Math Bite: Irrationality of √m (1999)

Re:Math Bite: Irrationality of √m (1999) by jfarmer on Wednesday December 23, 2015 @06:27AM (#10782889)

By "doesn't use divisibility" I assume the author means that the point of contradiction doesn't rest on the "divisibility properties" of the integer, not that division is never used. In particular, this proof doesn't rely on the fundamental theorem of arithmetic.

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Re:Math Bite: Irrationality of √m (1999) by vog on Wednesday December 23, 2015 @06:33AM (#10782899)

> I assume the author means that the point of contradiction doesn't rest on the "divisibility properties" of the integer

I see what the author meant by that. I just think it was stated in a slightly exaggerated way. The "divisibility properties" of the integers are still used. That part has just has been moved to another corner of the proof, by transforming fractional equations.

One of the comments in the article is quite interesting here:

| Theodor Estermann proved the irrationality of 2√ without relying on the prime factorization of m.

I believe that this statement is more correct. The "traditional" proof uses not just divisibility, but prime factorization, which is quite a strong property. And that is something the alternative proof doesn't make use of.

Maybe the introduction should have been stated that way.

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Re:Math Bite: Irrationality of √m (1999) by jfarmer on Wednesday December 23, 2015 @04:21PM (#10785746)

As near as I can tell, this proof would work in any well-ordered integral domain D where D's field of fractions would the role of the rationals. The analogue of the standard proof would require that D also be a unique factorization domain (or maybe the slightly weaker condition that any two elements have a GCD).

It might be the case that all these properties together "force" D to be a UFD or that the author snuck another property of the integers in there, but I've only taken a cursory look.

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Re:Math Bite: Irrationality of √m (1999) by jameshart on Wednesday December 23, 2015 @01:18PM (#10784704)

Indeed - I think the claim that alpha can be expressed as p/q with some 'lowest q' seems to rest on divisibility properties.

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Re:Math Bite: Irrationality of √m (1999) by jfarmer on Wednesday December 23, 2015 @04:03PM (#10785631)

No, it follows directly from the fact that the positive integers are well-ordered, i.e., any set of positive integers has a least element.

And in case one is tempted to think that well-ordering and divisibility are somehow equivalent, consider Presburger arithmetic[1]. It's not even possible to define a general notion of divisibility or primality in that context, but I'm almost positive the well-ordering principle holds (it's equivalent to the axiom schema of induction).

[1]: https://en.wikipedia.org/wiki/Presburger_arithmetic

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