Month: November 2014

Angry Birds and Snap Cubes: Using Bryna Kra’s MoMath public lecture to talk math with kids

Last night I watched Bryna Kra’s public lecture at the Museum of Math:

 

I’m not quite sure how to talk through some of the simple dynamical system ideas in the lecture, but the earlier material about patterns and the pigeonhole principle are definitely fun topics to talk about with kids. We used our collection of snap cubes and Angry Bird stuffed animals as props 🙂

In the first part of the talk we introduce the pigeonhole principle and talk through a simple pattern with only single blocks based on one of the elementary patterns Kra uses in her talk. This simple pattern allows us to get a little bit of practice identifying the “pigeons” and the “pigeonholes” in a problem:

 

In the second talk we look at a slightly more complicated pattern – patterns you get with two blocks rather than one. For this pattern we consider the order of the birds to be important – so a (red, blue) group is different than a (blue, red) group. The example we look at in the last part of today’s talk will consider those two groups to be the same.

The boys were able to see the four different patters that we could make with the two birds / blocks. My older son even noticed a connection with Pascal’s triangle which was fun to see. We then talked about how to count the different types of pairs by looking at the number of choices we had for the first bird and for the second bird. That led my younger son to wonder if there would be a total of 9 groups of two birds if we allowed three different birds in the pattern. Pretty fun discussion:

 

At the end of the last talk my younger son wondered what would happen if we used three different colors of blocks rather than two. I hadn’t planned on discussing that problem, but what the heck! It was interesting to see the kids figure out how to group the blocks to make the 9 pairs. They were also now able to see how the patterns would continue if we varied the colors and/or number of blocks in the pattern. Fun little exercise. Watching this again I wish I would have spent a little time responding to my older son’s comment that there was no connection to Pascal’s triangle pattern anymore – oh well, next time!

 

Our last project was a slightly different twist on the Pigeonhole principle. We looked at a tournament involving 4 birds in which each game involves 2 birds. The question I had the boys look at was this: If there are 7 total games played in this tournament, show that at least two of the games must involve the same two players.

I liked their approach to solving this problem. Their instinct was to solve the problem by listing out all of the types of games that could happen. If we were at our whiteboard I would have drawn a square with its sides and diagonals, but their list of all of the types of games was good enough for this project. They had a little difficulty identifying the pigeons and pigeonholes here, but that’s ok, it isn’t always so obvious how to make that identification.

 

So, a fun project based on another MoMath talk. See here for our last project based on a MoMath lecture:

Part 3 of using Terry Tao’s MoMath lecture to talk about math with kids

I think the public lectures at the Museum of Math are a great way for kids to see some amazing math. There will surely be some lectures that are too advanced for young kids, but many of these lectures have ideas in them that are not hard at all for kids to understand. With Bryna Kra’s lecture, the ideas about patterns and the pigeonhole principle are topics that kids can play with and really enjoy. I’m super glad that MoMath is making these lectures available to the public. It is really fun to show kids some ideas that professional mathematicians use in their research, and hopefully also a great way to inspire a new generation of mathematicians!

Problem Solving part 2 – an old AIME problem

[note: I’m up in Boston today and decided to take a trip to the MIT library to check out a book that Jordan Ellenberg mentions in his book “How not to be Wrong.” I didn’t know the MIT libraries were closed until 11:00 am today (they used to be open 24 hours!!), so I wrote this post fairly quickly in the hour I had to kill waiting for the library to open!. Sorry if it seems rushed – it sort of was!]

I few months ago I wrote about Tim Gowers live blogging his solution to one of the IMO problems:

Problem Solving and Tim Gowers’s live blogging an IMO problem

I thought that live blogging problem solving was a good idea because I think that kids (and everyone) needs to see that most solutions you get to problems aren’t the super perfect “official solutions” and don’t come to mind immediately. In that post I talked through my solution to a number theory problem that David Radcliffe had put on Twitter. Today’s blog is my second attempt at live blogging a problem. I saw today’s problem in Art of Problem Solving’s Precalculus book in the chapter 8 challenge problems.

