vanguard fund achieved can come from.

If we look at the 30 year coupon bond we can see a similar, though not exactly the same, move in prices. Again according to Bloomberg, the price of the official 30 year US government bone on December 31, 2013 was about \$96 and that bond paid an annual coupon of \$3.75. Roughly speaking that means you could have bought this bond for \$96 on 12/31/2013, then received \$3.75 every year for the next 30 years and then received \$100 at the end of 30 years.

One year later the price of the exact same bond was about \$120. So, at the end of the first year you could have sold the bond for \$120 and kept all of the \$3.75 in interest that you received during the year. You would have paid \$96 and received \$123.75 in interest and sales proceeds. That’s a return of about 29% on your \$96 investment! Again, that return comes from the combination of falling interest rates and the fact that the maturity of the bond is so far in the future.

So, understanding the math on bonds helps us see how it is at least plausible that a US government bond fund could have 46% returns in a year when interest rates were quite low by historical standards. It may seem surprising that a seemingly small movement in interest rates – that is rates falling from 4% to 2.75% – could produce a 46% return, but that more or less exactly how the math works out. It was a lucky year to be taking bets on the values of long-dated bonds!