Comments on: It’s a Threeven Day!

By: blag

blag — Tue, 12 Apr 2011 19:07:41 +0000

throdd.

By: Kate Nowak

Kate Nowak — Thu, 24 Mar 2011 21:18:43 +0000

I always thought a number was “even” because if you have that many objects you can line them up in two rows and they line up evenly. But if you don’t have an even number, there’s an odd one out. But I might have just made that up somewhere along the way.

But then you could say “threeven” is for lining them up in three rows. Anyway. What a good word.

By: Ξ

Mon, 07 Mar 2011 16:43:06 +0000

LOL Yaacov! And yes, you’re right about the pronunciation, especially since it leads to a punchline.

Adam, I don’t think I’ve seen that proof before, but I really like how it makes it almost transparent that square roots of integers are either integers or irrational. (Hmmm; does it generalize to roots of any order? I think it would.) Despite this I wouldn’t have thought automatically that it was an easier proof for students to understand, so I’m glad for your experience on it. [I think I'm done with this proof for the non-majors, but I'll be teaching it again next year both to majors and non-majors and might use this version instead.]

By: Yaacov

Yaacov — Sat, 05 Mar 2011 14:29:44 +0000

It should really be sixven and sevenven, and so on. Easier to say and easier to understand. And of course, expressions divisible by x would be X-Ven: The Last Stand.

By: Adam Glesser

Adam Glesser — Fri, 04 Mar 2011 15:27:55 +0000

I’ve had the same problem when teaching this. Recently, I started teaching a slightly different proof of the irrationality of 2 than I was taught and it seems to make a difference. I start in the usual way, assuming that sqrt{2} = a/b for relatively prime positive integers a and b. Squaring gives 2 = a^2/b^2. Here is where the change comes in: multiply through by b, leaving a single b in the denominator on the right. This gives 2b = a^2/b. This implies that b divides a^2—since the left side is an integer—which is only possible if b = 1 (as a and b are relatively prime). Therefore, 2 = a^2, which is clearly impossible since 2 is not a square. I find that students generalize this proof much more easily—even to higher roots—and they can see immediately why it doesn’t work for 4.
If you work just a little bit more with it—which I don’t think you would want to do with liberal arts majors—you can prove easily that any non-integer (real) root of a monic polynomial with integer coefficients is irrational.

By: Ξ

Fri, 04 Mar 2011 11:57:48 +0000

We think alike! With this class I was pleased that three people understood enough to laugh when I suggested it, and I only briefly outlined the proof and where it went wrong [there's a huge disparity in mathematical comfort in this class, probably larger than in any I've ever taught].

But when I’ve done that in our Intro to Proofs class for math majors, typically as HW right after the proof about 3, I’d say that most of them immediately realize that the question doesn’t make sense but only a few of them can find the spot where the proof fails.

By: Joshua Zucker

Joshua Zucker — Fri, 04 Mar 2011 05:56:45 +0000

To test whether they really understand it, after they have proved that the square root of 2 is irrational and the square root of 3 is irrational, ask them to use the same technique to prove that the square root of 4 is irrational.

What percentage of the students will realize that it’s not going to work? How many will be able to point at the specific step that works for 2 and 3 but not 4?