It also happens to be problem 15 from the 1991 AIME – see the bottom of the page here:

The 1991 AIME Problems hosted by Art of Problem Solving

Here is the problem:

For a positive integer n define S_n to be the minimum value of the sum:

\sum_{k = 1}^{n} \sqrt{ (2k - 1)^2 + a_{k}^{2} }

where a_1, a_2, \ldots, a_n are positive real numbers whose sum is 17. There is a unique positive integer n for which S_n is also an integer, find this n.

 

The “live blogging” of my solution is below:

Background:

I had a nice (and short) discussion on twitter with Patrick Honner about Art of Problem Solving’s Precalculus book. I personally like this book because of chapter 8 – the geometry of complex numbers. This chapter, and in particular section 8.5, has content that I’ve not seen anywhere else.

Following that discussion, I grabbed my copy of the book to take a fresh look at chapter 8. I ended up back in the challenge problems and the problem I’m writing about today caught my eye.

My first reaction, frankly, was fear. Back in high school I’d not really understood analysis all that well, and did not have a good grasp of theorems that helped with this type of summation problem – the Cauchy Schwarz inequality, for example: The Cauchy-Schwarz Inequality. An “analysis” approach to this problem probably requires an even more general inequality – Hölder’s inequality – though I would not have known that in high school.

Whenever I see this type of problem, even today, I remember the fear of the analysis ideas that I had in high school. Funny how those old high school fears never seem to go away!

But wait – why would the Art of Problem Solving folks put an analysis problem in the chapter about the geometry of complex numbers? There must be a geometric solution here somewhere? Where is it?

So, let’s look at a few terms:

(1) n = 1. The “sum” is just one term \sqrt{1^2 + 17^2}. Certainly not an integer, but also certainly reminiscent of the Pythagorean theorem and a right triangle with legs of length 1 and 17.

(2) n = 2. The sum is now \sqrt{1^2 + a^{2}} + \sqrt{3^2 + (17 - a)^2}. Now there are two right triangles – one with legs 1 and a, and another with legs 3 and (17 - a). The sum of the integer length legs is 4, and the sum of the other legs is 17.

Oh, wait, I see what’s going on. The general case is going to have a bunch of right triangles whose “heights” always adds up to 17 and whose “lengths” will sum up to 1 + 3 + 5 + (2n - 1) = n^2.

See here for the sum of the odd numbers:

 

How cool! The minimum value of the sum we are being asked to consider is going to come when we can make all of the hypotenuses of these triangles form a straight line, and that length will be \sqrt{ (n^2)^2 + 17^2} from a “large” right triangle with length n^2 and height 17.

Sweet – per the statement of the problem, there must only be one right triangle with integer side lengths where one of the side lengths is 17. That shouldn’t be too hard to find.

We know there’s a 3,4,5 triangle, and a 5, 12, 13 triangle, and a 7, 24, 25 triangle – which helps see the pattern fairly quickly. For an odd number, you’ll get a right triangle with integer sides by finding two consecutive numbers that add up to the square of the odd number. From the three examples above: 4 + 5 = 3*3, 12 + 13 = 5*5, and 24 + 25 = 7*7.

So . . . 144 + 145 = 289 = 17*17, so there must be a 17, 144, 145 triangle. By luck 144 is a perfect square and we indeed have a right triangle with integer sides having one leg equal to 17 and the other leg equal to a perfect square.

That means the minimum value of the sum we were looking for is 145, and the value n for that particular triangle is 12.

Fun and super clever problem. Love the connection between geometry and algebra here – especially because the algebra side of this problem still makes me nervous 🙂

Most of the thoughts above were in my head – my “work” for the problem is in the picture below. I’m glad I saw the geometric solution before trying to dive into all of the analysis which, I assume, is way more grungy and way less of a fun solution.

Solution Pic

The last 4 digits of Graham’s number

In the spring we had a lot of fun talking about Graham’s number. If you haven’t seen anything about Ggraham’s number before, you might enjoy checking out that prior blog post:

An Attempt to explain Graham’s number to kids

Also definitely check out the excellent series on Graham’s number that Numberphile did with Ron Graham!


Our talk today assumes just a tiny little bit of knowledge of Graham’s number – (1) mainly that it is an outrageously tall tower of powers of three, and (2) so large, in fact, that it is nearly impossible to even imagine how large the number actually is.

