Irrationality of sqrt (n^2 - 1)
qsymmetry
sqrt(n^2 - 1)
is irrational?
Bart Goddard
news:10194448.109348.1242424...@nitrogen.mathforum.org:
> If n is an integer and |n| > 1, is it true that
>
> sqrt(n^2 - 1)
>
>
> is irrational?
>
Yes.
B.
--
Cheerfully resisting change since 1959.
Roman B. Binder
for n=1 or n=-1:
n^2 -1 = 0 then sqrt(0) is 0 and rational;
else:
n^2 -1 = (n-1)(n+1) and not 0
Once let n-1 = m^2 so n+1 = m^2 +2
suppose n+1 is also square = a^2
then a^2 - m^2 = 2
(a-m)(a+m) = 2 possible only if:
a+m = 2 ; a-m = 1 :
giving 2a = 3 ; 2m = 1 :
also inconsistency for n integers
Again suppose n-1 = km^2 where k integer
then n+1 = ka^2
What is possible for n-1 and n+1 not of gcd=1
and especially for n odd number:
but again ka^2 - km^2 = 2
k(a^2 - m^2) = 2
will be inconsistent for n integers as above
from k=1 : a^2 - m^2 = 2 for integer a,b .
Does someone has another opinion ?
Regards
Ro-Bin
Arturo Magidin
Yes. Any integer that is not a perfect square has an irrational square
root; that is, the only rationals that have a square that is an
integer are the integers. Follows from unique factorization.
Since for |n|>1, n^2 - (n-1)^2 > 1, it follows that n^2 - 1 cannot be
a perfect square unless n=1 or -1.
--
Arturo Magidin
KMF
can easily generalize the proof that square root of
2 is irrational to show that sqr(n) is irrational
iff. n is a perfect square.
Arturo Magidin
That would be quite a feat. You mean, surely, that sqrt(n) is
*rational* if and only if n is a perfect square.
And I believe that is *exactly* what I said, and that when I mentioned
unique factorization, it was clear that this meant you were
generalizing the proof that sqrt(2) is irrational.
--
Arturo Magidin
Roman B. Binder
I can see factorization:
(n-n+1)(n+n-1) = 2n-1 > 1
but here just more clear is to write:
n^2 - 1 = (n-1)^2 the next smallest perfect square
what requires just 2n-1 = 1
what is possible only for n=1 .
similar is to calculate n^2 + 1 also not some
perfect square:
n^2 + 1 = (n+1)^2 the next smallest perfect square
(n+1)^2 - n^2 = (2n+1) = 1
what is not true for any n integer
Following this results only for certain natural k :
n^2 + k or n^2 - k will be some next perfect square:
For n^2 - k once 2n-1 = k and k > 1 once n > 1
so the smallest n = 2 ; k = 3 ; 2^2 - 1 = 1 ; etc.
For n^2 + k once 2n+1 = k so for smallest n=1
k = 3 ; 1^2 + 3 = 2^2 ; etc.
Now with growing number k we could find also not only
the first next perfect square but also some second
and so on in order perfect square: lets sign it with
some natural s:
Now for s-next in order smallest perfect squares:
n^2 - k = (n-s)^2
s(2n-s) = k
Or for s-next in order biggest perfect squares:
n^2 + k = (n+s)^2
s(2n+s) = k
etc. and etc. it will be possible to find
some generalization and identity allowing to
such number k to be difference of some perfect squares...
Regards
Ro-Bin
Bill Dubuque
>On May 15, 4:51�pm, qsymmetry <qsymme...@email.com> wrote:
>>
>
> Yes. Any integer that is not a perfect square has an irrational square
> root; that is, the only rationals that have a square that is an
> integer are the integers. Follows from unique factorization.
GCD's suffice for the proof (vs. the higher powered unique factorization).
Recall EUCLID'S LEMMA: (a,b) = 1 & a|bc => a|c. In particular
the special case c = b: (a,b) = 1 & a|bb => a|b.
Therefore n = bb/aa with (a,b) = 1 => a|bb => a|b, i.e. a/b in Z
i.e. sqrt(n) = b/a => b/a is an integer.
This thread appears to be a fork of the following Ask an Algebraist thread
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1601
--Bill Dubuque
Arturo Magidin
wrote:
> On May 15, 2009 8:06 PM ArturoMagidinwrote:
>
>
>
> > On May 15, 4:51 pm, qsymmetry <qsymme...@email.com>
> > wrote:
> > > If n is an integer and |n| > 1, is it true that
>
> > > sqrt(n^2 - 1)
>
> > > is irrational?
>
> > Yes. Any integer that is not a perfect square has an
> > irrational square
> > root; that is, the only rationals that have a square
> > that is an
> > integer are the integers. Follows from unique
> > factorization.
>
> > Since for |n|>1, n^2 - (n-1)^2 > 1, it follows that
> > n^2 - 1 cannot be
> > a perfect square unless n=1 or -1.
>
> > --
> > ArturoMagidin
>
> I can see factorization:
> (n-n+1)(n+n-1) = 2n-1 > 1
This is the trivial factorization, since n-n+1 = 1 and n+n-1 = 2n-1.
> but here just more clear is to write:
> n^2 - 1 = (n-1)^2
This is false for any n except n=0.
As usual, your usage of "more clear" seems to be "completely confusing
and unintelligible except perhaps to someone as confused as Roman B.
Binder." Kindly post your ramblings not in reply to me.
--
Arturo Magidin
Arturo Magidin
> ArturoMagidin<magi...@member.ams.org> wrote:
> >On May 15, 4:51 pm, qsymmetry <qsymme...@email.com> wrote:
>
> >> Is sqrt(n^2 - 1) irrational for integer |n| > 1 ?
>
> > Yes. Any integer that is not a perfect square has an irrational square
> > root; that is, the only rationals that have a square that is an
> > integer are the integers. Follows from unique factorization.
>
> GCD's suffice for the proof (vs. the higher powered unique factorization).
>
> Recall EUCLID'S LEMMA: (a,b) = 1 & a|bc => a|c. In particular
>
> the special case c = b: (a,b) = 1 & a|bb => a|b.
>
> Therefore n = bb/aa with (a,b) = 1 => a|bb => a|b, i.e. a/b in Z
You mean "b/a in Z", of course.
You are taking for granted that any rational may be put in "lowest
terms". How much of unique factorization is hidden in that assumption?
--
Arturo Magidin
Bill Dubuque
>
> For integer |n| > 1, is it true that sqrt(n^2 - 1) is irrational?
Here's a cute little-known proof that I discovered as a teenager:
THEOREM w = sqrt(n) is integral if rational (for integral n).
PROOF w = a/b, (a,b)=1 => ad-bc = 1 for some c,d via Bezout,
therefore 0 = (a-bw) (c+dw) = ac-bdn + w => w is an integer. QED
See also the parallel thread on Ask an Algebraist
Bill Dubuque
>On May 16, 2:11�pm, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>ArturoMagidin<magi...@member.ams.org> wrote:
>>>On May 15, 4:51�pm, qsymmetry <qsymme...@email.com> wrote:
>>>>
>>>> Is sqrt(n^2 - 1) irrational for integer |n| > 1 ?
>>>
>>> Yes. Any integer that is not a perfect square has an irrational square
>>> root; that is, the only rationals that have a square that is an
>>> integer are the integers. Follows from unique factorization.
>>
>>GCD's suffice for the proof (vs. the higher powered unique factorization).
>>
>>Recall EUCLID'S LEMMA: (a,b) = 1 & a|bc => a|c. In particular
>>
>>the special case c = b: (a,b) = 1 & a|bb => a|b.
>>
>>
>>i.e. sqrt(n) = b/a => b/a is an integer.
>
> terms". How much of unique factorization is hidden in that assumption?
I assume only that GCD's exist - which is only one half of
unique factorization. Nowhere do I assume that non0 nonunits
factor into atoms (the other half of unique factorization).
There are plenty of GCD domains not UFDs, i.e. that fail to
satisfy said finiteness assumption, e.g. the ring of all
algebraic integers, which has no atoms since x = /x /x.
--Bill Dubuque
KMF
> > As a quick followup/addition to Arturo's post, you
> > can easily generalize the proof that square root
> of
> > 2 is irrational to show that sqr(n) is irrational
> > iff. n is a perfect square.
>
> That would be quite a feat. You mean, surely, that
> sqrt(n) is
> *rational* if and only if n is a perfect square.
Yes, that's what I meant.
> And I believe that is *exactly* what I said, and that
> when I mentioned unique factorization, it was clear that this meant you were generalizing the proof that sqrt(2) is irrational.
>
I don't know it by that name/principle, nor the
connection between the two; surely you can conceive that
a given principle/thm/result can have different names.
I am neither an algebraist nor number-theorist, so I
don't know things as formally as you do in this area.
Anyway, if you prefer, I'll eliminate any relation
between your post and mine, and change my original
post. New post (to the OP):
Try generalizing the result that sqr2 is irrational
to sqr(n) , for any n not a perfect square.
And please ignore if it is redundant.
> --
> Arturo Magidin
Roman B. Binder
perfect square of the next smallest square as (n-1)^2 .
This is the fall for n integers. For n rationals
really it will be different and it was not this topic question.
>
> This is false for any n except n=0.
Then really for integers n<0 it should be written also:
n^2 -1 = (n+1)^2
What will be true only for n=-1
Also I am very sorry once overlooking here and so on
down the true cases for integers n<0 .
>
> As usual, your usage of "more clear" seems to be
> "completely confusing
> and unintelligible except perhaps to someone as
> confused as Roman B.
> Binder." Kindly post your ramblings not in reply to
> me.
>
> --
> Arturo Magidin
Any next time.
Now I can see the case of sqrt(n^2 -1) for n integers
could be solved with some elementary procedure but it
should be taken special care for integers n<0 .
I used to overlook this and I insist my faulty
here: "more clear"
Once more time sorry !
