>> 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 quit

>> 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.