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| Neil B.... |
 Posted: Sun Mar 15, 2009 8:10 pm |
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Guest
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Here is my proof that roots of integers have to be either whole or
irrational. I worked it out many years ago (pi day yesterday reminded
me of this exercise.) I figure someone thought of this or similar before
but I haven't seen it elsewhere, simple as it is. It's a reductio like
the very old, other proof you see around. It's interesting in any case,
due to its great power but simplicity.
If not an integer, assume that m^(1/n) is a fraction a/b. Consider that
all integers can be represented as a unique product series of powers of
primes (e.g., 637197957 = 3^2*11*23^5, etc.) We can represent a/b as
maximally reduced, so the fraction would have to be representable in the
form (5^3*13*29^7*...)/(7*17^4*61^2*...) That is, no prime on the top
can also be on the bottom.
Then, consider squaring a/b (or cubing etc.) to attempt to result in 2,
or 11, etc. We'd have by exponent laws the same product series above as
before, but with each exponent times two (or three, etc.) Below we have
the same series as before with each exponent times two (or three, etc.)
Yet again, we can't divide them out to get an integer because there are
no matching primes to cancel out. (Remember, there is only one unique
representation in product series of powers of primes, of any given
integer. Therefore you can't say, "Suppose the series on top can *also*
be represented as the series below multiplied by the number we want to
get, so it cancels out for the number we want.") Therefore there is not
any a/b that could be the square root of two, or the fifth root of 77,
etc. (This is a more powerful proof than simply for one integer or
exponent.) Has anyone seen this before? |
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| Robert Israel... |
| Posted: Sun Mar 15, 2009 8:32 pm |
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Guest
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"Neil B." <neil_delver at (no spam) caloricmail.com> writes:
[quote:ffdc9a51e4]Here is my proof that roots of integers have to be either whole or
irrational. I worked it out many years ago (pi day yesterday reminded
me of this exercise.) I figure someone thought of this or similar before
but I haven't seen it elsewhere, simple as it is. It's a reductio like
the very old, other proof you see around. It's interesting in any case,
due to its great power but simplicity.
If not an integer, assume that m^(1/n) is a fraction a/b. Consider that
all integers can be represented as a unique product series of powers of
primes (e.g., 637197957 = 3^2*11*23^5, etc.) We can represent a/b as
maximally reduced, so the fraction would have to be representable in the
form (5^3*13*29^7*...)/(7*17^4*61^2*...) That is, no prime on the top
can also be on the bottom.
Then, consider squaring a/b (or cubing etc.) to attempt to result in 2,
or 11, etc. We'd have by exponent laws the same product series above as
before, but with each exponent times two (or three, etc.) Below we have
the same series as before with each exponent times two (or three, etc.)
Yet again, we can't divide them out to get an integer because there are
no matching primes to cancel out. (Remember, there is only one unique
representation in product series of powers of primes, of any given
integer. Therefore you can't say, "Suppose the series on top can *also*
be represented as the series below multiplied by the number we want to
get, so it cancels out for the number we want.") Therefore there is not
any a/b that could be the square root of two, or the fifth root of 77,
etc. (This is a more powerful proof than simply for one integer or
exponent.) Has anyone seen this before?
[/quote:ffdc9a51e4]
Well, yes, this is the basic idea of the standard proof that an algebraic
integer that is rational is an (ordinary) integer.
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| William Elliot... |
| Posted: Sun Mar 15, 2009 8:42 pm |
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On Sun, 15 Mar 2009, Neil B. wrote:
[quote:4ce37de0bb]roots of integers have to be either whole or irrational.
[/quote:4ce37de0bb]
Assume r in N, r^(1/n) in Q. Thus some a,b in N with r^(1/n) = a/b.
rb^n = a^n; b^n | a^n; b | a (by simple theorem)
some k in N with a = kb; r = k^n, QED. |
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| Neil B.... |
| Posted: Sun Mar 15, 2009 8:45 pm |
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"Robert Israel" <israel at (no spam) math.MyUniversitysInitials.ca> wrote in message
news:rbisrael.20090316012607$63db at (no spam) news.acm.uiuc.edu...
[quote:db9b616bb0]"Neil B." <neil_delver at (no spam) caloricmail.com> writes:
Here is my proof that roots of integers have to be either whole or
irrational. I worked it out many years ago (pi day yesterday
reminded
me of this exercise.) I figure someone thought of this or similar
before
but I haven't seen it elsewhere, simple as it is. It's a reductio
like
the very old, other proof you see around. It's interesting in any
case,
due to its great power but simplicity.
