Day: March 17, 2015

A first look at square roots

I started in on a new section with my younger son yesterday – square roots.

We have, of course, talked about square roots informally through the various projects that we do, but we have not discussed properties of square roots in detail until now. It is always really interesting to me to see the initial reactions that the boys have to these new subjects.

The first discussion today was about some basic properties that he already knew, and then I threw in a few twists. The idea that a square root has to be non-negative caused a little confusion. We ended up with a short discussion of i.

 

Next up was today’s lesson about finding square roots of large integers. I enjoyed his approach to the specific problem below, and think that the time that we spent in the first half of this school year studying number theory has helped him become comfortable with this type of problem.

I found his explanation of why you take half of a power when you take a square root to be really fascinating.

 

So, a nice start to our new chapter on square roots, and nice to see a little bit of the work we did in the Introduction to Number Theory book paying off.

A fun and challenging geometry problem from twitter

Saw this problem posted on Twitter earlier today (via a John Golden retweet) –

Not the easiest problem in the world, but since my son and I are studying a new section about circles in our Introduction to Geometry book, I thought I’d give it a try.

If you are interested in watching the thoughts of a kid as he struggles through a tough problem, today is your lucky day 🙂

First, an introduction to the problem and maybe 5 minutes of his initial thoughts. He’s walking towards the solution the whole way – slowly to be sure, but steadily.

 

So, I just turned the camera on and off to break the last video at approximately 5 minutes. In this second video he continues working towards the solution. Eventually he sees that the circle is the circumcircle of the triangle he’s drawn. That plus the area formula:

Area = A * B * C / 4R, where A, B, and C are the side lengths and R is the radius of the circumscribed circle gets him to the finish line.

Finding the approximate value of (5/2) * \pi confused him a little at the end, but he eventually was able to conclude that this expression was less than 8.

 

So, a fun exercise for me watching my son work through this problem, and a pretty challenging problem for him. Made for a good night. Thanks, as always, Twitter!