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fishfry
science forum Guru Wannabe

Joined: 29 Apr 2005
Posts: 299

Posted: Mon Jul 17, 2006 3:05 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

"mariano.suarezalvarez@gmail.com" <mariano.suarezalvarez@gmail.com>
wrote:

 Quote: D'Alambert used to tell his students ``Allez en avant, et la foi vous viendra''...

"Push on and faith will catch up with you."

Sounds like JSH's motto!
mariano.suarezalvarez@gma

Joined: 28 Apr 2006
Posts: 58

Posted: Mon Jul 17, 2006 2:38 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

J. B. Kennedy wrote:
 Quote: [snip] I like the suggestion that the basic problem is that there are no integers between zero and one (etc.), and that this is the root cause of irrationality. Irrationality is so important and so influential in mathematics, it is surprising to me that there is no clear consensus about its nature.

You seem to attach to irrationality an importance and an influence
that I nor none of the matematicians I've ever met and worked with
attaches to that notion. Moreover, I would be surprised if there
were anything at all regarding this notion which is not the subject
of a wide consensus.

I can but wonder from where you get these impressions.

D'Alambert used to tell his students ``Allez en avant, et la foi
vous viendra''...

-- m

(*) That is, something similar to ``go forwards, faith
will come to you of its own''
Gene Ward Smith
science forum Guru

Joined: 08 Jul 2005
Posts: 409

Posted: Mon Jul 17, 2006 12:57 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

J. B. Kennedy wrote:

 Quote: Most (or all?) proofs of irrationality or transcendence are proofs by contradiction, and these are, however formally acceptable, always a cheat because they depend upon further arguments to pick out the offending assumption. What I would like is a `positive proof.'

If you define a rational number as a number whose continued fraction
terminates and an irrational number as a number where it does not
terminate, you can prove some irrationality results directly--including
not only sqrt(2), but also eg e.
Gerry Myerson
science forum Guru

Joined: 28 Apr 2005
Posts: 871

Posted: Mon Jul 17, 2006 12:37 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

In article
<24726914.1152871938353.JavaMail.jakarta@nitrogen.mathforum.org>,
"J. B. Kennedy" <john.b.kennedy@manchester.ac.uk> wrote:

 Quote: I like the suggestion that the basic problem is that there are no integers between zero and one (etc.), and that this is the root cause of irrationality. Irrationality is so important and so influential in mathematics, it is surprising to me that there is no clear consensus about its nature. I was satisfied with thinking it was a manifestation of the FTAr, itself a deep idea, and thus provoked by the Well-Ordering proofs which seem partially or fully to avoid it. Most (or all?) proofs of irrationality or transcendence are proofs by contradiction, and these are, however formally acceptable, always a cheat because they depend upon further arguments to pick out the offending assumption. What I would like is a `positive proof.' Irrationality is a negative property and thus deriving it requires some negative assumption (say, with an odd number of negations). I would like a proof which starts with something like `there are no integers between zero and one, etc.' and leads to `there is no rational root of two' but is not a proof by contradiction. Without that, I think we are missing something. We cannot formally anatomize the interaction between the structure of the natural numbers and the nature of multiplication which produces irrationality. The Well-Ordering proofs look so simple, they suggest to me that such a positive proof must be lurking nearby. I'll work on it.

There are no integers between zero and one. Therefore, the numbers
n sqrt 2, n sqrt 2 (sqrt 2 - 1), n sqrt 2 (sqrt 2 - 1)^2,
n sqrt 2 (sqrt 2 - 1)^3, etc., can't all be integers. Therefore,
none of them can be an integer (since it's easy to show that if any
one is an integer, so is the next one). In particular, n sqrt 2 can't be
an integer. Thus, there is no rational square root of 2.

--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
J. B. Kennedy
science forum beginner