Thank You for recalling my name:
Roman B. Binder
Ro-Bin
Arturo Magidin
> > On May 15, 8:47 pm, KMF <tee...@hotmail.com> wrote:
> > > As a quick followup/addition to Arturo's post, you
> > > can easily generalize the proof that square root
> > of
> > > 2 is irrational to show that sqr(n) is irrational
> > > iff. n is a perfect square.
>
> > That would be quite a feat. You mean, surely, that
> > sqrt(n) is
> > *rational* if and only if n is a perfect square.
>
> Yes, that's what I meant.
>
> > And I believe that is *exactly* what I said, and that
> > when I mentioned unique factorization, it was clear that this meant you were generalizing the proof that sqrt(2) is irrational.
>
> I don't know it by that name/principle, nor the
> connection between the two; surely you can conceive that
> a given principle/thm/result can have different names.
My question was, what is it that you believe you were adding to my
post in your reply?
If you were unfamiliar with "unique factorization", then perhaps the
thing to do would be to ask what that meant; or whether it was related
to showing that sqrt(2) is irrational; or what role it plays in that
proof, etc. But you labeled it an "addition", meaning you believed
that the information you added was not contained in mine; if, as you
now say, it is just that you had no idea what it was I said, perhaps
you might have started there...
--
Arturo Magidin
KMF
O.K; I did confuse things. It was not really a followup
to your post , other than in the sense below:
All I wanted to say is that the OP can show that
sqr(n) is rational iff n is a perfect square. I
thought you had suggested the OP
could do, i.e., prove that sqrn is irrational
only for n a square. It is in this that I referred
to your comment. I am sorry, I did not mean to confuse
anyone.
I thought this could be proved by extending
the proof of the irrationality of sqr.2 to sqrn
when n is a perfect square, i.e., your idea of showing
sqrn is irrational iff n is square could be
shown by a proof similar and more general
than that of the irrationality of sqr2. This is
what I meant by ading to your comment, no more,
no less, no offense meant.
If this is not what you said, than my bad, and
I confused things. Sorry.
Arturo Magidin
Two errors there: you again write "irrational" when you mean
"rational", and that is a biconditional (an "if and only if"), not a
mere conditional.
> I thought this could be proved by extending
> the proof of the irrationality of sqr.2 to sqrn
> when n is a perfect square,
You mean, when n is NOT a perfect square.
> i.e., your idea of showing
> sqrn is irrational iff n is square
NOT a square.
> could be
> shown by a proof similar and more general
> than that of the irrationality of sqr2. This is
> what I meant by ading to your comment, no more,
> no less, no offense meant.
>
> If this is not what you said, than my bad, and
> I confused things. Sorry.
My point is that you were not, in fact, "adding" anything. You were
simply ->repeating<- what I said. If the reason you believed what you
wrote was new information is because you did not, in fact, understand
what I wrote, then you should have asked what it was I meant. If the
reason you believed you were adding information is because you
misunderstood what I said, then that's more serious. On the other
hand, the use of "add" may simply have been just another example of
carelessness, as the carelessness above with rational vs. irrational,
square vs. not a square.
--
Arturo Magidin
KMF
"
I actually used 'iff.', and not 'if'. In my
math environment, iff is used to mean if and only if.
Arturo Magidin
I know that; and that is not what I referred to.
Here is what you wrote:
" All I wanted to say is that the OP can show that
sqr(n) is rational iff n is a perfect square. I
thought you had suggested the OP
could do, i.e., prove that sqrn is irrational
only for n a square."
The first time, you say it correctly, and I did not comment on -
>that<-.
The second time, however, you say "irrational only for n a square". It
is that clause in which you made both mistakes: 'irrational' should be
'rational', and "only" should be "if and only if".
What you wrote was equivalent to "if sqrt(n) is [ir]rational, then n
is a square." That is, a single implication. I'll wager even in your
"math environment" 'only' does not mean 'if and only if'.
--
Arturo Magidin
cbr...@cbrownsystems.com
> On May 15, 4:51 pm, qsymmetry <qsymme...@email.com> wrote:
>
> > If n is an integer and |n| > 1, is it true that
>
> > sqrt(n^2 - 1)
>
> > is irrational?
>
> Yes. Any integer that is not a perfect square has an irrational square
> root; that is, the only rationals that have a square that is an
> integer are the integers. Follows from unique factorization.
>
I'm a bit rusty on this, but can we then generalize to: Suppose R is a
UFD. Let Q be the rational field over R. Then does the following (or
something like it) hold: for strictly positive integer n, if q in Q
such that q^n in R, then q in R?
Cheers - Chas
Ludovicus
Never.
For sqr(x) = rational, x must be a perfect square.
But if x = n^2 -1 = (n-1)(n+1), the two factors must be
perfect squares. Let be n-1 = a^2 / b^2 and n+1 = c^2 / d^2
then c^2/d^2 = a^2/b^2 + 2
2.(d.b)^2 = (c.b)^2 - (d.a)^2
But this is impossible because the difference of two integer
squares is always an odd number.
Ludovicus
Dave Seaman
> On 15 mayo, 17:51, qsymmetry <qsymme...@email.com> wrote:
>> sqrt(n^2 - 1)
>> is irrational?
> Never.
You misspelled "always".
> For sqr(x) = rational, x must be a perfect square.
And n^2-1 is never a perfect square under the stated conditions. Therefore
sqrt(n^2 - 1) is always irrational in that case.
> But if x = n^2 -1 = (n-1)(n+1), the two factors must be
> perfect squares. Let be n-1 = a^2 / b^2 and n+1 = c^2 / d^2
> then c^2/d^2 = a^2/b^2 + 2
> 2.(d.b)^2 = (c.b)^2 - (d.a)^2
> But this is impossible because the difference of two integer
> squares is always an odd number.
> Ludovicus
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>
Arturo Magidin
Yes; this follows because a UFD is always integrally closed (so any
root to a monic polynomial that is in the field of fractions must in
fact be in the domain.
As Bill Dubuque pointed out, though, you don't even need all of that;
you only need GCD's and Euclid's Lemma: a|xy and (a,x)=1, then x|y.
--
Arturo Magidin
KMF
> > "Two errors there: you again write "irrational"
> when you mean "rational", and that is a biconditional
> (an "if and only if"), not a mere conditional.
> > "
> >
> > I actually used 'iff.', and not 'if'. In my
> > math environment, iff is used to mean if and only
> if.
>
> I know that; and that is not what I referred to.
Listen: I recognized I was wrong by referring to your
post carelessly.
But now you are doing something similar
yourself: you claim I am wrong on something and you do
not bother to specify what it is. Yes, you are explaining
it now, but when you claimed I was wrong, you did not
make clear just what that is. And I beleive that I what
bothered you about my post.
You should have made
_specific_ reference to where I was wrong. If you do
not want others making unfounded or unclear references
to your post, you should not do so either.
>
> Here is what you wrote:
>
> " All I wanted to say is that the OP can show that
> sqr(n) is rational iff n is a perfect square. I
> thought you had suggested the OP
> could do, i.e., prove that sqrn is irrational
> only for n a square."
>
> The first time, you say it correctly, and I did not
> comment on -
> >that<-.
>
> The second time, however, you say "irrational only
> for n a square". It
> is that clause in which you made both mistakes:
> 'irrational' should be
> 'rational', and "only" should be "if and only if".
>
> What you wrote was equivalent to "if sqrt(n) is
> [ir]rational, then n
> is a square." That is, a single implication. I'll
> wager even in your
> "math environment" 'only' does not mean 'if and only
> if'.
>
> --
Your wrong on two different counts here:
i) The statement:"if sqrt(n) is rational,
then n is a perfect square" is correct.
It does not negate the double implication
iff. At best it is incomplete, but, when
transcribed into "if sqrt(n) is rational,
then n is a perfect square" , this is ,
as a conditional, a true statement.
And no, only does not mean iff, but iff
detaches into only. And this is what I
did. So using 'only' is not incorrect
by any meaning I can think of.
ii) Now, since i) is not factually wrong, you
may claim that it is wrong in the context in
which I wrote it. But you are not quoting that
context. So you are making reference to my post
--and claiming that I am wrong -- carelessly again.
> Arturo Magidin
KMF
Again I say: I should not have made reference to your
post if I was not clear on what you meant by unique
factorization. I was wrong. Having said that:
I disagree with your claim that my suggestion does
not add anything. I went over this proof years
ago, without making any mention of unique factorization.
You basically make some assumptions about real numbers
( this proof was done in an analysis class), including
--I can see now, but I did not know back then -- that
the reals are a UFD.
Only someone who does this proof from an algebraic
perspective would agree with you. If done within
analysis, where there is no focus on algebraic properties of real numbers being a Euclidean ring=>UFD,
etc. it is not necessary to explicitly address the
issue of unique factorization.
Now, just to examine the claim of the fact that there
is no necessary connection between the two, look up
books containing the proof of the irrationality
of sqr.2, and tell me if there is mention in those
books about the unique factorization properties of
the real numbers. I doubt it. Since you are claiming that there _is_ a(n) (almost) necessary connection,
it should be your burden to provide these examples.
So if this proof is done within analysis, then your
reference to unique factorization is a broken link
in people's minds. If it is done in an algebra class,
then it is not. You, as an algebraist, have this bias/
inclination. I doubt every other mathematician makes
a connection between these two. Convince me I am wrong
and I will take it back.
So you have no way of guaranteeing that your
suggestion will be helpful to people in the first
group of analysts, among which I am. So I am _not_ necessarily repeating, for this reason: I am appealing to a simpler-minded crowd that for some reason does not delve in the properties of the real numbers as a ring.
And I don't think there is a way of determining which
crowd the OP belongs to. so your claim is unsupportable.
To the simpler crowd, I am saying something new.
Surely you would not want to delve into the depths
of every specialty when proving a result outside of
algebra. would you?
