Conway's "calculus" proof of the irrationality of the square root of 2 (and more).
(www.youtube.com)https://www.youtube.com/watch?v=wNOtOPjaLZsThis is a nicely paced 13 minute presentation. It goes slowly and carefully over root two, before going faster over root N, then finishing off rapidly with monic polynomials. Suitable for a variety of audiences. But I still cannot figure out how it avoids fancy properties of the integers. I expect that you must at least use Euclid's Algorithm, but it doesn't. Here is how I prove that square roots are either integers or irrationals:
Suppose that √n∈ℚ. Write √n = a/b where a and b are coprime. Then b√n=a. Also, by Bezout's Theorem there are r,s∈ℤ such that ra+sb=1. Multiply by √n
ra√n + sb√n = √n
We already have b√n=a so sb√n is an integer. Does anything similar happen for ra√n ? Yes, just multiply b√n=a by √n obtaining bn=a√n, from which we deduce that ra√n must also be an integer. Thus √n is the sum of two integers and itself an integer.
That is awkward; a rational square root always turns out to be a whole number! Since 1<2<4 we know that 1<√2<2. But there is no integer between one and two so √2 is irrational.
But the approach in the video has neither Euclid's Algorithm, nor Bezout's Theorem (which I view as a corrollary of Euclid's Algorithm).