We returned to Graham’s number today because my younger son just started a new chapter about last digits in his number theory book. He’ll be learning about how to find the last digit of numbers like 3^{1000} and some other similar numbers. It is a neat subject and a fun way to continue to build number sense.

Right at the beginning, though, he asked me why we were only taking about the last digit – why not the tens digit, or hundreds digit? Well . . . today we’ll talk about the last 4 digits of Graham’s number just for fun.

Our first talk is a quick review of Graham’s number. If you want to understand the “up arrow” notation check out the links above, but that notation isn’t important for today. All you really need to know is that Graham’s numbers is a huge tower of powers of 3:

With the review out of the way we turn our attention to the last digit of Graham’s number. After looking at the first few powers of 3 we see that the last digit appears to repeat every 4th number. Quite surprisingly that pattern gives us enough information to infer that the last digit of Graham’s number is either 3 or 7. We spend probably half of the movie arriving at that fact and then we perform a more detailed calculation to see what the last digit actually is. One point that caused a little bit of confusion is that we need to look at the power itself in cycles of 4 (or what the remainder is when you divide by 4) even though we are looking for the last digit (so remainder when divided by 10):


Next we moved to the computer to get a little help from Mathematica! We essentially repeat the calculations that we just did on the whiteboard, but looking at the last two digits rather than looking only at the last digit. When you look at the last two digits you see a pattern that repeats every 20 powers – hence why a computer is helpful! Once we know that there is a pattern that repeats every 20 numbers we can use the computer to perform the same computation that we did by hand in the last movie to find the last two digits of Graham’s number:


The next step was looking for the last 3 digits. It is essentially the same process. We found that the last digit of the powers of 3 repeat every 4 powers and the last two digits of powers of 3 repeat every 20, so I asked the boys what they thought the pattern in the last 3 digits would be. They both guessed the repetition would be every 100 powers, which turns out to be right. Again, the computer is your friend here!

Also, we made a little mistake in this video and got confused between 100 and 1,000 when the pattern was repeating. Luckily that just made more work for the computer to do rather than for us (which is probably why we didn’t notice), but the result is unchanged (luckily).


We wrapped up by wondering why we are seeing powers of 5 in the way the digits repeat. The units digit repeats every 4 powers, the last two digits repeat every 20, the last three digits repeat every 100 powers, and the last 4 repeat every 500 powers – why are we multiplying by 5 every step? We didn’t arrive at an answer for this problem, but rather left it as something to wonder about.

The last thing we did was check out the Wikipedia page about Graham’s number to see if we got the last three digits right. That page gives the last 500 digits and our last 3 do actually match! We also now have a procedure to use to (perhaps) find all 500 digits.


So, a fun little project. Kicking myself for the 100 vs 1,000 mistake, but I guess that happens. The project kept the kids engaged all the way through – both the math and the computer results are really interesting. It is amazing (especially for kids) to see that even though you can’t really say anything at all about the number itself, you can compute some of the final digits.

Proof in math

There are been three recent pieces about mathematical proof that have caught my attention in the last couple of months:

(1) Most recently Keith Devlin’s piece here:

(2) Evelyn Lamb’s coverage of Leslie Lamport’s talk at the Heidelberg Laureate Forum (including, importantly, a link to a paper in which Lamport takes a critical look at a proof in Michael Spivak’s Calculus textbook)

A Computer Scientist Tells Mathematicians How To Write Proofs

(3) Numberphile’s “All Triangles are Equilatleral” video featuring Carlo Séquin whose Art and Math collection is always a pleasure to look through:



These three pieces have kept me thinking about proof in math for a while now. I was sick over the weekend and spent a little time browsing through the Museum of Math’s public lectures and found this nice one from Steven Strogatz where, by happy coincidence, the topic of proof in mathematics also comes up:

In this lecture Strogatz discuses a proof of the fact that the area of a circle is equal to \pi r^2. He also wrote an article about the same proof in the New York Times:

Steven Strogatz discusses the proof that the area of a circle is \pi r^2

The combination of the recent writing on proofs in mathematics and watching Strogatz’s lecture gave me an idea for a fun way to talk about proofs in mathematics with my kids. Not in any formal way, but (1) just to show them some easy “proofs,” and (2) to show them that it is ok to question something even if it is supposedly a mathematical proof.