"If the reason you believed what you
wrote was new information is because you did not, in fact, understand what I wrote, then you should have asked what it was I meant. If the
reason you believed you were adding information is because you
misunderstood what I said, then that's more serious. On the other
hand, the use of "add" may simply have been just another example of carelessness, as the carelessness above with rational vs. irrational,
square vs. not a square.
"
See my first paragraph.
Arturo Magidin
> > On May 17, 6:05 pm, KMF <tee...@hotmail.com> wrote:
> > > "Two errors there: you again write "irrational"
> > when you mean "rational", and that is a biconditional
> > (an "if and only if"), not a mere conditional.
> > > "
>
> > > I actually used 'iff.', and not 'if'. In my
> > > math environment, iff is used to mean if and only
> > if.
>
> > I know that; and that is not what I referred to.
>
> Listen: I recognized I was wrong by referring to your
> post carelessly.
> But now you are doing something similar
> yourself: you claim I am wrong on something and you do
> not bother to specify what it is.
I said "you wrote 'irrational' when you mean 'rational'..."
Since the sentence you defended did *not*, in fact, contain the word
"irrational", why would you think that I was refering to that sentence
with my comment, rather than the very last sentence before my comment,
which *did* contain the word I was quoting?
> Yes, you are explaining
> it now, but when you claimed I was wrong, you did not
> make clear just what that is.
I specified the word in question, and how to change it, and that the
implication in that sentence was only in one direction. Exactly what
is it you think that I did not make clear?
> And I beleive that I what
> bothered you about my post.
>
> You should have made
> _specific_ reference to where I was wrong.
I *did*. And you misunderstood it *anyway*.
> > Here is what you wrote:
>
> > " All I wanted to say is that the OP can show that
> > sqr(n) is rational iff n is a perfect square. I
> > thought you had suggested the OP
> > could do, i.e., prove that sqrn is irrational
> > only for n a square."
>
> > The first time, you say it correctly, and I did not
> > comment on -
> > >that<-.
>
> > The second time, however, you say "irrational only
> > for n a square". It
> > is that clause in which you made both mistakes:
> > 'irrational' should be
> > 'rational', and "only" should be "if and only if".
>
> > What you wrote was equivalent to "if sqrt(n) is
> > [ir]rational, then n
> > is a square." That is, a single implication. I'll
> > wager even in your
> > "math environment" 'only' does not mean 'if and only
> > if'.
>
> > --
>
> Your wrong on two different counts here:
>
> i) The statement:"if sqrt(n) is rational,
> then n is a perfect square" is correct.
Oh, for crying out loud. Can't you even *read*? What you wrote and I
addressed, as was clear from my explicitly noting that it was the word
"irrational" that was incorrect, was "sqrtn is IRRATIONAL only for n a
square [emphasis added]".
> It does not negate the double implication
> iff.
Since the clause is preceded by "i.e.", meaning "that is", it means
that you are announcing that you are about to give a statement which
is equivalent to the one preceding the "i.e." (the correct statement
that "sqrt(n) is rational iff n is a perfect square"). Since the
statement that goes after the i.e. is not, in fact, equivalent to the
one preceding the i.e., that means that you are making a mistake,
either in the statement, or in the assertion that you are about to
give an equivalent statement.
> At best it is incomplete, but, when
> transcribed into "if sqrt(n) is rational,
> then n is a perfect square" , this is ,
> as a conditional, a true statement.
>
> And no, only does not mean iff, but iff
> detaches into only. And this is what I
> did. So using 'only' is not incorrect
> by any meaning I can think of.
>
> ii) Now, since i) is not factually wrong,
Now, since (i) *is*, in fact, incorrect in its assertions about what
you wrote, and that what you wrote *is*, in fact, factually incorrect
(notice the IRRATIONAL word there)...
> you
> may claim that it is wrong in the context in
> which I wrote it. But you are not quoting that
> context. So you are making reference to my post
> --and claiming that I am wrong -- carelessly again.
Do try reading all the letters in your words, please. You wrote
"IRRATIONAL only for n a square". Are you claiming this assertion is
"not factually wrong"? Really?
--
Arturo Magidin
Rob Johnson
This is similar to a proof that the only rational algebraic integers
are integers, <http://www.whim.org/nebula/math/ratalint.html>, that I
posted long ago. I like using Bezout much more than GCDs or prime
factorization, since it is a much more basic theorem. The others are
like swatting flies with sledgehammers to me.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>qsymmetry <qsym...@email.com> wrote:
>>>
>>> For integer |n| > 1, is it true that sqrt(n^2 - 1) is irrational?
>>
>> Here's a cute little-known proof that I discovered as a teenager:
>>
>> THEOREM w = sqrt(n) is integral if rational (for integral n).
>>
>> PROOF w = a/b, (a,b)=1 => ad-bc = 1 for some c,d via Bezout,
>>
>> therefore 0 = (a-bw) (c+dw) = ac-bdn + w => w is an integer. QED
>
> are integers, <http://www.whim.org/nebula/math/ratalint.html>, that
> I posted long ago.
I see no similarity whatsoever - except for the fact that both proofs
use Bezout's Lemma. But that's not saying much since every such proof
must employ that or a closely related property - namely that the
domain Z is a Euclidean => PID => GCD.
Moreover, as I've mentioned to you here on a few prior occasions [1]
the proof in your link is just an obfuscated version of perhaps the
most well-known version of this proof. Namely, it simply takes the
usual proof of the Rational Root Theorem and instead of using the
usual final inference: (p,q)=1, q|p^n => q|1 (by Euclid's Lemma)
it instead makes that inference by Bezout's Lemma, essentially:
ap+bq = 1, q|p^n => q | (ap)^n = (1-bq)^n => q|1
But it goes even further and instead of abstracting out that inference
as a lemma it instead substitutes the Bezout identity throughout all
its equations - which greatly obfuscates the essence of the matter.
Of course one can always "inline" lemmas, abstractions... to produce
completely "elementary" versions of theorems. A well-known case of
such is the completely self-contained proof of unique factorization
of integers by Klappauf, Lindemann, Zermelo [2]. This inlines the
Euclidean/Division algorithm descent into a direct inductive proof.
By eliminating all the internal structure supporting the argument
such proofs tremendously obfuscate the real essence of the matter.
In fact it was the discovery of such hidden structure (esp. ideals
and modules) in elementary number theory that led Dedekind to
invent most of the major classical algebraic structures. Without
the introduction of such abstraction there would be little hope
of conquering the complexity inherent in modern number theory.
> I like using Bezout much more than GCDs or prime factorization,
> since it is a much more basic theorem. The others are like
> swatting flies with sledgehammers to me.
Not true. In this case using gcds is simpler and more concise,
e.g. it eliminates the extraneous objects a, b that arise
in the Bezout identity. Moreover, it is also more general.
I can't imagine why anyone would consider it a sledgehammer.
--Bill Dubuque
[1] http://google.com/groups?selm=y8zzmf2xg5q.fsf%40nestle.csail.mit.edu
[2] http://www.math.uiuc.edu/~dan/ShortProofs/uf.html
cbr...@cbrownsystems.com
(You meant of course not x|y but a|y). Thanks.
Cheers - Chas
Arturo Magidin
<cbr...@cbrownsystems.com> wrote:
>
> > As Bill Dubuque pointed out, though, you don't even need all of that;
> > you only need GCD's and Euclid's Lemma: a|xy and (a,x)=1, then x|y.
>
> (You meant of course not x|y but a|y). Thanks.
Yup. Thank you.
--
Arturo Magidin
KMF
False. Here is a proof of the irrationality of
sqr2 that makes no explicit mention of any
unique factorization. It uses it implicitly,
but there is no need to make every implict assumption
explicit. Just take a,b below to be integers and
use basic number-theoretic properties of integers.
:
Assume sqr2 is rational.
Then a/b=sqr2 .
Assume we can have a/b in its lowest form , so that
(a,b)=1 . No need to bring in a tank to kill a fly
and talk about unique factorization. Make some
assumptions that seem reasonable, if you are not
doing algebra, as this is proven in analysis
or beginning number theory classes. So assume
(a,b)=1
now square both sides to get:
a^2/b^2=2
then a^2=2b^2
Now 2 is a prime. So 2|a^2 means 2|a .
No need to mention generalities about rings, even
if these are being implicitly used. Just use a
simple argument for a,b as natural numbers, not
as elements of a UFD:
if p|ab and p|a, then p|b . Otherwise, the factors
of p are somehow spread between a and b. But p
has only p and 1 as factors. It follows that
p|a or p|b . Simple. No UFD's.
In our case we have 2|a^2 , or 2|a*a
Then 2|a . So a =2k , for some integer k
Substitute back in original:
a^2=(2k)^2=2b^2 , so 4k^2=2b^2
Simplify, without stating that R is a field
(so it is a UFD, but that is not necessary)
Then 2k^2=b^2
By an argument similar to the previous, it follows
that 2|b .
Now we get a contradiction: we have 2|a and 2|b,
when we assumed (a,b)=1 . Then sqr2 is irrational.
Now, take any n not a perfect square, and
express n=p_1^e_1*p_2^e_2...p^ke_k
Just take the fundamental thm. of arithmetic
and assume this is possible for integers. No
need for unique factorization.
Consider sqrn , with n as above. Since n is
not a perfect square, you may be able to factor
out some terms from the expression for n, but
I think it is clear that you will be left with
at least one term in the root, a prime number
with power 1.
so you will have something like:
sqrn= p_1e^1/2*..p_je^j sqr(p_j+1..P_k)
An argument similar to that for sqr2 will
show that sqrn is not rational. Let me know
if you want me to expand on this.
If you want a more general argument using
algebra, you can do that. But I made no
explicit statement of UFactorization, and
my proof works. So there is _no_ necessary
connection between UFactorization and the
argument I suggested. My argument _does_
add.
