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.
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.
> 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.
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.
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).
>> 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.
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).
Not necessarily. Strictly mathematically a (non-negative) fraction q/p is just the equivalence class of all non-negative integer pairs (s, t), t != 0, such that
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.
By the same reasoning, divisibility does not use divisibility. Instead of asking if x is divisible by y you create a set of all products of the form ny where n is an integer, then check to see if x is a member of that set.
I don't buy it. qt = sp looks too much like x = ny to me.
I don't understand. The end of the proof says r/p is a fraction, presumably because r must be an integer. But why must r be an integer?
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".
I'm trying to work this out in my head, but I believe that this holds if m is a rational such that its numerator and denominator are not both perfect squares (e.g., √(2/3) is irrational but √(4/9) is rational). Am I wrong?
When this proof has been presented to me in the past, I believe we assume that m is a positive integer. This is mentioned in one of the comments, but I agree that it should be more clear in statement of the theorem.
It's pixelated because the paper is a scan of a 17 year old paper. The Fermat's Library site is designed to allow people to annotate and comment on old classic papers, most of them from before digitisation of journals.
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
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!