I’ve used this format for a talk previously after learning about Jordan Ellenberg’s concept of “algebraic intimidation” (and you’ll see in our first video from today how much that talk still bothers my younger son!)

Jordan Ellenberg’s “Algebraic Intimidation” and the series 1 + 2 + 3 + 4 + . . . = -1/12

The two topics for today were (1) the proof that Strogatz used to show that the area of a circle is \pi r^2 and (2) the proof that \pi = 4. The proofs are actually quite similar and it turns out that one of them ends up with an incorrect result (though I won’t say which one!).

At the end of each proof I asked my kids what they thought. Funny enough, my younger son does not believe either of them because he is uncomfortable with the use of infinity in the proofs. My older son believes the first one but not the second one.

Here are those two talks – the first is about the area of a circle:


and the second is about the value of \pi:

So, a fun morning with the boys talking through a few “proofs.” I really like the lessons in both Devlin’s piece and in Strogatz’s lecture about using proofs to tell stories and to illuminate. Lamb’s piece about Lamport reminds us that details are important, though, and the two pretty similar proofs I went through with the boys this morning serve as a reminder that the details can actually be pretty subtle (but fun, of course).

Difficult and not difficult

As I wrote this morning on twitter:

The outcome isn’t really surprising at all since:

(1) I’ve never taught young kids before,
(2) I’ve never taught elementary math before,
(3) The experience with one kid probably doesn’t translate in any way to the other kid,

and . . . well, I could probably get to (100) without much difficultly.

Still, despite maybe not being intellectually surprising, it still surprises me. All. the. time.

We had friends staying overnight and got started with school a little late this morning. My younger son was starting a new section in his number theory book today – “Units Digits.” We went through a few examples of finding units digits. The problems we did were mainly introductory problems like: find the units digit of 4*23, 214*23, and 492*5137. Then we did the same exercise for a few powers of 4.

With those examples out of the way I wanted to explain how looking at patterns can help you find the last digit of large powers. I thought the large powers would make for a fun movie project so I started off our movie by asking him to find the last digit of 3^{1000}. He worked through the problem as if we’d been studying this subject for weeks. I expected the transition from finding the pattern to evaluating what the pattern would be at the 1000th step to be much more difficult. The approach that I walked through in the 2nd half of the video is what I was expecting to be talking about (in pieces) for the entire discussion:

While I was working with my younger son, my older son was working on a few old math contest problems. The one linked here gave him quite a bit of trouble:

2003 AMC 10B Problem 5

Here’s the problem w/o the link:

“Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath he cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that no grass is missed. He walks at the rate of 5000 feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn: (a) 0.75 (b) 0.8 (c) 1.35 (d) 1.5 (e) 3.”

Contributing to my surprise was that first the first time ever last weekend he helped me mow the lawn. Ha!

 

If it wasn’t 35 degrees and raining this morning, I would have gone out in the back yard with him and mowed for a bit just for context! More to the math points, though, I was genuinely surprised at how much difficulty this problem gave him. Two of the larger struggles came from:

(1) The interaction of the 28 inch wide cut and the 4 inch wide overlap.

And, yes, this part of the problem is gimmicky, no question. However, even drawing a picture of what was going on in this problem was hard for him. We probably talked about it for 15 minutes before he had the “aha” moment of realizing you were essentially moving across the lawn in 2 ft steps.

(2) The transition from knowing you were moving in 2 ft steps to figuring out how long it would take to mow the lawn.

We had a picture on our whiteboard of a rectangle chopped up into a bunch of thin (90 foot long) strips. Maybe the long talk about the first half of the problem sort of used up all of his mental energy, who knows, but this part took a lot longer to work through than I would have guessed. Even when we got to the last bit and just needed to evaluate (75 strips) * (90 ft / strip ) / ( 5000 ft / hour ) he told me that he was worried that we hadn’t solved the problem correctly because the units were wrong.

I don’t mind struggles – in fact I want struggles. I’m just surprised (constantly) when I don’t see them coming. In fact, this morning when I looked at the problems he was going to be working on I was sure this was the one that we’d be talking about:

2003 AMC 10B Problem 7

I wonder how many years it will be until I get good at seeing these struggles ahead of time?