Arturo seems to think everyone is an
algebraist. I _never_ thought of UFactorization
being related to a generalization of the irra-
tionality of sqr2. And I think my proof works,
without any mention of UFD's. Just some basic
number theory results are enough.
And I don't believe that the average person
that hears or reads UFactorization will think
of this generalization either. It is a bogus
claim that my comment did not add anything.
KMF
squares is odd. Otherwise take 100 and 16.
This gives you a nice way of showing that the
sum of odds beginning at 1 is a perfect sqaure:
you start with n=1 and then add the differences
between consecutive squares, i.e:
1+3=4 , and 2^2-1^2=3
4+5=9 , and 3^2-2^2=5
So you are adding to n^2 the difference between
(n+1)^2 and n^2 , so you get:
n^2+((n+1)^2-n^2))= (n+1)^2
so no need for induction to show:
1+3+5+..+(2n-1)=n^2 .
> Ludovicus
pubkeybreaker
> In article <y8zpre87srw....@nestle.csail.mit.edu>,
>
> Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >qsymmetry <qsymme...@email.com> wrote:
>
> >> For integer |n| > 1, is it true that sqrt(n^2 - 1) is irrational?
>
> >Here's a cute little-known proof that I discovered as a teenager:
>
> >THEOREM w = sqrt(n) is integral if rational (for integral n).
>
> >PROOF w = a/b, (a,b)=1 => ad-bc = 1 for some c,d via Bezout,
>
> >therefore 0 = (a-bw) (c+dw) = ac-bdn + w => w is an integer. QED
>
> >See also the parallel thread on Ask an Algebraist
>
> This is similar to a proof that the only rational algebraic integers
> are integers, <http://www.whim.org/nebula/math/ratalint.html>, that I
> posted long ago. I like using Bezout much more than GCDs or prime
> factorization, since it is a much more basic theorem. The others are
> like swatting flies with sledgehammers to me.
>
> Rob Johnson <r...@trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font
One can also use the rational roots theorem for polynomials.
If r is a rational root of x^2 - (n^2-1), then r divides n^2 - 1.
It then follows that r^2 can not be sufficiently close enough to
n^2.
Arturo Magidin
> > On May 15, 8:47 pm, KMF <tee...@hotmail.com> wrote:
> > > As a quick followup/addition to Arturo's post, you
> > > can easily generalize the proof that square root
> > of
> > > 2 is irrational to show that sqr(n) is irrational
> > > iff. n is a perfect square.
>
> > That would be quite a feat. You mean, surely, that
> > sqrt(n) is *rational* if and only if n is a perfect square.
> > And I believe that is *exactly* what I said, and that
> > when I mentioned unique factorization, it was clear that this meant you were
> > generalizing the proof that sqrt(2) is irrational.
>
> > --
> > Arturo Magidin
>
> False. Here is a proof of the irrationality of
> sqr2 that makes no explicit mention of any
> unique factorization. It uses it implicitly,
So then, it has nothing to do with unique factorization, except
insofar as it actually uses unique factorization. Got it.
> Arturo seems to think everyone is an
> algebraist.
KMF seems to think he can read minds. Good for him.
> It is a bogus
> claim that my comment did not add anything.
Indeed. Your comments added a number of mistakes (continued miswriting
of "irrational" when "rational" was meant, over and over and over
again). I didn't want to bring it up, but since you insist on it...
And they added to a generally pointless argument which I am happy to
quit at this point.
--
Arturo Magidin
Bill Dubuque
>On May 18, 6:54�pm, r...@trash.whim.org (Rob Johnson) wrote:
>>In article <y8zpre87srw....@nestle.csail.mit.edu>,
>>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>>qsymmetry <qsymme...@email.com> wrote:
>>>>
>>>> For integer |n| > 1, is it true that sqrt(n^2 - 1) is irrational?
>>>
>>> Here's a cute little-known proof that I discovered as a teenager:
>>>
>>> THEOREM w = sqrt(n) is integral if rational (for integral n)
>>>
>>>
>>
>> This is similar to a proof that the only rational algebraic integers
>> are integers, <http://www.whim.org/nebula/math/ratalint.html>, that
>
> One can also use the rational roots theorem for polynomials [...]
As I mentioned in prior replies, it's of course obvious that his theorem
is simply the monic case of the Rational Root Test. Moreover, his proof
is simply an obfuscation of the standard proof of this standard theorem.
Essentially he inlines a Bezout-based proof of (a,b)=1, b|a^n => b|1
instead of using the standard simpler proof based on Euclid's Lemma.
See my prior reply here for further details and references.
The (monic) Rational Root Test is essentially equivalent to the fact
that the domain of integers is integrally-closed. If you examine the
elementary direct proofs from this more general standpoint they're
essentially all equivalent. This is evident and trivial to anyone
familiar with notion of conductor ideals (dating back to Dedekind).
Alas, apparently not all number-theorists are familiar with this
basic notion, e.g. see Gerry Myerson's post [1] where he rediscovers
the well-known elementary inductive proof that obtains from unwinding
the elegant one-line conductor-based proof. Further, in the same
thread, he seems to lend support to the absurd claim by Estermann
(and Niven?) that this proof is new. Nothing could be further from
the truth. Such methods are merely elementary manifestations of the
raison d'etre for the abstraction incorporated in the conductor ideal.
This fundamental object is introduced very early in any study of
algebraic number theory and the perceptive student should easily
recognize the fundamental role it plays in more concrete instances.
--Bill Dubuque
[1] Gerry Myerson, sci.math post on 2006/7/12
Thread: Irrationality and the Fundamental Theorem of Arithmetic
http://google.com/groups?selm=gerry-6EFCF3.10502013072006%40sunb.ocs.mq.edu.au
Rob Johnson
There is nothing in what you say that I disagree with.
I agree that using unique factorization makes most complex proofs
easier to understand. However, I like using Bezout when proving
simpler theorems to make sure not to use what I am trying to prove.
Furthermore, it is difficult to know what prerequisites that a
poster is working with, and so for simple theorems, it again seems
better to work with basics.
Euclid's Lemma is usually proven using Bezout's Theorem and the
Rational Root Theorem is usually proven using either Euclid's Lemma
or Bezout's Theorem. So if a proof is not complicated and can be
completed easily with Bezout, especially if I am unsure of the
reader's prior knowledge, I will opt for starting with Bezout.
>> I like using Bezout much more than GCDs or prime factorization,
>> since it is a much more basic theorem. The others are like
>> swatting flies with sledgehammers to me.
>
>Not true. In this case using gcds is simpler and more concise,
>e.g. it eliminates the extraneous objects a, b that arise
>in the Bezout identity. Moreover, it is also more general.
>I can't imagine why anyone would consider it a sledgehammer.
Not true? I have only presented my preferences and opinions, and I
think I know them pretty well. However, I was speaking too broadly
when I said that I considered GCDs and unique factorization to be
like a sledgehammer. I meant that sometimes they are used sometimes
when more basic methods would suffice (and not make the proof too
complex).
I do not wish to start a flame here, but I did not want to appear
to ignore your post.
Arturo Magidin
> Now, take any n not a perfect square, and
> express n=p_1^e_1*p_2^e_2...p^ke_k
>
> Just take the fundamental thm. of arithmetic
> and assume this is possible for integers. No
> need for unique factorization.
The Fundamental Theorem of Arithmetic ->is<- unique factorization for
the integers. They are one and the same.
--
Arturo Magidin
Gerry Myerson
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
Readers may be interested in my paper, Irrationality via well-ordering,
Australian Math. Soc. Gazette 35 (2008) 121-125, available on the web
at http://www.austms.org.au/Publ/Gazette/2008/May08/Myerson.pdf
In that paper, you'll find,
Theorem 2. If m is an integer, and sqrt m is not an integer, then
sqrt m is irrational.
Proof. Assuming sqrt m is rational but not an integer, let k be the
integer such that k < sqrt m < k+1, and let n be the smallest positive
integer whose product with sqrt m is an integer; then n(sqrt m - k) is
a smaller positive integer whose product with sqrt m is an integer,
contradiction.
You'll also find the theorem stating that the integers are integrally
closed in the rationals;
Theorem 5. Let f be a monic polynomial with integer coefficients. Let
alpha be a solution of f(alpha) = 0. Then if alpha is not an integer,
it is irrational.
Proof. Assume the hypotheses, assume alpha is rational, and let n be
the smallest positive integer whose product with alpha^j is an integer
for all j less than the degree of f. Then n times the fractional part
of alpha is a smaller positive integer with the same property,
contradiction.
I ask in the paper for the earliest publications of these (and other)
proofs in the form in which I relate them. This form may well be
merely an elementary manifestation of the conductor ideal, but I
reckon there's a lot to be said for elementary manifestations. In
particular, students who are still years away from learning about
conductor ideals can still understand, maybe with a little help, the
proofs above.
The earliest publication I've seen of the proof given above for
Theorem 2 is the 1975 paper of Estermann. I haven't seen any
print publication of the proof of Theorem 5 given above, but it was
posted to this newsgroup in July 2000, and I've seen it on the web.
There may well have been earlier appearances of these proofs
in the form given here, and I'd like to see them.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
<3c0e4c58-95ee-4a66...@s16g2000vbp.googlegroups.com>,
Arturo Magidin <mag...@member.ams.org> wrote:
> The Fundamental Theorem of Arithmetic ->is<- unique factorization for
> the integers. They are one and the same.
Agreement on this is not universal. E.g., Uspensky and Heaslet,
Elementary Number Theory, use the phrase "fundamental theorem
of arithmetic" to describe the result that says that if a divides bc
and a is relatively prime to b then a divides c.
Arturo Magidin
wrote:
> In article
> <3c0e4c58-95ee-4a66-8704-bd1c4254f...@s16g2000vbp.googlegroups.com>,
> Arturo Magidin <magi...@member.ams.org> wrote:
>
> > The Fundamental Theorem of Arithmetic ->is<- unique factorization for
> > the integers. They are one and the same.