Joined: 10 Jul 2006
Posts: 9

 Posted: Fri Jul 14, 2006 10:11 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic Thanks for both of your illuminating posts. May I respond to both with this one message? I will take a look at the Burger and Tubbs and the Maths Mag article. They look promising. I like the suggestion that the basic problem is that there are no integers between zero and one (etc.), and that this is the root cause of irrationality. Irrationality is so important and so influential in mathematics, it is surprising to me that there is no clear consensus about its nature. I was satisfied with thinking it was a manifestation of the FTAr, itself a deep idea, and thus provoked by the Well-Ordering proofs which seem partially or fully to avoid it. Most (or all?) proofs of irrationality or transcendence are proofs by contradiction, and these are, however formally acceptable, always a cheat because they depend upon further arguments to pick out the offending assumption. What I would like is a `positive proof.' Irrationality is a negative property and thus deriving it requires some negative assumption (say, with an odd number of negations). I would like a proof which starts with something like `there are no integers between zero and one, etc.' and leads to `there is no rational root of two' but is not a proof by contradiction. Without that, I think we are missing something. We cannot formally anatomize the interaction between the structure of the natural numbers and the nature of multiplication which produces irrationality. The Well-Ordering proofs look so simple, they suggest to me that such a positive proof must be lurking nearby. I'll work on it. Thanks again for your refs and the thoughtful response. J. K.
Gerry Myerson
science forum Guru

Joined: 28 Apr 2005
Posts: 871

Posted: Thu Jul 13, 2006 1:18 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

In article <gerry-62AB1D.11154012072006@sunb.ocs.mq.edu.au>,
Gerry Myerson <gerry@maths.mq.edi.ai.i2u4email> wrote:

 Quote: In article , Bill Dubuque wrote: Gerry Myerson wrote: "J. B. Kennedy" wrote: But I then came across a simpler, seemingly more basic proof ascribed to Niven (which doesn't invoke the FTAr): If r is rational then there is a least integer b such that br is an integer but (br-b) is less than b (since r is btwn 1 and 2) and then (br-b)r also is an integer, which contradicts the assumption that b was the least and so r is irrational. According to Jorn Steuding, Diophantine Analysis, page 15, this idea for proving the irrationality of sqrt 2 is due to Estermann. A reference is Math Gazette 59 (1975), page 110. That proof is much, much older than 1975. I'm shocked any number theorist could believe it so new. Are you sure the reference is not to a different proof? The London Mathematical Society published a long obituary for Estermann. It's on the web, and one page of it, http://www.numbertheory.org/obituaries/LMS/estermann/page11.html is relevant here. It says, "In retirement, Estermann discovered [1975] an elegant proof of Pythagoras's theorem which is actually simpler than the original and is sufficiently short to be included verbatim. [note - from what follows, it's evident that they don't mean the theorem about the square of the hypotenuse, they mean the irrationality of the square root of two] They then give Estermann's proof, which is essentially the one ascribed to Niven up near the top of this post. I wouldn't be a bit surprised to learn that Niven (or someone) got there before Estermann, but I don't think anyone in this thread has cited a reference predating Estermann.

I've had a bit of a look for the proof.

In a thread on sci.math.research in September, 1998, Jim Propp
attributed the proof to Niven, without any references. Michael
Hardy claimed that the proof was actually older than the odd/even
and unique factorization proofs, but I think he was talking
about a geometric proof. The Well-Ordering proof may be nothing
more than the algebraic formulation of that geometric proof, but
I don't accept that the geometric proof *is* the well-ordering
proof.

You can find the Propp and Hardy posts under the Subject header,
The Book.

I didn't find the proof in Niven's book, Numbers: Rational and
Irrational, in the New Mathematical Library series, nor in his
book, Irrational Numbers, in the Carus Mathematical Monographs
series. I searched Math Reviews for anything that had both
Niven and irrational in it, and found nothing that looked like
a reference to this proof.

David Bloom published (a version of) the proof in Mathematics
Magazine (A one-sentence proof that sqrt2 is irrational, Math
Mag 68 (1995) 286). He writes that it is an algebraic version
of a geometric argument in a math history book by Eves, and it
was presented by Niven at a lecture in 1985.

In short, I still haven't found any reference to Niven or any-
one else preceding Estermann's paper in 1975.

--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
science forum Guru

Joined: 28 Apr 2005
Posts: 871

Posted: Thu Jul 13, 2006 12:50 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

In article <gerry-9B6BB4.10241712072006@sunb.ocs.mq.edu.au>,
Gerry Myerson <gerry@maths.mq.edi.ai.i2u4email> wrote:

 Quote: In article 14183393.1152611628375.JavaMail.jakarta@nitrogen.mathforum.org>, "J. B. Kennedy" wrote: Thanks for the lovely references and links. I like the Estermann proof because it also does not seem to rely on the FTAr, and therefore does seem simpler than the usual proofs. However, I do not think it is the same as the Niven proof. At least I cannot massage the algebra to make them look the same. The Niven proof uses the property that the square root of two is between one and two; the Estermann proof uses geometry to motivate an identity. Both proofs leave me in a quandry. I used to tell my students that the irrationality was a consequence of the FTAr, now I will just have to say it is a `fact' with no deeper conceptual diagnosis. You can still tell your students that the irrationality of the square root of two is a consequence of the Unique Factorization Theorem, because it is - the proof using UFT doesn't become invalid, just because there are proofs that don't use UFT. Moreover, the proof using UFT generalizes to a proof that if n and k are integers and the k-th root of n isn't an integer then it is irrational - it's a bit harder to get this out of the other proofs, nice as they are. You can even go farther; if a polynomial with integer coefficients is not satisfied by any rational with numerator a factor of the constant term and denominator a factor of the leading coefficient, then all of its roots are irrational. And you prove this using UFT.

