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That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angl

by vog 11y ago

That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles. However, I slightly disagree with the introduction of the proof: | The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. That's quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility. However, I don't see that. (Does anyone else see how this proof could work on such a generalized setting?) Instead, it seems that this proof introduces the concept of "divisibility" through the backdoor, by calculating with fractions and making use of the basic fraction laws, which are based on divisibility. In particular, the equation 1 / (sqrt(m) - n) = (sqrt(m) + n) / (m - n^2) makes use of the fact that you can extend and cancel fractions. Moreover, these fractions are not even fractions in Z, but in R, or at least in Z[sqrt(m)]. So there's still a lot of specialized structure involved in this proof. Nevertheless, great proof!

deleted 11y ago
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jfarmer 11y ago
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.
- vog 11y ago
  > 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.
  - jameshart 11y ago
    Indeed - I think the claim that alpha can be expressed as p/q with some 'lowest q' seems to rest on divisibility properties.
    - jfarmer 11y ago
      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 https://en.wikipedia.org/wiki/Presburger_arithmetic
  - jfarmer 11y ago
    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.
contravariant 11y ago
>> The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. >That's quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility. It doesn't, for the somewhat obvious reason that divisibility makes sense in any set that has a notion of 'multiplication'. The proof itself doesn't use any statements of the form 'x is divisible by y' though, not even implicitly as far as I can tell. Even fraction laws don't really count as 'using' divisibility. Since fractions can be defined in a way that doesn't directly use divisibility (although the concepts are obviously related). Being well ordered is a far more restrictive property than 'having a notion of divisibility'. > Moreover, these fractions are not even fractions in Z, but in R, or at least in Z[sqrt(m)]. So there's still a lot of specialized structure involved in this proof. According to the proof the fractions are in Q not R.
- vog 11y ago
  > According to the proof the fractions are in Q not R. That's a circular argument. During the proof this is not given, only after the proof.
  - contravariant 11y ago
    During the proof it is assumed, this assumption leads to a contradiction, hence the assumption 'sqrt(m) is in Q' is false. The reason this is not a circular argument is that they don't assume the thing they're trying to prove, but rather it's negation.
    - vog 11y ago
      You are right. My bad.

by vog 11y ago

deleted 11y ago

[deleted]

jfarmer 11y ago

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.

vog 11y ago
> 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.
- jameshart 11y ago
  Indeed - I think the claim that alpha can be expressed as p/q with some 'lowest q' seems to rest on divisibility properties.
  - jfarmer 11y ago
    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 https://en.wikipedia.org/wiki/Presburger_arithmetic
- jfarmer 11y ago
  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.

contravariant 11y ago

>> The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. >That's quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility. It doesn't, for the somewhat obvious reason that divisibility makes sense in any set that has a notion of 'multiplication'. The proof itself doesn't use any statements of the form 'x is divisible by y' though, not even implicitly as far as I can tell. Even fraction laws don't really count as 'using' divisibility. Since fractions can be defined in a way that doesn't directly use divisibility (although the concepts are obviously related). Being well ordered is a far more restrictive property than 'having a notion of divisibility'. > Moreover, these fractions are not even fractions in Z, but in R, or at least in Z[sqrt(m)]. So there's still a lot of specialized structure involved in this proof. According to the proof the fractions are in Q not R.

vog 11y ago
> According to the proof the fractions are in Q not R. That's a circular argument. During the proof this is not given, only after the proof.
- contravariant 11y ago
  During the proof it is assumed, this assumption leads to a contradiction, hence the assumption 'sqrt(m) is in Q' is false. The reason this is not a circular argument is that they don't assume the thing they're trying to prove, but rather it's negation.
  - vog 11y ago
    You are right. My bad.