Irrationality of the Square Root of 2

During the week I saw this neat post on Twitter from Evelyn Lamb:

The 2nd link in the tweet is to an amazing list of 27 different proofs of the irrationality of the square root of 2 on Cut the Knot’s website.

I thought it would be fun to review of few of these proofs with the boys this weekend.  Though obviously some of the details will be over their heads, I think it is still interesting to share ideas like these with younger kids every now and then.  It helps communicate some of the beauty of math and also shows that problems can have many different and surprising solutions.

We started with the “standard” proof.  In this proof we assume that \sqrt{2} can be written in the form A / B where A and B are integers and the fraction is in lowest terms, and then arrive at the contradiction that both A and B have to be even.

Both boys have some familiarity with this proof since we talked about it before (probably more than once, actually). Before jumping into the review of this proof, though, we talked about what the boys knew about \sqrt{2} and what they knew about irrational numbers.

The next proof is similar to proofs 3 and 3′ on Cut the Knot’s page. This proof uses a little bit of number theory and a little bit of combinatorics. Since my younger son is studying basic number theory right now (from this book: http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt&%20) and since we studied some basic combinatorics over the summer, I thought this proof would be fun for them to see.

The idea in this proof is once again to assume that \sqrt{2} is equal to A / B where A and B are integers.  This time we count the number of factors in A^2 and B^2 to arrive at a contradiction:

[Post publication note – a former student of mine, Andrew Gacek, has point out that the video below has an error.  I’ve corrected the argument in the video below this one, but will leave the original video – see if you can spot the error, too! ]

Here’s the correction:

Next up is a neat geometric proof attributed to Tom Apostol and is proof #7 on the Cut the Knot list (though the comments on the Cut the Knot page indicate that this proof was known prior to Apostol talking about it).  This proof has a neat geometric idea – if \sqrt{2} is rational, say A / B,  we can find a right triangle with legs of length A and hypotenuse of length B.  Among all possible such triangles there will be a smallest one  (this statement is essentially the same as the idea that we can write the fraction A / B in lowest terms).   Given this smallest triangle, we show that there’s actually a smaller one which contradicts the idea that the first one was the smallest.  Thus, the idea that \sqrt{2} is irrational!

Finally, proof #21 on the Cut the Knot list – using continued fractions to show that \sqrt{2} is irrational.  I’ve loved continued fractions since learning about them for the first time in high school.  My high school math teacher was also really fond of them and I’ll always remember him standing in front of our class telling us to “split, flip, and rat”!

The idea in this proof is that the continued fraction for \sqrt{2} goes on forever, but continued fractions for rational numbers have only finitely many terms.  The other neat thing in this proof is that we get to see a surprisingly simple representation of \sqrt{2} which is a nice contrast to the idea that irrational numbers are simply infinite, non-repeating decimals.  I love the part in this video where my youngest son suddenly recognizes the pattern!

So, though some of the details in these proofs are a little difficult for kids to follow completely, it was still a lot of fun to go through this exercise this morning.  We got a little practice with logical reasoning, basic number theory, basic geometry, and even a little bit of work in with fractions!  We also saw some nice examples showing that there are many different ways to look at a problem, even a problem involving only a single number.  Definitely a fun morning.

4 thoughts on “Irrationality of the Square Root of 2

    1. yes, but I wasn’t surprised to see it take a few extra step for the boys.

      The same argument works for Sqrt( n^2 + 1). The continued fraction will be n + 1 / 2n + 1 / 2n + 1 / 2n (hopefully that formatting makes sense).

      The Apostol geometric proof also extends for Sqrt(n^2 + 1) also.

  1. I don’t think your argument in the second proof is correct. When you are counting the factors of 2B^2 you take the factors of B^2 and say that you can multiply them by 2^0 or by 2^1 and then you add up the number of elements in those two sets. But you don’t know that the two sets are disjoint, and in fact when B is even they are not disjoint. Trying to rid 2 from B probably leaves you doing proof #1 instead. I think this is why the Cut the Knot version focuses on prime factors instead.

    1. Oh, dang, you are absolutely right.

      The argument I should have used is that the power of 2 in B^2 is even, adding the extra 2 makes the power of 2 in (2B^2) odd, and that means you have an even number of factors. Will add a correction

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