A great piece on Grothendieck by Ed Frenkel and a nice problem for students interested in math

[note: home sick with some stomach bug for the last two days – sorry for what is surely a bit of a rambling post]

Ed Frenkel published a nice piece in the New York Times today on the life and work of Alexander Grothendieck.

The Lives of Alexander Grothendieck, a Mathematical Visionary

In addition to Frenkel’s perspective on Grothendieck, what caught my attention was an almost off-hand observation about complex numbers that is really fascinating. I know it would have been quite a head scratcher for me in high school so I thought it would be fun to write about. Here’s the comment about the equation x^2 + y^2 = 1:

“One can show that the solutions of [x^2 + y^2 = 1] in complex numbers are points of an entirely different space; namely, a plane with one point removed.”

Students familiar with the equation x^2 + y^2 = 1 probably have only thought about this equation when both of the variables x and y are real numbers (when the solution is the familiar unit circle). The extension to complex numbers is a nice mathematical surprise.

So how can you think about Frenkel’s example? An excellent starting point is Richard Rusczyk’s sample solution for problem #25 of the 2013 AMC 12. The video below is a great way for students to see the power of geometric reasoning with complex numbers:

An approach similar to what Rusczyk outlines above is also a good way to start thinking about Frenkel’s equation. Try a few examples first – if x = 10i, for example, what values of y will satisfy the equation x^2 + y^2 = 1 (remember that both x and y are complex numbers)?

Now, if you have a generic value of x, what values of y will solve the equation? You’ll find that there are 2 values of y for most values of x, though importantly, not all.

Next is a real geometric leap – if every point x in the complex plane paired with exactly two points in Frenkel’s equation, seems as though the solution to the equation would be equivalent to two copies of the complex plane (possibly glued together in some strange way). Though it is challenging for sure, it is fun to think about what’s different from the situation I just described – in what way is the situation Frenkel describes similar to a plane with a point missing?

Away from this fun example of geometry with complex numbers, it was nice to see Grothendieck’s work described to the public. Another recent article about mathematicians written for the public was Michael Harris’ piece in Slate about the Breakthrough Prizes in math:

Michael Harris on the Breakthrough Prizes in Math

One of Harris’ points caught me off guard:

“Tao—the only math laureate with any social media presence (29,000-plus followers on Google Plus)—was a guest on The Colbert Report a few days after the ceremony. He is articulate, attractive, and the only one of the five who has done work that can be made accessible to Colbert’s audience in a six-minute segment.”

I was surprised to hear that Harris thought that the work of Jacob Lurie, Richard Taylor, Maxim Kontsevich, and Simon Donaldson really could not be made accessible to the public. Surprised enough, actually, to ask Jordan Ellenberg on twitter if he agreed with the statement:

Though his answer was not really a shock, it still disappoints me a little that work of these researchers is so inaccessible to the general public. Hopefully Frenkel, or other mathematics writers, can find a way to bring the beauty of their work to the public. I’d love to know more about Lurie’s work, or any of their work, frankly.

More public lectures like the one Terry Tao gave at the Museum of Math would be great, too. I’ve already done three projects with my kids already based on that lecture. It is amazing for them to be able to learn from Terry Tao!

Terry Tao’s MoMath lecture Part 1 – The Moon

Terry Tao’s MoMath lecture Part 2 – Clocks and Mars

Terry Tao’s MoMath lecture Part 3 – the Speed of Light and Paralax

It would be wonderful if there were more opportunities like Tao’s public lecture to introduce kids to research mathematicians and more article’s like Frenkel’s, too. Despite being home sick, Frenkel’s article made my day today.

Amazing how many things connect to the number 11.

A few weeks ago David Wees asked the following question on Twitter:

The there were many fun answers to David’s question if you click through to the thread on twitter.  There’s another one in this super blog post that Patrick Honner linked last week, too:

The ideas in Vincnet Knight’s post have been kicking around in the back of my mind for several days.  Today I stumbled on an old favorite Ben Orlin post that sort of connected the dots for me:

A Teaching Philosophy I’m Not Ashamed Of

This quote, in particular, speaks volumes:  “Math is big ideas, approached from as many angles as possible.”

Funny enough the dots that got connected led to nearly the opposite conclusion in this specific case.  For me the number 11 is a simple, or in Orlin’s words, small idea, but this small idea allows you to approach many different areas of mathematics directly.  It really is an incredible how useful this simple idea is.

