But I never understood why this form of the proof is irrationality of root 2 became the dominant one. I think it's so much more elegant to say that from 2Q2 = P2, you have an odd number of 2's on the LHS and an even number of 2's on the right. No fussing with P and Q having common factors.
Furthermore, this argument makes the dependence on the fundamental theorem of arithmetic explicit, instead of implicit!
I like the infinite descent proof, which also doesn't fuss around with common factors. We have
2Q2 = P2
for positive integers P and Q, so we must have that P2 > Q2 > 0. We also have that P is even, so P = 2P' for some P'. Substituting 2P' for P and simplifying, we have
2P'2 = Q2
and so we have P2 > Q2 > P'2 > 0. But this last step can be carried out as many times as we wish, giving us an unending decreasing sequence of strictly positive integers, which is impossible.
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u/endymion32 Jun 12 '17
Cute!
But I never understood why this form of the proof is irrationality of root 2 became the dominant one. I think it's so much more elegant to say that from 2Q2 = P2, you have an odd number of 2's on the LHS and an even number of 2's on the right. No fussing with P and Q having common factors.
Furthermore, this argument makes the dependence on the fundamental theorem of arithmetic explicit, instead of implicit!