This one caught my eye because I felt that the wrong choices weren’t selected very well and the correct answer was obvious from the choices:

Despite not liking the question so much for a calculus exam, I thought it would make a pretty neat estimation problem for kids. My older son was looking at exponentials and logarithms this week anyway, so using this question as an estimation problem sort of fit in naturally with what he was doing anyway this week.

I started off today’s project by introducing the problem and asking the kids how they would approach it.

My younger son had a bit of a difficult time understanding the problem – which wasn’t a big surprise. My older son wanted to start by estimating what the square root of e was. Not the starting idea I was expecting, but it made for a good estimating problem:

Now that we we’d guessed that would look a lot like we decided to draw the region.

My younger son thought that the region would be a trapezoid – my older son thought it would be more of a curved shape.

Now that we had the shape drawn, we could estimate the area. We actually used the idea that it was nearly a quadrilateral to make that estimate.

Finally, we used our estimate of the area (3.71 square units) to see if we could identify the correct answer from the choices given in the original problem:

]]>I started by introducing the problem and then with a proof by contradiction example that he already knows -> the square root of 2 is irrational:

Now we started down the path of proving that e is irrational. We again assumed that it was rational and then looked to find a contradiction.

The general idea in the proof is to find an expression that is an integer if e is irrational, but can’t be an integer due to the definition of e.

In this part we find the expression that is forced to be an integer if e is irrational.

Now we looked at the same expression that we studied in the previous video and showed that it cannot be an integer.

I think my favorite part of this video is my son not remembering the formula for the sum of an infinite geometric series, but then saying that he thinks he can derive it.

This is a really challenging proof for a kid, I think, but I’m glad that my son was able to struggle through it. After we finished I showed him that some rational expressions approximating e did indeed satisfy the inequality that we derived in the proof.

]]>An attempt to explain Graham’s number to kids, and

The last 4 digits of Graham’s number

These projects were inspired by this fantastic Evelyn Lamb article:

Graham’s numbers is too big for me to tell you how bit it is

Today I thought it would be fun to revisit the calculation of a few of the last digits of Graham’s number.

So, with no review, I asked my older son what he remembered about Graham’s number and then we talked about the surprising fact that you could calculate the last few digits even though you really couldn’t say much else about the number:

Next I asked my son about how he would approach calculating the last digit. He gravitated to the right idea -> modular arithmetic. The ideas were a little confusing to him, but I let him work mostly on his own.

We didn’t get to the end in this video, but you can see how the ideas start coming together.

In the last video he had made some progress on finding the last digit, but one piece of the argument kept giving him trouble.

BUT, he did have a correct argument – it just took him a minute to realize that he was on the right track.

Again, this is a nice example of how a kid works through some advanced mathematical ideas.

Next we went to the computer to begin looking at the last two digits of Graham’s number. The last two digits of powers of 3 repeat every 20 powers, so it was easier to use Mathematica to find the cycle than it was to do it by hand.

Here I just explain the short little computer program I wrote to him.

Finally, we tried to see if we could use the idea that the powers of 3 repeat their last two digits every 20 steps to see if we could find the last 2 digits of Graham’s number.

As we started down the path here, I didn’t know if we’d find those last two digits. But we did! It was a nice way to end the project.

I found a series that gives the solution, but don’t know a closed form for the solution. Still, though, I thought sharing this problem with kids would be fun – some of the basic ideas you need to start down the road to the solution are accessible to kids. I’m really happy with how the discussion went today.

We started out by reading the problem and I asked the boys a few questions to make sure they understood how the game worked:

Next we started talking about the solution. My younger son had a little bit of a misconception initially, but the problem started to make a bit more sense for him after my older son suggested a different approach.

Still, it was interesting to hear the boys discuss whether the probability of winning in exactly 3 tosses was 1/4 or 1/8.

Now we tried to figure out how likely it would be to win with a string of exactly three heads at the end (so your final four flips would be THHH).

Finally, we tried to count the ways that you could win with final flips of THHHH – so winning by flipping four heads in a row. Their guess in the last video was that there would be 3 cases – but they realized fairly early on that there would be more than 3.

After we finished I showed them my code in Mathematica that calculated the sum that gives the final answer. To 10 decimal places the answer is 0.7112119049.

Thank to Amy Hogan for sharing this problem – even though the exact answer is a little out of reach for young kids, it is still a terrific problem for them to study.

