Math Bite: Irrationality of √m (1999)
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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)
Nevertheless, great proof!
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.
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).
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.
>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.
That's a circular argument. During the proof this is not given, only after the proof.
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.
By the way, this particular note appears to be taken from "Biscuits of Number Theory". Compare with this scan from Google Books:
https://books.google.si/books?id=_g5TvMCJQB4C&pg=PA109&lpg=P...
Making the denominator as small as possible uses divisibility.
qt = sp
As such we may make q as small as possible by finding a representative in the equivalence class where the numerator is smallest (using the well ordering principle of the of the natural numbers). You can identify this pair uniquely (when the fraction is not 0) by using divisibility but that isn't required here.
I don't buy it. qt = sp looks too much like x = ny to me.
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).
It's finding the set of all possible q that uses divisibility, not finding the least member of that set.
There seems to be an implicit assumption that m is an integer, but the explicit assumptions only give the much weaker statement that "m is not a perfect square".
r is a shorthand for (m-n^2)q-2np, which is an integer because m,n,p,q are all integers.
> There seems to be an implicit assumption that m is an integer, but the explicit assumptions only give the much weaker statement that "m is not a perfect square".
Yes, it would have been more explicit to say "m is an integer that is not a perfect square."
On the other hand, this is pretty clear from the context. This is like looking at a computer program and saying "foo has no side-effect" without stating that "foo is a function".