## A Conversation for Irrational Numbers

### Improvement to the proof

Jim Lynn Started conversation Nov 21, 2000

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

Joe aka Arnia, Muse, Keeper, MathEd, Guru and Zen Cook (business is booming) Posted Nov 21, 2000

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

Mark Moxon Posted Nov 22, 2000

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.

### Improvement to the proof

Bagpuss Posted Nov 22, 2000

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

Joe aka Arnia, Muse, Keeper, MathEd, Guru and Zen Cook (business is booming) Posted Nov 22, 2000

Key: Complain about this post

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