>
> Agreement on this is not universal. E.g., Uspensky and Heaslet,
> Elementary Number Theory, use the phrase "fundamental theorem
> of arithmetic" to describe the result that says that if a divides bc
> and a is relatively prime to b then a divides c.
Since the OP used the name to refer specifically to expressing n as a
product
of prime powers...
--
Arturo Magidin
KMF
>
> > Now, take any n not a perfect square, and
> > express n=p_1^e_1*p_2^e_2...p^ke_k
> >
> > Just take the fundamental thm. of arithmetic
> > and assume this is possible for integers. No
> > need for unique factorization.
>
I thought you did not want to discuss this anymore.
Still, Ill reply:
Please don't assume from the onset that my position
is ridiculous. I may be wrong, but I have tried to state this clearly. I think I gave your argument a
fair shot:
(This is absurdly long and explicit, but this seems
necessary, since we seem to otherwise be unable to
understand each other)
> The Fundamental Theorem of Arithmetic ->is<- unique
> factorization for the integers. They are one and the same. .
But my point is that even if/when I am making use of this fact, the fact does not need to be made explicit in an intro number theory class, and it was not done when I
saw this proof, just like there is no mention of the lub
property of the reals when teaching basic integration
in R, or R^n ( it may be a different story when one
studies general measure theory, where one needs to be
more aware of the assumptions being made), and were one
is able to do proofs at this level which do not state
explicitly the fundamental properties of R, so that these proofs are made at a lower level of resolution
than the same proofs would be made in an analysis class.
In this analogy, your proof was a measure-theoretical one,mine was a Calc proof. I may be able to do a measure
theory proof 'P' by just declaring that every open set in
R is the countable union of open intervals, without
explicitly stating the result that rationals are dense
in R and are countable that is used in this proof.
Then someone who only knows my proof 'P', would not
think of it as a proof that makes use of basic properties of rational numbers.
Again, the proof 'P' has a lower level of resolution
or zoom than other more sofisticated proofs 'P_2 'posible.
My argument is a 'P' argument, yours is a 'P_2'
arguement. Resolution. Any mention on your part
of your proof as the proof using properties of
rationals in R would trigger no link or response
in my knowledge base, even if I am making explicit
use of these properties. Again, level of resolution.
If one just wants to show that sqrn is irrational
for n not a perfect square, it is not necessary to
explicitly state that one is making use of unique
factorization, just like I am able to calculate
integral Int( xdx ) without explicitly being aware
that I am using somehow the lub property, or I am
assuming that the Riemann sum of f(x)=x converges,
nor issues and conditions of convergence of Riemann
sums in general. It is the difference between analysis and calculus.
I can also use the MVTheorem without
explicitly mentioning nor being aware of the basic
properties of reals that I am implicitly using/assuming
in proving the MVT.
Again: your suggested proof is an analysis proof, mine was a calculus one. There is a difference in the "zooming" in our arguments: you are doing a proof that is more adavanced than mine, in that your proof is more general,and makes explicit the assumptions made. Mine does not,and it is more quick and dirty, with a lesser "zoom".
My proof is suited for an intro. class, your is
for a more advanced crowd. Unique factorization, in
this respect, is not openly present in my proof, so
I was not even aware that I was using it until I took
an abstract algebra class. In addition, your mention
of unique factorization did not link in my mind to
this proof at first, until I stopped and reflected
upon it. It is in this sense that, at first, your
use of the term unique factorization meant nothing
to me, and I did not associate it with the proof
I used, which, again , is a proof of less "resolution"
or "zoom" than yours.
Like the bad old joke, I was able to speak in
prose all my life without being aware of it.
This is the difference in the proofs each of us
came up with: I did a 'calculus' proof , and you did a more refined analysis-style proof, which is more abstract and more general, where all the assumptions
are stated. Mine is more 'down and dirty', and it does
Let's just consider ourselves lucky that we
do not have to work or live together, given that
we seem to be completely unable to communicate with
each other. I think you misunderstood what I was
saying, and I likely misunderstood ( or misunderestimated :) ) what you were saying.
>
> --
> Arturo Magidin
Gerry Myerson
<24809421.122908.1242797...@nitrogen.mathforum.org>,
KMF <tee...@hotmail.com> wrote:
> Like the bad old joke, I was able to speak in
> prose all my life without being aware of it.
It was actually a pretty good joke. It comes from Moliere's play,
Le Bourgeois Gentilhomme.
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>
>> The (monic) Rational Root Test is essentially equivalent to the fact
>> that the domain of integers is integrally-closed. If you examine the
>> elementary direct proofs from this more general standpoint they're
>> essentially all equivalent. This is evident and trivial to anyone
>> familiar with notion of conductor ideals (dating back to Dedekind).
>> Alas, apparently not all number-theorists are familiar with this
>> basic notion, e.g. see Gerry Myerson's post [1] where he rediscovers
>> the well-known elementary inductive proof that obtains from unwinding
>> the elegant one-line conductor-based proof. Further, in the same
>> thread, he seems to lend support to the absurd claim by Estermann
>> (and Niven?) that this proof is new. Nothing could be further from
>> the truth. Such methods are merely elementary manifestations of the
>> raison d'etre for the abstraction incorporated in the conductor ideal.
>> This fundamental object is introduced very early in any study of
>> algebraic number theory and the perceptive student should easily
>> recognize the fundamental role it plays in more concrete instances.
>>
>> Thread: Irrationality and the Fundamental Theorem of Arithmetic
>> http://google.com/groups?selm=gerry-6EFCF3.10502013072006%40sunb.ocs.mq.edu.au
>
> Readers may be interested in my paper, Irrationality via well-ordering,
> Australian Math. Soc. Gazette 35 (2008) 121-125, available on the web
> at http://www.austms.org.au/Publ/Gazette/2008/May08/Myerson.pdf
A better title would be "Irrationality via the Division Algorithm"
(or "via the Euclidean Algorithm") since your proofs there are simply
well-known proofs of PID results that've been pulled-back to Euclidean
domains by eliminating ideal-theoretic notions. More on this below.
It is misleading to say the the proofs are "via well-ordering".
Rather, the proofs depend crucially on a domain D being Euclidean.
This requires not only a "size" map from D into a well-ordered set,
but also, crucially, a Division Algorithm with respect to the map.
Euclideaness is used implicitly in your proofs when you take the
integer or fractional part of a fraction to help achieve a descent
on possible denominators for w = sqrt(m) (or w an algebraic integer).
> In that paper, you'll find,
>
> THEOREM 2. If m is an integer, and sqrt m is not an integer,
> then sqrt m is irrational.
>
> Proof. Assuming w = sqrt m is rational but not an integer, let k be the
> integer such that k < w < k+1, and let n be the smallest positive
> integer such that nw is an integer; then n(w-k) is a smaller positive
> integer whose product with sqrt m is an integer, contradiction.
That's very well-known and very widely published.
> You'll also find the theorem stating that the integers
> are integrally closed in the rationals;
>
> THEOREM 5. Let f be a monic polynomial with integer coefficients.
> Let w be a solution of f(w) = 0. Then if w is not an integer,
> it is irrational.
>
> Proof. Assume the hypotheses, assume w is rational, and let n be
> the smallest positive integer whose product with w^j is an integer
> for all j less than the degree of f. Then n times the fractional part
> of w is a smaller positive integer with the same property, contradiction.
Again that's very-well known and I do recall seeing it published at
least a handful of times if not more. Notice that your n is simply
a least common denominator for the fractions w^j. More structurally
consider instead the denominator set C<Z for the whole ring Z[w],
i.e. C Z[w] < Z, i.e. C = {n in Z: n f(w) in Z, all f(w) in Z[w]}
This yields an _ideal_ C known as the _conductor_ of Z[w] into Z.
C satisfies the crucial property that n in C => nw in C, simply
because w is integral over Z (i.e. w^k is a Z-linear combination
of w^j, for j < k = deg w). This then implies that C is not only
an ideal of Z, but also of Z[w] (in fact the largest such ideal).
But since Z is a PID, the conductor ideal is principal C = nZ
so nw in C = nZ => w in Z. So we've proved any fraction integral
over a PID actually lies in it, i.e. PIDs are integrally closed.
Your proofs essentially proceed the same way except you have
eliminated the lemma that Z is a PID by "inlining" it in your
descent-based proof on denominators, i.e. n in C => nw in C
=> n(w-k) in C, for k = integer_part(w). But 0 < n(w-k) < n,
via 0 < w-k < 1, so n(w-k) is a smaller denominator than n.
This is equivalent to descent-based proof that the denominator
ideal C is principal. It's essentially a special case of the
the standard Euclidean descent proof that Euclidean => PID,
i.e. if n is the smallest elt of C then n must divide every
other elt m of C, else Euclidean division yields a smaller
elt m mod n of C, contra minimality of n. Above is just the
special case m = nw, which yields n|nw in Z, i.e. w in Z.
Here is how I like to present this proof in structural form:
The CONDUCTOR ideal of domains D < E is C = {d in D: dE < D}
One easily sees C = CD = CE, i.e. C is ideal of both D and E
C>0 & D PID => C = cD = cE => D = E via cancel c != 0 [*]
Said proof is just the special case C,D,E = n,Z,Z[w]. Here
C>0 since w = a/b, deg f = k => b^(k-1) in C, i.e. b^(k-1)
is a denominator for all elts of Z[w] = Z<1,w,w^2,..,w^(k-1)>
Notice how simple the proof has become when expressed in the
appropriate abstractions: it is a trivial one-line proof [*]
based on the conductor ideal, being principal, is cancellable.
More generally, cancellable/invertible ideals (esp. Dedekind
domains) play a fundamental role in algebraic number theory.
