Which proof is better?

I don’t know how many people here have any interest in math. I know BonusQuestion and Finkelitis did, but I have not seen either of them here for some time.

There is an historically important proof that the square root of two is irrational that was given by Euclid. It is the first published proof that a number is irrational, though the original discovery was made by one the followers of Pythagoras. It created a bit of controversy at the time since Pythagoras ran what might be considered a mathematical cult which had as its premise that everything in the world could be described in terms of ratios of integers.

Anyway, Euclid’s proof goes as follows:
Assume that sqrt(2) is rational and therefore there are integers p and q such that (p/q)^2 = 2. Further assume that p and q are chosen so they have no common divisor. We get p^2 = 2q^2. p^2 is therefore an even number and since odd times odd gives odd, it follows that p must be even. Then p=2r for some r. Substituting, we get (2r)^2 = 2q^2, or 4r^2 = 2q^2. Dividing both sides by 2 we get q^2 = 2r^2 and conclude that q must be even, but this contradicts the assumption that p and q have no divisor, so we can conclude that no p/q exists for sqrt(2) and that it is therefore irrational.

My problem with this proof is that it seems like a court case where the defendant gets out on a technicality. It doesn’t seem to involve the essence of irrational numbers.

Now consider this proof, which uses a technique that can be used in a lot of cases to show that a number is irrational.

It is easy to show that sqrt(2) – 1 < 0.5. Then (sqrt(2) – 1)^n < (.5)^n. But (sqrt(2) – 1)^n has the form a*sqrt(2) – b. Suppose sqrt(2) = p/q. Then (a*p + -b*q)/q < (.5)^n. But 1/q < (a*p + -b*q)/q, so 1/q < (.5)^n for all n. But this is impossible since (.5)^n can be made arbitrarily small. So therefore we get a contradiction and the proof follows as before.

This proof is longer, but it uses a more applicable technique that gets at the heart of the properties of rational numbers.

So what do you think?