Although I've worked out how to do a Well-Ordering Principle proof
of the polynomial result.

Theorem: Let f be a monic polynomial with integer coefficients.
Let r be a solution of f(r) = 0. Then if r is not an integer,
it is irrational.

Proof. Assume the hypotheses, assume r is rational, and let n
be the smallest positive integer such that n r^j is an integer
for all j less than the degree of f. Then n {r} (where {z}
means the fractional part of z) is a smaller positive integer
with the same property, contradiction, QED.

Justification of this proof is a good exercise for the reader,
and the reader who wants some good exercise should skip the rest
of this post & figure out the justification on her own.

First of all, if r is rational, say, r = a / b with a and b
integers, b > 0, then b^(d - 1), where d is the degree of f, has
the property that multiplication by r^j gives an integer for all
j less than the degree of f. Thus, the set of all such positive
integers is not empty. By Well-Ordering, there is a least such
positive integer, which we call n.

As r is not an integer, we have 0 < {r} < 1, so 0 < n {r} < n,
so n {r} is a smaller positive integer. We write {r} = r - k
for some integer k.

[Aside: we're using the mathematician's definition of fractional
part, where, e.g., {- 0.3} = 0.7]

Now n {r} r^j = n r^(j + 1) - k n r^j is clearly an integer so
long as j + 1 is less than the degree of f. But the equation
f(r) = 0 tells us how to write r^d as a sum of integer multiples
of smaller powers of r, so even in the case j = d - 1 we get
that n {r} r^j is an integer.

This completes the justification of the proof.

--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
science forum Guru

Joined: 28 Apr 2005
Posts: 871

Posted: Thu Jul 13, 2006 12:22 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

In article
<20434773.1152690104218.JavaMail.jakarta@nitrogen.mathforum.org>,
"J. B. Kennedy" <john.b.kennedy@manchester.ac.uk> wrote:

 Quote: Yes, I now agree that the proof I thought was Niven's was actually Estermann's. Thanks for the direct link to that last page of the obituary. Does this show that the FTAr is not a necessary ingredient of proofs of irrationality? If so, what conceptual diagnosis of irrationality can we offer our students? JK

Did you see the part where I wrote
that you can use the Unique Factorization Theorem
to prove that the base-10 logarithm of 2
is irrational?

There may be a way to prove that without the UFT,
but I've never seen it done.

I'm not entirely sure what you mean by
a "conceptual diagnosis of irrationality,"
On the first page of Chapter 1 of Burger & Tubbs,
Making Transcendence Transparent,
it says,

The Fundamental Principle of Number Theory: There are no integers
between zero and one.

Time after time, proofs of irrationality and/or transcendence
come down to this simple principle.
Is that a good enough "conceptual diagnosis of irrationality"
for you?

What do you want?

--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
mariano.suarezalvarez@gma

Joined: 28 Apr 2006
Posts: 58

Posted: Wed Jul 12, 2006 3:33 pm    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