If not an integer, assume that m^(1/n) is a fraction a/b. Consider
that
all integers can be represented as a unique product series of powers
of
primes (e.g., 637197957 = 3^2*11*23^5, etc.) We can represent a/b as
maximally reduced, so the fraction would have to be representable in
the
form (5^3*13*29^7*...)/(7*17^4*61^2*...) That is, no prime on the top
can also be on the bottom.
Then, consider squaring a/b (or cubing etc.) to attempt to result in
2,
or 11, etc. We'd have by exponent laws the same product series above
as
before, but with each exponent times two (or three, etc.) Below we
have
the same series as before with each exponent times two (or three,
etc.)
Yet again, we can't divide them out to get an integer because there
are
no matching primes to cancel out. (Remember, there is only one unique
representation in product series of powers of primes, of any given
integer. Therefore you can't say, "Suppose the series on top can
*also*
be represented as the series below multiplied by the number we want
to
get, so it cancels out for the number we want.") Therefore there is
not
any a/b that could be the square root of two, or the fifth root of
77,
etc. (This is a more powerful proof than simply for one integer or
exponent.) Has anyone seen this before?
Well, yes, this is the basic idea of the standard proof that an
algebraic
integer that is rational is an (ordinary) integer.
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
[/quote:db9b616bb0]
Robert, thanks for the prompt reply. I checked around, and found the
following from Wikpedia article
http://en.wikipedia.org/wiki/Irrationals#General_roots:
"
General roots
The proof above for the square root of two can be generalized using the
fundamental theorem of arithmetic which was proved by Gauss in 1798.
This asserts that every integer has a unique factorization into primes.
Using it we can show that if a rational number is not an integer then no
integral power of it can be an integer, as in lowest terms there must be
a prime in the denominator which does not divide into the numerator
whatever power each is raised to. Therefore if an integer is not an
exact kth power of another integer then its kth root is irrational.
"
That is indeed the proof I independently discovered and described above.
(I'm still proud of doing that, it's no big deal but lets me know I have
savvy for such things however humbly.)
So I still wonder a few things:
Did someone, like Gauss, think of this proof about the irrationality of
roots soon after discovery of the FTOA? Who, when, etc. - note Wikipedia
says "can be ..." not who actually first said so. (I suppose someone
needs to say so and get credit thereby, however "obvious" of an
implication follows from something?) That leads into my second question,
why doesn't this (apparently, to me anyway) "get around" more in
discussions of the history of math? They talk a lot about the old proof,
but not IIRC the better one referred in Wikipedia. |
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| achille... |
| Posted: Sun Mar 15, 2009 11:22 pm |
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On Mar 16, 4:40 pm, David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
[quote:38f3117663]On Sun, 15 Mar 2009 18:42:08 -0700, William Elliot
Anyway:
rb^n = a^n; b^n | a^n; b | a (by simple theorem)
Do you know a complete proof of this "simple
theorem"?
What do you mean? I though these are well known facts:[/quote:38f3117663]
If factorization into prime factors EXIST and is UNIQUE,
then matching the prime factors of b^n with those in a^n
immediately give b | a.
AND factorization is unique over Z because of the
existence of Euclid algorithm. |
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| achille... |
| Posted: Mon Mar 16, 2009 12:14 am |
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On Mar 16, 6:08 pm, David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
[quote:b192d03ae2]On Mon, 16 Mar 2009 02:22:58 -0700 (PDT), achille
achille_... at (no spam) yahoo.com.hk> wrote:
On Mar 16, 4:40 pm, David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Sun, 15 Mar 2009 18:42:08 -0700, William Elliot
Anyway:
rb^n = a^n; b^n | a^n; b | a (by simple theorem)
Do you know a complete proof of this "simple
theorem"?
What do you mean?
A _question_ doesn't "mean" anything. I was curious
whether William knew how to prove the result he
called a "simple theorem". (Because he seemed to be
saying the result was entirely trivial; I suspect it's
less trivial than he thought, because it depends on
unique factorization...)
I though these are well known facts:
I _wasn't_ curious whether _you_ knew how to prove it...
If factorization into prime factors EXIST and is UNIQUE,
then matching the prime factors of b^n with those in a^n
immediately give b | a.
AND factorization is unique over Z because of the
existence of Euclid algorithm.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
[/quote:b192d03ae2]
Okay, sorry to alienate you, Dave. |
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| Dave L. Renfro... |
| Posted: Mon Mar 16, 2009 3:27 am |
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Neil B. wrote (in part):
[quote:64cc6e49c4]Did someone, like Gauss, think of this proof about the
irrationality of roots soon after discovery of the FTOA?