> I ask in the paper for the earliest publications of these (and other)
> proofs in the form in which I relate them. This form may well be
> merely an elementary manifestation of the conductor ideal, but I
> reckon there's a lot to be said for elementary manifestations. In
> particular, students who are still years away from learning about
> conductor ideals can still understand, maybe with a little help,
> the proofs above.
The same might be argued for analogous "elementary" proofs that've
had their higher-order structure removed, e.g the famous proof [2]
of unique factorization by Klappauf, Lindemann, Zermelo that I
cited earlier in this thread. It isn't published too frequently
(except as a curiosity) because it obfuscates the underlying
algebraic structure and hence has little pedagogical value.
Someone ignorant of the relevant algebraic theory might stumble
upon such proofs with great effort. But someone familiar with
the theory can _mechanically_ churn out these monstrosities by
simply rewriting the higher-order proofs to use only the most
primitive notions - eliminating structure (rings, ideals, etc.)
Indeed, the process is so mechanical that one could easily
program a computer to perform such proof transformations.
> The earliest publication I've seen of the proof given above for
> Theorem 2 is the 1975 paper of Estermann. I haven't seen any
> print publication of the proof of Theorem 5 given above, but it was
> posted to this newsgroup in July 2000, and I've seen it on the web.
> There may well have been earlier appearances of these proofs
> in the form given here, and I'd like to see them.
There's little point in doing such for the reasons enumerated
above. The conductor-based ideas are over a century old - dating
back to the ground-breaking work of Dedekind. It's highly likely
that the proof is already in Dedekind's work (perhaps implicitly
since there are various ideal-theoretic forms of such proofs).
Alas, no one has yet tackled the enormous task of writing the
definitive history of unique factorization - where many of these
topics would be discussed. I once thought I might do such and
had many files of papers of historical significance to aid in
this endeavor. However I no longer have these available since
they were among items that were stolen while in storage. But
nowadays such a task should be easier since many older papers
are now much more easily accessible in electronic archives.
--Bill Dubuque
Arturo Magidin
> > On May 19, 1:55 am, KMF <tee...@hotmail.com> wrote:
>
> > > Now, take any n not a perfect square, and
> > > express n=p_1^e_1*p_2^e_2...p^ke_k
>
> > > Just take the fundamental thm. of arithmetic
> > > and assume this is possible for integers. No
> > > need for unique factorization.
>
> I thought you did not want to discuss this anymore.
> Still, Ill reply:
>
> Please don't assume from the onset that my position
> is ridiculous. I may be wrong, but I have tried to state this clearly. I think I gave your argument a
> fair shot:
I think you have little or no idea what "my argument" was. I did not
then, nor did I ever, insinuate the use of "high power algebra" or
abstract algebra, just basic properties of the integers. When I said
"unique factorization", in the context of integers, I *meant* exactly
what I said: unique factorization. You now seem to think I was
invoking high-falluting algebraic concepts, "assuming everyone is an
algebraist". Your analogy about measure theory is simply misguided,
and shows that you still have little or no clue what it was I was
saying; the proof I alluded to was, SURPRISE! exactly the one you
wrote: use unique factorization of integers into primes to write n as
a product of prime powers, deduce that it has a rational square root
if and only if it is a perfect square. Hence, my original point: you
did not, in fact, add anything to the discussion except careless
errors.
> (This is absurdly long and explicit, but this seems
> necessary, since we seem to otherwise be unable to
> understand each other)
I think I understand exactly what you were and are saying. What you
seem to not be able to understand is exactly where and why you were
wrong, even to the point of claiming that the assertion "sqrtn is
irrational only if n is a perfect square" was 'factually correct'.
> Again: your suggested proof is an analysis proof, mine was a calculus one.
Again: no, you clearly misunderstood and continue to misunderstand
what I wrote.
> There is a difference in the "zooming" in our arguments:
There is a difference between knowing what one is talking about and
not really knowing much about what one is talking about. I'll let you
figure out which is which.
> Let's just consider ourselves lucky that we
> do not have to work or live together, given that
> we seem to be completely unable to communicate with
> each other.
There certainly is one of us who is completely unable to understand
the other. It ain't me, though.
--
Arturo Magidin
Bill Dubuque
It took me only about 5 minutes to find an earlier publication [1]
of Theorem 2. I recall seeing much older publications of both Theorem 1
and Theorem 5. As I mentioned in a prior post here they're probably both
in Dedekind's work in their more natural ideal-theoretic form in terms
of the conductor ideal. In particular, Estermann's boasting that it
is "the first new proof since Pythagoras" [2] is ludicrous. However,
I'm not surprised that Estermann (and Niven) supported this claim
since they both seem to have little knowledge of algebraic number
theory (based on a quick perusal of their publications listed in
the AMS Math Reviews database). Anyone who has basic knowledge of
algebraic number theory is familiar with the fundamental notion of
conductor ideals and, armed with such, should quickly notice its
crucial hidden role in such "elementary" irrationality proofs. Alas,
this is one of the weak spots in most number theory expositions.
When new abstractions are introduced it is expected that the reader
has the mathematical maturity to take the effort to consider how
the abstractions specialize in well-known concrete instances. One
will not go very far in mathematics without often exercising such.
--Bill Dubuque
[1] Vaughan, Herbert E. On the Irrationality of Roots
Amer. Math. Monthly, Vol. 67, #6 pp. 576-578
http://www.jstor.org/stable/2309183
[2] Derbyshire J. Prime obsession. p. 379
Bernhard Riemann and the greatest unsolved problem in mathematics
(Natl.Acad.Press, 2003) ISBN 0309085497
KMF
> > > On May 19, 1:55 am, KMF <tee...@hotmail.com>
> wrote:
> >
> > > > Now, take any n not a perfect square, and
> > > > express n=p_1^e_1*p_2^e_2...p^ke_k
> >
> > > > Just take the fundamental thm. of
> arithmetic
> > > > and assume this is possible for integers.
> No
> > > > need for unique factorization.
> >
> > I thought you did not want to discuss this anymore.
> > Still, Ill reply:
> >
> > Please don't assume from the onset that my
> position
> > is ridiculous. I may be wrong, but I have tried to
> state this clearly. I think I gave your argument a
> > fair shot:
>
> I think you have little or no idea what "my argument"
> was. I did not then, nor did I ever, insinuate the use of "high power algebra" or abstract algebra, just basic properties of the integers. When I said
> "unique factorization", in the context of integers, I
> *meant* exactly what I said: unique factorization.
But I think that _existence_ of a factorization
is what matters the most, and not uniqueness.
Since to me, taking a naive
approach to integers satisfying a bunch of nice
properties, I had no idea of where uniqueness
would be used in this proof. I took it for granted,
as I think anyone at a basic level would, once
existence of the factorization is given. How would
uniqueness then help?.
So I find the reference of uniqueness to be
obscure, and it does nothing to help me construct
a proof, while a reference to the irrationality
of sqr2 does help me. Now I don't know if this is
general, and applies to the OP, but I thought my
hint was less obscure than yours, maybe because
I tutor undergrads, and I don't get paid if they
don't understand what I say.
So I think my hint does add.
You now seem to think I was
> invoking high-falluting algebraic concepts, "assuming
> everyone is an algebraist". Your analogy about measure theory is simply misguided,
> and shows that you still have little or no clue what
> it was I was saying; the proof I alluded to was, SURPRISE! exactly the one you wrote: use unique factorization of integers into primes to write n as
> a product of prime powers,
O.K. Just to clarify: by unique factorization of n
you mean that n is expressible in the given form as
a product of primes with respective exponents, _and_
that this expression is unique, correct?.
Then uniqueness is somewhat high-fallutin to
a beginner, while existence goes closer to the
heart of it. And the ref. to the proof of the
irrationality of sqr2 (I haven't mixed up the
two for a while now!) gets even closer to it
in my view. You may be equating uniqueness
of factorization with existence. I am not, and
I did not when I read it.
I believe your initial position, in your first
response to me, is that your hint of "unique factorization" , was enough to lead someone to
our common proof, and that my hint of generalizing the
proof of the irrationality of sqr2 was then
redundant, correct?.
You believe that your hint of unique factorization
is enough to lead anyone to our common proof, and
that my comment does not give any additional hint
or guidance towards constructing such proof.
True that I should not have made ref. to your
hint, but I _do_ belive my "co-hint" was helpful,
and not redundant.
If this is not your position, please state so, (or pull a hair out: Serenity Now! :)).
Otherwise, I _have_ understood your position,
and I disagree with it: I will refer to this
position from now on, so please ignore if I
am not addressing your position effectively.
Talking about uniqueness when referring to integers
_does_ in my opinion seem to be overkill and obscure, when talking about basic properties of integers that most take for granted. This uniqueness of expression is something that anyone at a basic level would take for granted and that if the existence is assumed, uniqueness is something of a curiosity for the beginner (and I
have no way of knowing the level of the OP).
So uniqueness is just remotely associated to the
heart of the proof, unlike with existence of factorization. Now _you_ may consider these two
to be equivalent, but I am not so sure someone
else would; it didn't to me, and I would like to
believe I am not the dumbest one around.
Now, do you think that your term UF triggers this
argument in _any_ reader?. I would believe the FTA
would do it, if anything does it at all. I don't
believe it myself, but I may be wrong.
Try it with your students, see what happens,
how/if it helps.
It seems an unsoportable claim that my suggestion
for this proof will help no one that has read yours.
And you have not convinced me yet.
> rational square root if and only if it is a perfect square. Hence, my original point: you
> did not, in fact, add anything to the discussion
> except careless errors.
>
I insist that it is not necessary to invoque unique
factorization , even when it is being used.
At the level at which I encountered this proof
(beg. undergrad), I did not even have the training
to spot the issue of uniqueness of this expression
as mattering.
I knew little , if anything at all
about general rings, so I could not even conceive
that there could be a different expression. Maybe
I was/am just a simpleton. Who knows.