We’ve had a lot of discussions where the number 11 proved to be a surprisingly useful starting point.  One interesting connection is Pascal’s triangle.  We discussed how Pascal’s triangle relates to powers of 11 here:

Pascal’s Triangle and Powers of 11

Here’s a sample of that discussion:

 

Another fun discussion that related to the number 11 came when we were talking about converting numbers between bases:

All about that base: A fun exercise from Art of Problem Solving

Here’s a sample of that discussion:

 

Finally, although this technically relates to the number 12 (oh so close!) you could easily replicate this exercise for the number 11 and the polynomial (x + 1)^n:

Complete This Sentence: Math is _______

A sample of this discussion is here:

 

So, a lot of thinking on my end inspired by Vincent Knight’s post.  Who would have thought that this one little number could be so interesting!

Terry Tao’s MoMath lecture part 3: The speed of light and parallax

[sorry if this doesn’t read too well.  woke up sick today and am getting sicker.  boo 😦 Didn’t have the energy for too much editing. ]

In the last few weeks I’ve been writing about Terry Tao’s incredible public lecture delivered at the  Museum of Math over the summer and how that lecture provides many great examples you can use to talk about math with kids.  The first two posts are here:

Part 2 of using Terry Tao’s MoMath lecture to talk about math with kids – Clocks and Mars

Part 1 of using Terry Tao’s MoMath lecture to talk about math with kids – the Moon and the Earth

for ease, the direct link to the Terry Tao lecture  is here:

Today we were talking about the piece of the talk starting around 1:04:30 – how physicists obtained the first estimate for the speed of light and also  how astronomers measured the distance to nearby stars.

We began by watching Terry Tao’s presentation and then discussing the boy’s reaction to the video.  They seemed to have a reasonably good understanding of how the measurement of the speed of light was done.  Their ability to understand the talk is why I think Tao’s lecture is so great for kids to see – his explanations are incredibly easy to follow.  We had to clarify a few points, but after those clarifications we were able to repeat the calculation.

Also, working through the calculation is a nice exercise in place value and division for kids.

Following that conversation we moved on to the discussion of how astronomers measured the distance to the nearest stars.  The portion of Tao’s lecture that discusses parallax is amazing, but one really interesting thing that his pictures don’t really illustrate are all of the distances involved.  I draw a similar picture to the one Tao used in the talk and then mention what the proper proportions would be at the end of the video.

Also, at the beginning of this video my younger son was confused by the distances I had written down for the radius of the Earth and the radius of the Sun.  I’m not sure exactly what was bothering him, but since a critical point for understanding parallax is understanding distances, we spent a few minutes at the beginning of the video making sure he understood those distances properly.

Finally, we went out to the back yard to demonstrate the relative distances involved in the measurement of the distance to Alpha Centauri.  We used a small balloon with radius about 2.5 inches for the sun and a grain of salt for the Earth.  At this scale the radius of the Earth’s orbit around the sun is about 50 ft.    Also at this scale, if we were standing in New York City Alpha Centauri would be in Los Angeles!   Sorry for all the coughing in this one – I’m a little sick today 😦

So, one more neat project for kids coming from Terry Tao’s lecture.  It is a little hard to go into the details of how the angles were measured since you need trigonometry for that, but the geometry is easy enough to understand.   Attempting to “draw” the picture to scale in our back yard was really fun, too.  The calculation of the speed of light really just requires a little arithmetic and is a nice example to show to help build up number sense.

Definitely a fun morning!

The Collatz conjecture and John Conway’s “amusical” variation

I enjoy looking at unsolved math problems with the boys.  Although there aren’t too many of these problems that are accessible to kids, some of the ones that they can understand are pretty neat.  Today I decided to return to the Collatz conjecture and present a “new to me” twist on the problem described by John Conway in his essay “On Unsettleable Arithmetical Problems.”  I saw the essay in the book pictured below, which I cannot recommend highly enough:

Math Book

We began by taking a fresh look at the Collatz Conjecture.  The statement of this particular unsolved problem involves only a little bit of arithmetic and talking about this problem is a really great way to show some interesting math and build up a little number sense at the same time.  Here is the problem:

Step 1:  Pick any positive integer n.
Step 2:  If n is even divide it by 2, and if n is odd multiply it by 3 and then add 1.
Step 3:  Repeat step 2 with the number you ended up with after step 2, but stop if you ever get to 1.