Rob Johnson wrote:
 Quote: In article <1152682261.789920.187840@s13g2000cwa.googlegroups.com>, "mariano.suarezalvarez@gmail.com" wrote: Rob Johnson wrote: In article <2825276.1152533007084.JavaMail.jakarta@nitrogen.mathforum.org>, "J. B. Kennedy" wrote: Since the usual reductio proof of the irrationality of the square root of two is conceptually opaque, I thought that the sophisticate's explanation was that it was due to the FTAr, since If r is the square root of two and p=r q (i.e., suppose its rational), then p squared = 2 q squared but then the prime factors on both sides won't match up (because the LHS is all primes squared and 2 is not) which contradicts the FtAr and so r is irrational. But I then came across a simpler, seemingly more basic proof ascribed to Niven (which doesn't invoke the FTAr): If r is rational then there is a least integer b such that br is an integer but (br-b) is less than b (since r is btwn 1 and 2) and then (br-b)r also is an integer, which contradicts the assumption that b was the least and so r is irrational. There are many proofs of the irrationality of various constants, but can we offer our students any simple conceptual answer to the question of why the square root of two is irrational beyond exhibiting the formal proofs? Another proof, using Bezout's formula, can be found at http://www.whim.org/nebula/math/ratalint.html>. There it is shown that any algebraic integer which is also a rational number must be an integer. Since the square root of 2 is an algebraic integer, but not an integer, it cannot be rational. The argument in that link is slightly more complicated than necessary: Assume p/q, with q>0 and gcd(p, q) = 1, is a root of f(x) = x^n + sum_{i=0}^{n-1} a_i x^i with the a_i integral. Then 0 = q^n f(p/q) = p^n + sum_{i=0}^{n-1} a_i q^{n-i} p^i Of course the sum is divisible by q so p^n is divisible by q. This, together with gcd(p,q)=1 implies that q = 1, so that p/q is an integer. It is definitely true that if gcd(p,q) = 1 and p^n is divisible by q, then q = 1. The shorter proof above cites this fact, but leaves it unproven. This fact can pretty quickly be proven by looking at the primes dividing q and using the Fundamental Theorem of Arithmetic. Another way of proving this fact is using Bezout's formula. My proof linked above also avoids using the Fundamental Theorem of Arithmetic by using Bezout's formula.

Of course you need to prove that. But prove it as a separate
lemma, at the point where it is needed: it becomes much more
clear than using your substitution p/q -> (1-bq)/aq, which IMO
obscures what is happening.

-- m
mariano.suarezalvarez@gma

Joined: 28 Apr 2006
Posts: 58

Posted: Wed Jul 12, 2006 3:23 pm    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

J. B. Kennedy wrote:
 Quote: Thanks. Your proof is the tradtional one that does not seem to connect the result to any deep mathematical principle, and is therefore conceptually opaque. I claim it does assume more than that an even number is a multiple of two. It assumes that one number (q^2), which is equal to another which is a multiple of two, is also a multiple of two. This slightly different idea is an expression of the FTAr.

Hm? The definition of `multiple' I was given in elementary
school was: a number n is a multiple of k if there is
a number m such that n = km. Your lemma that a number
which is equal to a another which is a multiple of 2 is itself
even follows from the transitivity of equality.

I'm quite sure you can prove that sqrt(2) is not rational
using the theory of quadatic reciprocity or even class field
theory. Maybe Brauer group theory can be used to. You can
probably pick a prime p with respect to which p-adic
analysis can frame a cool proof. You can certainly use
Dirichlet's principle that uses aproximability by rational
numbers to characterize irrationality. Or the theory of
values of algebraic functions.

Using tools that are unrelated to and overengineered
with respect to the fact you want to prove is what I call
opaque.

-- m
Rob Johnson
science forum Guru

Joined: 26 May 2005
Posts: 318

Posted: Wed Jul 12, 2006 11:54 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

"mariano.suarezalvarez@gmail.com" <mariano.suarezalvarez@gmail.com> wrote:
 Quote: Rob Johnson wrote: In article <2825276.1152533007084.JavaMail.jakarta@nitrogen.mathforum.org>, "J. B. Kennedy" wrote: Since the usual reductio proof of the irrationality of the square root of two is conceptually opaque, I thought that the sophisticate's explanation was that it was due to the FTAr, since If r is the square root of two and p=r q (i.e., suppose its rational), then p squared = 2 q squared but then the prime factors on both sides won't match up (because the LHS is all primes squared and 2 is not) which contradicts the FtAr and so r is irrational. But I then came across a simpler, seemingly more basic proof ascribed to Niven (which doesn't invoke the FTAr): If r is rational then there is a least integer b such that br is an integer but (br-b) is less than b (since r is btwn 1 and 2) and then (br-b)r also is an integer, which contradicts the assumption that b was the least and so r is irrational. There are many proofs of the irrationality of various constants, but can we offer our students any simple conceptual answer to the question of why the square root of two is irrational beyond exhibiting the formal proofs? Another proof, using Bezout's formula, can be found at http://www.whim.org/nebula/math/ratalint.html>. There it is shown that any algebraic integer which is also a rational number must be an integer. Since the square root of 2 is an algebraic integer, but not an integer, it cannot be rational. The argument in that link is slightly more complicated than necessary: Assume p/q, with q>0 and gcd(p, q) = 1, is a root of f(x) = x^n + sum_{i=0}^{n-1} a_i x^i with the a_i integral. Then 0 = q^n f(p/q) = p^n + sum_{i=0}^{n-1} a_i q^{n-i} p^i Of course the sum is divisible by q so p^n is divisible by q. This, together with gcd(p,q)=1 implies that q = 1, so that p/q is an integer.