All that seemed to matter is that this expression
of n --just anyone expression -- was possible, and
this seemed feasible.
If I had forgotten how to do this proof after
encountering it , even if encountering many times,
a hint of " use unique factorization" , would have
done _nothing_ to jog my memory, since I did not
consider this assumption to be central; I may
not even have paid attention to the relevance of
factorization being unique to this proof. In fact,
it is only now that I link these two. I do not link
UF to FTA.
>
> > (This is absurdly long and explicit, but this
> seems necessary, since we seem to otherwise be unable
> to understand each other)
>
> I think I understand exactly what you were and are
> saying. What you seem to not be able to understand is exactly where and why you were wrong, even to the point of claiming that the assertion "sqrtn is
> irrational only if n is a perfect square" was
> 'factually correct'.
I do understand why "," is incorrect and that it
is incorrect. Just look at my correct
proof of the corrected fact " sqrn is rational
iff. n is a perfect square".
I _never_ claimed -- after admitting I carelessly
mixed 'rational' with 'irrational'-- that I believed this was factually correct. This was, as I said, careless on my part, not a result of a conceptual confusion.
So I _do_ understand where I was wrong in _carelessly
mixing_ , but not conceptually confusing these two. Again, my correct proof of the corrected
statement " sqrn rational iff. n a perfect square" backs me on this, and I doubt you can argue otherwise.
>
> > Again: your suggested proof is an analysis
> proof, mine was a calculus one.
>
> Again: no, you clearly misunderstood and continue to
> misunderstand what I wrote.
>
> > There is a difference in the "zooming" in our
> arguments:
>
> There is a difference between knowing what one is
> talking about and not really knowing much about what one is talking about. I'll let you figure out which is which.
>
I think my correct proof shows that I _do_
know what I am talking about, despite my carelessness in
exchanging of irrational for rational.
And so does my claim that it is the existence of this
factorization that matters, and that is more relevant
to the proof than the uniqueness, and that the uniqueness may be above the heads of those who have only worked with integers up to this point, for which
uniqueness will not likely ring a bell, however much
you seem to believe that anyone who hears UF will
consider the FTA .
If you had stated "Fund. Thm. of Arithmetic", that
would be a different story. However FTA may be linked
to UF in your mind, this is not so in everyone's mind.
> > Let's just consider ourselves lucky that we
> > do not have to work or live together, given that
> > we seem to be completely unable to communicate
> with
> > each other.
>
> There certainly is one of us who is completely unable
> to understand the other. It ain't me, though.
>
> --
> Arturo Magidin
Serenity Now!!
Gerry Myerson
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> >
> > Theorem 2. If m is an integer, and sqrt m is not an integer, then
> > sqrt m is irrational.
> >
> > Proof. Assuming sqrt m is rational but not an integer, let k be the
> > integer such that k < sqrt m < k+1, and let n be the smallest positive
> > integer whose product with sqrt m is an integer; then n(sqrt m - k) is
> > a smaller positive integer whose product with sqrt m is an integer,
> > contradiction.
> >
> > Theorem 5. Let f be a monic polynomial with integer coefficients. Let
> > alpha be a solution of f(alpha) = 0. Then if alpha is not an integer,
> > it is irrational.
> >
> > Proof. Assume the hypotheses, assume alpha is rational, and let n be
> > the smallest positive integer whose product with alpha^j is an integer
> > for all j less than the degree of f. Then n times the fractional part
> > of alpha is a smaller positive integer with the same property,
> > contradiction.
> >
> > The earliest publication I've seen of the proof given above for
> > Theorem 2 is the 1975 paper of Estermann. I haven't seen any
> > print publication of the proof of Theorem 5 given above, but it was
> > posted to this newsgroup in July 2000, and I've seen it on the web.
> > There may well have been earlier appearances of these proofs
> > in the form given here, and I'd like to see them.
>
> It took me only about 5 minutes to find an earlier publication [1]
> of Theorem 2. I recall seeing much older publications of both Theorem 1
> and Theorem 5.
>
> [1] Vaughan, Herbert E. On the Irrationality of Roots
> Amer. Math. Monthly, Vol. 67, #6 pp. 576-578
> http://www.jstor.org/stable/2309183
Thanks. If you come across older publications of this form of proof
of Theorems 2 and/or 5, I'd like to know.
Bill Dubuque
>
> False. Here is a proof of the irrationality of sqr2 that makes no
> explicit mention of any unique factorization. It uses it implicitly,
> but there is no need to make every implict assumption explicit.
> Just take a,b below to be integers and use basic number-theoretic
> properties of integers. :
>
> its lowest form, so that (a,b)=1 . No need to bring in a tank to
> kill a fly and talk about unique factorization. Make some assumptions
> that seem reasonable, if you are not doing algebra, as this is proven
> in analysis or beginning number theory classes. So assume (a,b)=1
> now square both sides to get: a^2/b^2 = 2 then a^2 = 2b^2
> Now 2 is a prime. So 2|a^2 means 2|a .
>
> No need to mention generalities about rings, even if these are being
> implicitly used. Just use a simple argument for a,b as natural numbers,
> not as elements of a UFD:
>
> spread between a and b. But p has only p and 1 as factors. It follows
> that p|a or p|b . Simple. No UFD's.
No, you're implicitly assuming atoms (irred's) p are prime in Z, i.e.
p|ab => p|a or p|b. That's trivially equivalent to unique factorization
in domains (like Z) that are atomic (i.e. elts factor into atoms).
> In our case we have 2|a^2 , or 2|a*a. Then 2|a . So a = 2k, for some
> integer k. Substitute back in original:
>
> a^2 = (2k)^2 = 2b^2, so 4k^2 = 2b^2
>
> Simplify, without stating that R is a field (so it is a UFD, but that
> is not necessary). Then 2k^2 = b^2. By an argument similar to the
> previous, it follows that 2|b . Now we get a contradiction: we have
> 2|a and 2|b, when we assumed (a,b)=1 . Then sqr2 is irrational.
That is the oldest and most well-known irrationality proof.
Since only one atom p = 2 is involved one can easily check that
p|a^2 => p|a by brute force checking the p cases a=0,1,...,p-1.
It's more difficult when one requires this property for _all_ p
as in the more general irrationality proof for sqrt(n) below.
> Now, for n non perfect square, express n = p1^e1 p2^e2...p_k^e_k
>
> Just take the fundamental thm. of arithmetic and assume this is
> possible for integers. No need for unique factorization.
> Consider sqr n, with n as above. Since n is not a perfect square, you
> may be able to factor out some terms from the expression for n, but I
> think it is clear that you will be left with at least one term in the
> root, a prime number with power 1. so you will have something like:
>
> sqr n = p1^(e1/2)...p_j^(e_j/2) sqr(p_{j+1}...p_k)
>
> An argument similar to that for sqr2 will show that sqr n is not
> rational. Let me know if you want me to expand on this.
But here one needs: for _all_ p, atoms p are prime. As I explained
above, this fact is trivially equivalent to unique factorization
> But I think that existence of a factorization is what matters the
> most, and not uniqueness. Since to me, taking a naive approach to
> integers satisfying a bunch of nice properties, I had no idea of
> where uniqueness would be used in this proof. I took it for granted,
> as I think anyone at a basic level would, once existence of the
> factorization is given. How would uniqueness then help?.
You seem confused because you didn't actually attempt to carry
out the proof for the general case - you barely even started it.
If you carry out such a proof you will soon see that it ends up
employing either atoms are prime or, equivalently, uniqueness.
Theorems require formal proof, not fuzzy hand-waving like your
above "I think it is clear... an argument similar... shows" etc.
In fact it's not existence but uniqueness that is crucial for the
proof. Indeed, as I mentioned elsewhere in this thread the proof
works for any gcd domain, and there are plenty of gcd domains not
atomic, i.e. where existence fails. E.g. consider the ring A of all
algebraic integers. A has no atoms since every elt splits x = /x /x
Thus no elt in A has a factorization into atoms. However, since A
admits gcds, the Rational Root Test holds in A, so the irrationality
proof works. Hence here existence plays absolutely no role at all.
--Bill Dubuque
KMF
Yes. But my position is that these observations are clear to you _after_ having had a lot of training in number theory and algebra. Without this training,
I don't think this would be trivial at all. Since
integers are relatively familiar, we can just
implicitly assume p|ab implies p|a or p|b.
When I first saw this proof, I never thought
twice about this fact (p|ab..) .
It just seemed to make
sense to me. I did not have the background then--
I had not taken yet any abstract algebra class--to understand that this may not be true in other rings.
I took it for granted, as it seemed believable. I
never considered it was necessary to address the
possible non-uniqueness of the expression.
Even today, p|ab implies p|a or p|b is not,
in my knowledge base, closely related to unique
factorization. Now, I don't doubt you when you tell
me that it is, it is just that, not being an algebraist, I don't make these connections.
I don't know, maybe it is different for you, or
for others.
I think this is trivial to you, but not to me. I
think you need advanced training to see these connections. Do you think that you would have made
this connection as a college freshman (assuming an
average math background)?.
I was able to go thru this proof, and remember it
up to today, without any awareness of this connection
with UFactorization.
Yes, your perspective is deeper , and more general,
but, my position is that I think it is possible to present this proof at a somewhat superficial level
without addressing all the assumptions being made,
and without trying to generalize to UFD's. , and that
a beginner may grasp. I intended to present this
type of proof, where p|ab -> p|a or p|b is just
stated and taken for granted, as it agrees with
one's (superficial) experience dealing with integers.
> > But I think that existence of a factorization is
> what matters the
> > most, and not uniqueness. Since to me, taking a
> naive approach to
> > integers satisfying a bunch of nice properties, I
> had no idea of
> > where uniqueness would be used in this proof. I
> took it for granted,
> > as I think anyone at a basic level would, once
> existence of the
> > factorization is given. How would uniqueness then
> help?.
