
Math Bite: Irrationality of √m (1999) - luisb
http://fermatslibrary.com/s/irrationality-of-square-root-of-m
======
vog
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!

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

~~~
jfarmer
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)

------
2sk21
Discovering wonderful sites like this is why I read Hacker News!

~~~
avian
Yes, such wonderful sites that must put annoying bouncing "Click here to see
more" and "Join our newsletter" things around everything.

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...](https://books.google.si/books?id=_g5TvMCJQB4C&pg=PA109&lpg=PA109&dq=harley+flanders+irrationality&source=bl&ots=ShlKfjMAQg&sig=jBwurdoANm2e5cCgo-
jn0Fa0DGg&hl=sl&sa=X&ved=0ahUKEwiq2Nys5PHJAhXFFSwKHWr2C-YQ6AEIJjAA#v=onepage&q=harley%20flanders%20irrationality&f=false)

------
brlewis
> We may assume that q is _as small as possible_ (Estermann's key idea)

Making the denominator as small as possible uses divisibility.

~~~
jfarmer
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)

~~~
brlewis
alpha = p/q where p and q are positive integers.

It's finding the set of all possible q that uses divisibility, not finding the
least member of that set.

------
jamesfisher
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".

~~~
vog
_> why must r be an integer?_

 _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".

------
JerTheRipper
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?

~~~
tribe
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.