It is definitely true that if gcd(p,q) = 1 and p^n is divisible by q,
then q = 1. The shorter proof above cites this fact, but leaves it
unproven. This fact can pretty quickly be proven by looking at the
primes dividing q and using the Fundamental Theorem of Arithmetic.
Another way of proving this fact is using Bezout's formula. My proof
linked above also avoids using the Fundamental Theorem of Arithmetic
by using Bezout's formula.

Rob Johnson <rob@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 12, 2006 8:05 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

In article
<20434773.1152690104218.JavaMail.jakarta@nitrogen.mathforum.org>,
"J. B. Kennedy" <john.b.kennedy@manchester.ac.uk> wrote:

 Quote: Yes, I now agree that the proof I thought was Niven's was actually Estermann's. Thanks for the direct link to that last page of the obituary. Does this show that the FTAr is not a necessary ingredient of proofs of irrationality? If so, what conceptual diagnosis of irrationality can we offer our students? JK

We have just seen that the set of rationals whose squares are less than
2 does not have a rational LUB, but the set of reals must contain a LUB
(least upper bound) for that set.

So the completeness (LUB and GLB properties) of the field of reals
requires the presence of irrationals as well as rationals among the
reals.
J. B. Kennedy
science forum beginner

Joined: 10 Jul 2006
Posts: 9

 Posted: Wed Jul 12, 2006 7:41 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic Yes, I now agree that the proof I thought was Niven's was actually Estermann's. Thanks for the direct link to that last page of the obituary. Does this show that the FTAr is not a necessary ingredient of proofs of irrationality? If so, what conceptual diagnosis of irrationality can we offer our students? JK
J. B. Kennedy
science forum beginner

Joined: 10 Jul 2006
Posts: 9

 Posted: Wed Jul 12, 2006 6:55 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic Thanks. Your proof is the tradtional one that does not seem to connect the result to any deep mathematical principle, and is therefore conceptually opaque. I claim it does assume more than that an even number is a multiple of two. It assumes that one number (q^2), which is equal to another which is a multiple of two, is also a multiple of two. This slightly different idea is an expression of the FTAr.
mariano.suarezalvarez@gma

Joined: 28 Apr 2006
Posts: 58

Posted: Wed Jul 12, 2006 5:31 am    Post subject: Re: Irrationality and the Fundamental Theorem of Arithmetic

Rob Johnson wrote:
 Quote: In article <2825276.1152533007084.JavaMail.jakarta@nitrogen.mathforum.org>, "J. B. Kennedy" wrote: Since the usual reductio proof of the irrationality of the square root of two is conceptually opaque, I thought that the sophisticate's explanation was that it was due to the FTAr, since If r is the square root of two and p=r q (i.e., suppose its rational), then p squared = 2 q squared but then the prime factors on both sides won't match up (because the LHS is all primes squared and 2 is not) which contradicts the FtAr and so r is irrational. But I then came across a simpler, seemingly more basic proof ascribed to Niven (which doesn't invoke the FTAr): If r is rational then there is a least integer b such that br is an integer but (br-b) is less than b (since r is btwn 1 and 2) and then (br-b)r also is an integer, which contradicts the assumption that b was the least and so r is irrational. There are many proofs of the irrationality of various constants, but can we offer our students any simple conceptual answer to the question of why the square root of two is irrational beyond exhibiting the formal proofs? Another proof, using Bezout's formula, can be found at http://www.whim.org/nebula/math/ratalint.html>. There it is shown that any algebraic integer which is also a rational number must be an integer. Since the square root of 2 is an algebraic integer, but not an integer, it cannot be rational.

The argument in that link is slightly more complicated
than necessary:

Assume p/q, with q>0 and gcd(p, q) = 1, is a root of

f(x) = x^n + sum_{i=0}^{n-1} a_i x^i

with the a_i integral. Then

0 = q^n f(p/q) = p^n + sum_{i=0}^{n-1} a_i q^{n-i} p^i

Of course the sum is divisible by q so p^n is divisible
by q. This, together with gcd(p,q)=1 implies that q = 1, so that
p/q is an integer.

-- m

PS: This generalizes to the well-known lemma that says that
a rational root to a polynomial with integer coefficients has a
denominator which divides the leading coefficient, and a
numerator that divides the constant term.

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