For example, if you start with the number 3 you get the sequence:   3, 10, 5, 16, 8, 4, 2, 1.

For any number that has ever been tested, whether by hand or by computer, the sequence always ends up at 1.  The Collatz conjecture states that every integer will end up at 1 eventually.  Although many people believe that this conjecture is indeed true, it has not been proven to be true.

Here is our introductory discussion of the problem:

In the second part of our project today I introduced the twist on the Collatz conjecture that Conway describes in his essay:

Step 1:  Start with any positive integer n.

Step 2:  If the integer from the previous step is of the form:

2.a:  2k, make a new integer 3k,
2.b  4k + 1, make a new integer 3k + 1,

2.c  4k – 1, make a new integer 3k – 1.

Step 3:  Repeat step 2 until you encounter a number you found previously.

For example, if we start with the number 4, we get the sequence:  4, 5, 9, 7, 5, and then back to 4.

As with the Collatz problem, there’s not much that mathematicians have been able to say about this new problem.  There are a few known cycles, like the example above, but for most integers the sequence you get from following Conway’s process seems to go on forever without repeating.

In our introductory talk about this new problem we go through one example and then spend a little bit of time talking about why all positive integers are either even (step 2a above), one more than a multiple of 4 (step 2b), or one less than a multiple of 4 (step 2c).  This discussion hints at a basic idea in number theory called modular arithmetic.

Next we talked about why Conway’s problem is a little bit different than the Collatz conjecture.  One of the things that’s really interesting is that you can run the computations in Conway’s problem both forwards and backwards.  You can’t do that in the Collatz problem because multiple numbers can map to the same number.  For example, in the Collatz problem the number 10 can come from dividing 20 by 2 or by taking 3 and multiplying it by 3 and adding 1.  That means you can’t go backwards from 10.
In his article Conway plots a few of the sequences that arise in the new problem (actually, he plots the logarithm of the sequence, but that’s a detail I’ll put to the side for now).  The plots look surprisingly similar to the letter V.  The shape of those graphs was really surprising to me.

Because there is a connection between the number 12 and Conway’s version of the Collatz problem, he calls the pattern in the numbers you get by following the new process “amusical.”    I’ll leave the description of why to him, save for this bit of mathematical comedy:

“However, since the series always ascends by a fifth modulo octaves, it does not sound very musical, and it has amused me to call it amusical.”

The connection, or lack of a connection, to music seemed to be a neat thing to illustrate for the boys, so I used Mathematica’s Sound function to create a sound for each number in a particular sequence.  This part was just for fun, but watching it just now I wish I would have described the process a little better.  At least you can see the Mathematica code on the screen and copy it if you want.  The boys really liked hearing the music, though, and played around with the different sounds a little bit after we finished:

So,  a  really fun project going through two unsolved math problems.  Conway uses these problems as examples of problems that may actually be no way to solve.  That’s an interesting topic, too, though more than I wanted to get into today.  For now the goal was to show some fun math and continue to build up their number sense.  I really enjoyed this project.

Surprises you get watching kids do math

My older son is finishing up a chapter about right triangles in his Geometry book.  The last section in the chapter discusses how to construct perpendicular lines.  This topic has come up a few times previously, so I dedicated a bit more time to the previously section – Heron’s formula – before starting in with the new section this morning.  I also left one of the examples in the book for him to work through as part of his homework.

One of the other homework questions was to construct a segment with length \sqrt{3} given a segment of length 1.  I thought this might be a little bit of a challenge and was pretty interested to see how he would approach it.

When I got home from work this afternoon I got a nice surprise:

Me: Were you able to construct the 30-60-90 triangle?

Him:  That wasn’t one of the homework questions.

Me:  Oh, I thought there was one about constructing a length equal to \sqrt{3}.

Him:  I made a 1 – \sqrt{2} – \sqrt{3} triangle for that one.

Construction

Funny how focused your mind gets when you “know” how to do something.  Wouldn’t have thought of this approach in a million years.  Nice little construction!