Month: August 2018

Introducing the basic ideas behind quadratic reciprocity to kids

We are heading out for a little vacation before school starts and I wanted a gentle topic for today’s project. When I woke up this morning the idea of introducing the boys to quadratic reciprocity jumped into my head. The Wikipedia page on the topic gave me a few ideas:

Wikipedia’s page on quadratic reciprocity

I started the project by showing them the chart on Wikipedia’s page showing the factorization of n^2 - 5 for integers 1, 2, 3, and etc . . .

What patterns, if any, would they see?

Next we moved to a second table from Wikipedia’s page – this table shows the squares mod p for primes going going from 3 to 47.

Again, what patterns, if any, do they notice?

Now I had them look for a special number -> for which primes p could we find a square congruent to -1 mod p?

Finally, we wrote short program in Mathematica to test the conjecture that we had in the last video.  The conjecture was that primes congruent to 3 mod 4 would have no squares congruent to -1 mod p, and for primes congruent to 1 mod 4 would, -1 would  always be a square.

Sorry for the less than stellar camera work in this video . . . .

Trying out Edmund Harriss’s puzzle with kids

Saw a neat puzzle posted by Edmund Harriss last night:

I thought it would be fun to try it out with the boys this afternoon.

I didn’t give them much direction after introducing the puzzle – just enough to make sure that my younger son understood the situation:

After the first 5 minutes they had the main idea needed to solve the puzzle. In this video they got to the solution and were able to explain why their solution worked:

Definitely a fun challenge problem to share with kids. You really just have to be sure that they understand the set up and they can go all the way from there.

Using an AB Calculus question as a estimation problem for kids

Last night I was curious what the AP calculus questions looked like and was flipping through the multiple choice questions from the 2012 AB calculus exam.

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:

Screen Shot 2018-08-19 at 9.05.42 AM.png

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 e^{x/2} would look a lot like 1.65^x 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:

Walking through the proof that e is irrational with a kid

My son is finishing up a chapter on exponentials and logs in the book he was working through this summer. The book had a big focus on e in this chapter, so I thought it would be fun to show him the proof that e is irrational.

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.

Revisiting the last digits of Graham’s number

Several years ago we did a bunch of projects on Graham’s number.

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.

Trying out a challenging problem shared by Amy Hogan

Saw a neat tweet from Amy Hogan yesterday:

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.