>
> You seem confused because you didn't actually attempt
> to carry out the proof for the general case - you barely even started it.
> If you carry out such a proof you will soon see that
> it ends up employing either atoms are prime or, equivalently, uniqueness.
DO you mean showing it for any n non-square?
> Theorems require formal proof, not fuzzy hand-waving
> like your above "I think it is clear... an argument similar...shows" etc.
>
O.K. I did produce a general proof for any n.
I think it is not controversial to show that
assuming n has k prime factors, each to the first power, is enough; k=1,..,j , j a finite value.
Let me just do
k=2 , which I think generalizes the argument for any finite j. I'll do a similar proof, where I take certain
things for granted.
assume n=p_1p_2. Let a,b be integers with (a,b)=1
assume
a/b =sqr(p_1p_2), then
a^2/b^2 =p_1p_2
then a^2 =(p_1p_2)b^2
So p_1p_2|a^2
then p_1|a and p_2|a
proof of last statement:
if ~(p_1|a) , then the factors of p_1 are split
among the different copies of a. But only factors
of p_1 are p_1 and 1.
So p_1|a , and p_2|a . Then we show that it
follows that p_1|b , and p_2|b the same way as
with the proof of sqr2.
This arguemnt can no doubt be made more general,
and can be looked at more closely. I am doing handwaving, but this is how you learn at first.
But still, at an intro. level,
if you introduce all these details, you may lose your
audience. It is a tricky thing, I admit, to effectively
popularize mathematics, without trivializing it,
but I believed my proof was somewhat of a popularization. After the popular version is understood,
then any interested person may dig deeper, but I think
that dealing with every detail at once is too much to
swallow at once by the uninitiated. I was trying to
lay out a "training proof" that may be accessible
(while somewhat superficial), without being overwhelming.
I was tayloring my proof to beginners, which is
what I was when I first saw it.
Don't get me wrong; I think everyone should aim
to understand your proofs, and I prefer yours to
mine, but I believe mine is more accesible.
> In fact it's not existence but uniqueness that is
> crucial for the proof.
But this does not come up in my proof. I do not
explicitly state this, and I think that the overall
argument sounds convincing -- to the beginner.
Do you think you could take the perspective of
a beginner?.
Indeed, as I mentioned elsewhere in this thread the proof
> works for any gcd domain, and there are plenty of gcd
> domains not
> atomic, i.e. where existence fails. E.g. consider the
> ring A of all
> algebraic integers. A has no atoms since every elt
> splits x = /x /x
> Thus no elt in A has a factorization into atoms.
> However, since A
> admits gcds, the Rational Root Test holds in A, so
> the irrationality
> proof works. Hence here existence plays absolutely no
> role at all.
>
I don't doubt that, but I only intended to show
the truth of the statement for the integers, taking certain things for granted, and did
not intend for it to be a more general proof.
I think the OP was interested in the case for
n integer. Anyone more advanced would have
preferred yours and other posters' arguments,
no doubt. Cheep beer, vs. Champaign. Your
arguments get to the crux. Mine don't.
Do you think that a beginner would have understood your argument?.
It may just be a matter of tayloring the proof
to a certain audience. I agree that it would be ridiculous to present my proof to a grad.-level crowd,
wether a crowd of algebraists or people of other specialties.
I tried to give what I thought was a popularized
version of a subtle argument, purposefully ignoring
some of the subtleties, to avoid overwhelming the
beginners. Other, more advanced readers
would benefit more from yours and other poster's
presentation.
> --Bill Dubuque
Gerry Myerson
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> >
> > Readers may be interested in my paper, Irrationality via well-ordering,
> > Australian Math. Soc. Gazette 35 (2008) 121-125, available on the web
> > at http://www.austms.org.au/Publ/Gazette/2008/May08/Myerson.pdf
>
> A better title would be "Irrationality via the Division Algorithm"
> (or "via the Euclidean Algorithm") since your proofs there are simply
> well-known proofs of PID results that've been pulled-back to Euclidean
> domains by eliminating ideal-theoretic notions. More on this below.
You say this as if it's a bad thing. There is a lot to be said for
introducing ideal-theoretic notions, but there is also something to be
said for eliminating them, in order to present proofs to those not yet
ready for them.
> It is misleading to say the the proofs are "via well-ordering".
> Rather, the proofs depend crucially on a domain D being Euclidean.
> This requires not only a "size" map from D into a well-ordered set,
> but also, crucially, a Division Algorithm with respect to the map.
> Euclideaness is used implicitly in your proofs when you take the
> integer or fractional part of a fraction to help achieve a descent
> on possible denominators for w = sqrt(m) (or w an algebraic integer).
Granted. But people who are years away from learning about
Euclidean domains have no trouble accepting fractional parts.
Logically equivalent, but pedagogically not.
> > In that paper, you'll find,
> >
> > THEOREM 2. If m is an integer, and sqrt m is not an integer,
> > then sqrt m is irrational.
> >
> > Proof. Assuming w = sqrt m is rational but not an integer, let k be the
> > integer such that k < w < k+1, and let n be the smallest positive
> > integer such that nw is an integer; then n(w-k) is a smaller positive
> > integer whose product with sqrt m is an integer, contradiction.
>
> That's very well-known and very widely published.
Yes. In case I failed to make this clear, I don't and didn't claim any
originality here.
> > You'll also find the theorem stating that the integers
> > are integrally closed in the rationals;
> >
> > THEOREM 5. Let f be a monic polynomial with integer coefficients.
> > Let w be a solution of f(w) = 0. Then if w is not an integer,
> > it is irrational.
> >
> > Proof. Assume the hypotheses, assume w is rational, and let n be
> > the smallest positive integer whose product with w^j is an integer
> > for all j less than the degree of f. Then n times the fractional part
> > of w is a smaller positive integer with the same property, contradiction.
>
> Again that's very-well known and I do recall seeing it published at
> least a handful of times if not more.
I wouldn't be surprised (although I do wonder exactly how many proofs
there are in a handful) and would be pleased to have a reference.
I like your presentation of the proof that PIDs are integrally closed.
The ring of integers in a number field is integrally closed, but of
course need not be a PID - is there a similar proof for this theorem?
> > I ask in the paper for the earliest publications of these (and other)
> > proofs in the form in which I relate them. This form may well be
> > merely an elementary manifestation of the conductor ideal, but I
> > reckon there's a lot to be said for elementary manifestations. In
> > particular, students who are still years away from learning about
> > conductor ideals can still understand, maybe with a little help,
> > the proofs above.
>
> The same might be argued for analogous "elementary" proofs that've
> had their higher-order structure removed, e.g the famous proof [2]
> of unique factorization by Klappauf, Lindemann, Zermelo that I
> cited earlier in this thread. It isn't published too frequently
> (except as a curiosity) because it obfuscates the underlying
> algebraic structure and hence has little pedagogical value.
> Someone ignorant of the relevant algebraic theory might stumble
> upon such proofs with great effort. But someone familiar with
> the theory can _mechanically_ churn out these monstrosities by
> simply rewriting the higher-order proofs to use only the most
> primitive notions - eliminating structure (rings, ideals, etc.)
> Indeed, the process is so mechanical that one could easily
> program a computer to perform such proof transformations.
I'll have a look at the KLZ proof. Whether proofs of the type given in
my paper are mostrosities is, I think, a matter of taste. Certainly the
reaction of most people of my acquaintance to the well-ordering proof
of the irrationality of sqrt 2 is one of pleasure, not the revulsion you
might expect as a reaction to a monstrosity.
Bill Dubuque
>
> I like your presentation of the proof that PIDs are integrally closed.
> The ring of integers in a number field is integrally closed, but of
> course need not be a PID - is there a similar proof for this theorem?
As I mentioned, the proof by cancelling the conductor ideal
works also in a Dedekind domain because ideals are invertible
so cancellable. However, one can also proceed directly, e.g.
THEOREM Dedekind domains D (so PIDs) are integrally closed.
PROOF Suppose a fraction w over D is integral over D, i.e.
suppose w^n in (w^(n-1),...,w,1) [*]
Now (w,1)^n = (w^n,w^(n-1),...,w,1) as fractional ideals
= (w^(n-1),...,w,1) by [*]
Thus (w,1)^n = (w,1)^(n-1)
So (w,1) = (1) by cancelling invertible ideal (w,1)^(n-1)
=> w lies in (1) = D. Therefore D is integrally closed. QED
REMARK A common similar proof uses I^2 = I, I = (w^n,...,w,1).
COROLLARY GCD domains (so UFDs) are integrally closed.
PROOF The prior proof also applies here if one interprets
(a,b,...) not as an ideal but instead as gcd(a,b,...). QED
Those unfamiliar with _fractional_ ideals may clear denominators
from the prior proof by scaling it by b^n, where w = a/b, i.e.
(w,1)^n = (w,1)^(n-1) which, upon scaling through by b^n
=> (a,b)^n = b(a,b)^(n-1)
=> (a,b) = (b) by cancelling invertible ideal (a,b)^(n-1)
=> b|a in D => a/b in D
--Bill Dubuque
Martin Michael Musatov
V Forwarded conversation
Subject: Re: Injection and surjection
------------------------
From: Arturo Magidin <mag...@member.ams.org>
Date: Fri, Jun 12, 2009 at 7:07 PM
To: Musatov <marty....@gmail.com>
I have neither interest nor desire to exchange any messages with you. Please do not send me any e-mail again; I will consider future messages to constitute harassment, and will report them as such to your service provider.
----------
From: Martin Musatov <marty....@gmail.com>
Date: Fri, Jun 12, 2009 at 7:15 PM
To: Arturo Magidin <mag...@member.ams.org>
Yes, Sir. I am sorry if my mathematics upset you.
Sent from my Verizon Wireless BlackBerry
+++++++++++++++++++++++++++++++++++++++++++++++++++++