A Conversation for Irrational Numbers

Improvement to the proof

Post 1

Jim Lynn

It struck me while reading that the proof presented here kind of gets lost. It's correct, but not stated clearly enough. That's because it's missing an important condition at the start - that, because only the ratio of the two numbers are important, you remove any common factors from p and q *at the start* so that p and q cannot both be even numbers (otherwise they share a common factor - 2). You then continue the proof until you get to the point at which you've proved that both p and q must be even numbers - at which point you've reached a contradiction because we already said that p and q can't both be even. Hence, reductio ad absurdum, root 2 must be irrational.

That way, the proof doesn't tail off into the misleading continual simplification, which seems much clearer to me.

Improvement to the proof

Post 2

Joe aka Arnia, Muse, Keeper, MathEd, Guru and Zen Cook (business is booming)

Oops... I really should have stated that. I had made the erroneous assumption that p and q were in their simplest form in the fraction. Any chance of the line being added that this is so?

Improvement to the proof

Post 3

Mark Moxon

Well spotted Jim - I've rewritten the proof to reflect that.

If there are any more problems then it's definitely my fault and I'll fix 'em straight away.

smiley - smiley

Improvement to the proof

Post 4


It's not really a fault; I've seen books give proof by infinite descent, but I do agree assuming the simplest form to start with looks nicer.

Improvement to the proof

Post 5

Joe aka Arnia, Muse, Keeper, MathEd, Guru and Zen Cook (business is booming)


Minimal criminal! smiley - winkeye

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