]]>That shape was just “discovered” and is discussed on this New Scientist article:

oops – that tweet gives me a good picture, but the article itself is behind a paywall. Here are two free articles:

Gizmodo’s article on the Scutoid

The article in Nature introducing the shape

Last night I had the boys play with the shape (and I did not tell them what it was).

Here’s what my older son thought about it – sorry that it is a little hard to see the shape in the beginning. I add more light around 1:00 in:

Here’s what my younger son thought:

I thought it was interesting to hear that both boys thought that this shape would not appear in nature. I’ll have them

]]>“Although the series 1 + 2 + 3 + … diverges, zeta function regularization gives this sum the curious value of -1/12. I will discuss two ways of making sense of summing similar divergent series such as … + -2 + -1 + 0 + 1 + 2 + … that are doubly infinite rather than singly infinite. I will then explain how to generalize to sums over lattices of higher rank, as well as how to interpret parts of this as giving a summation formula for certain finite sums.”

The details from the lecture are too advanced for my kids (and probably for me, to be honest!) but I still thought it would be fun to revisit the old video from Numberphile about the series.

Since we’ve talked about the series before, I thought the best way to get started today was to get some thoughts from the boys about this strange result.

Their ideas are fascinating – I was surprised that both of them seem to have come around to believing the result. The first time we viewed the Numberphile video (when it first came out) they were very skeptical:

Next we watched the video again – so here it is:

Having watched the video, I asked they’ve changed their mind or had any extra suspicions. Again, to my surprised they were now even more convinced.

They did think a few things were strange, though, so at least they were thinking carefully about the result.

Next we tried to recreate the “proof” from Numberphile’s video. This section went pretty well.

Finally, I mentioned some of the ideas from the lecture at MIT on Thursday. There were many things in the lecture that I’d not seen before – the one new idea I introduced here was the doubly infinite series . . . + 1 + 1 + 1 + 1 + . . . .

I wanted to use that series to illustrate some of the even stranger than strange properties these divergent infinite series have. This was a really fun part of our project today.

]]>I was interested to see how the boys would approach this problem – and that was fun to hear. Unfortunately this project got way off the rails at the end when I tried to explain my solution to them. One of those unfortunate moments where something was clear to me but I was not communicating it well at all. I say it a lot in these blog posts – but they don’t all go well.

Here’s my older son’s solution:

Here’s my younger son’s version:

Next we move to me explaining my solution. This part did not go well at all:

Here’s the second part of the explanation of my solution. Neither kid likes my approach. Given the presentation, I’m not surprised at all

]]>The problem asks you to prove that the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.

Yesterday he worked through the solution in the book – today I wanted to talk through the problem with him. We started by introducing the problem and having my son talk through a few of the ideas that gave him trouble:

Next he talked through the first part of the solution that he learned from the book. We talked through a few steps of the algebra, but there were still a few things that weren’t clear to him.

Now we dove into some of the algebraic ideas that he was struggling with. One main point for him here, I think, was labeling the important unknowns in the problem.

For the last part, I wrote and he talked. I did this because I wanted him to be able to refer to some of our prior work. The nice thing here was that he was able to recognize the main algebraic connection that allowed him to finish the proof.

]]>That project is here:

Sharing a problem from the Julia Robinson math festival with the boys

Last night I got an interesting comment on twitter in response to my Younger son suggesting that we write the numbers in a circle – a suggestion that we didn’t pursue:

So, today we revisited the problem and wrote the numbers in a circle:

Next I asked them to try to find another set of numbers that would lead us to be able to pair all of the numbers together with the sum of each pair being a square. The discussion here was fascinating and they eventually found

This problem definitely made for a fun weekend. Thanks to Michael Pershan for sharing the problem originally and to Rod Bogart for encouraging us to look at the problem again using my younger son’s idea.

]]>This problem seemed like a nice one to use to get back in to our math project routine.

Here’s the introduction to the problem and the full approach the boys used to work through it the firs time:

When they solved the problem the first time around, they started by pairing 16 and 9. I asked them to write down their original pairs but to go through the problem a second time without starting with 16 and 9 and see if the choices really were forced. Here’s how that went:

This is a really nice problem for kids. It is easy to understand, so kids can jump right into it. There’s also lots of different ways to approach it. Definitely a fun way to get back into our math projects.

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