Who, when, etc. - note Wikipedia says "can be ..." not who
actually first said so. (I suppose someone needs to say so
and get credit thereby, however "obvious" of an implication
follows from something?) That leads into my second question,
why doesn't this (apparently, to me anyway) "get around"
more in discussions of the history of math? They talk a lot
about the old proof, but not IIRC the better one referred
in Wikipedia.
[/quote:64cc6e49c4]
For what it's worth, I've got several folders at home full
of several hundred publications (this includes letters to
the editor, short notes, book reviews, etc. in addition to
actual papers) relating to irrationality proofs of n'th roots
of integers, going back to the early 1800s, so this is a topic
that's been mined quite extensively in the various "elementary
journals" (the 3 MAA journals, Mathematical Gazette, Mathematics
Teacher, Nouvelles Annales de Mathematiques, Journal de Mathematiques
Elementaires, Mathematical Intelligencer, Elemente der Mathematik,
Mathesis Recueil Mathematique, etc. to name just a few). However,
these types of publications seem to have a "life-span/memory"
in the mathematical community of no more than a decade or two
(and this pertains to the small subset of the mathematical
community that reads/read any of these journals), so the same
ideas keep coming up over and over again (not just your idea,
which is one of the more elegant and mathematically interesting
proofs and thus "well known" to most mathematicians).
Incidentally, Theodoros (ancient Greece) proved that each of
the non-square integers from 2 to 17 were irrational. I don't
recall if his method has come down to us, but I do remember
reading that he supposedly treated each of these one-by-one
(this being, I think, from others living during Theodoros
time whose writings have come down to us), and I've got several
papers (at home, so I can't cite reference information right
now) that offer possible explanations, the most common of which
is (I believe) is that the Greeks didn't have the concept of
unique factorization (or if they did, they didn't have what
they considered an air-tight proof -- remember, Greek mathematics
was almost entirely geometrical; they didn't view numbers like
sqrt(2) the same way that we do, but rather as something akin
to lengths (and their ratios) in geometrical figures). In fact,
now that I think about it, I believe there is a discussion of
Theodoros and unique factorization somewhere in Hardy/Wright's
"The Theory of Numbers".
Dave L. Renfro |
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| David C. Ullrich... |
| Posted: Mon Mar 16, 2009 3:40 am |
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Guest
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On Sun, 15 Mar 2009 18:42:08 -0700, William Elliot
<marsh at (no spam) rdrop.remove.com> wrote:
[quote:afc3aa2cdb]On Sun, 15 Mar 2009, Neil B. wrote:
roots of integers have to be either whole or irrational.
Assume r in N, r^(1/n) in Q. Thus some a,b in N with r^(1/n) = a/b.
[/quote:afc3aa2cdb]
I really wish you'd either write in complete sentences or
stop complaining to others about their grammar.
Anyway:
[quote:afc3aa2cdb]rb^n = a^n; b^n | a^n; b | a (by simple theorem)
[/quote:afc3aa2cdb]
Do you know a complete proof of this "simple
theorem"?
[quote:afc3aa2cdb]some k in N with a = kb; r = k^n, QED.
[/quote:afc3aa2cdb]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| David C. Ullrich... |
| Posted: Mon Mar 16, 2009 4:30 am |
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On Sun, 15 Mar 2009 21:45:35 -0400, "Neil B."
<neil_delver at (no spam) caloricmail.com> wrote:
[quote:00cf085571]
"Robert Israel" <israel at (no spam) math.MyUniversitysInitials.ca> wrote in message
news:rbisrael.20090316012607$63db at (no spam) news.acm.uiuc.edu...
"Neil B." <neil_delver at (no spam) caloricmail.com> writes:
Here is my proof that roots of integers have to be either whole or
irrational. I worked it out many years ago (pi day yesterday
reminded
me of this exercise.) I figure someone thought of this or similar
before
but I haven't seen it elsewhere, simple as it is. It's a reductio
like
the very old, other proof you see around. It's interesting in any
case,
due to its great power but simplicity.
If not an integer, assume that m^(1/n) is a fraction a/b. Consider
that
all integers can be represented as a unique product series of powers
of
primes (e.g., 637197957 = 3^2*11*23^5, etc.) We can represent a/b as
maximally reduced, so the fraction would have to be representable in
the
form (5^3*13*29^7*...)/(7*17^4*61^2*...) That is, no prime on the top
can also be on the bottom.
Then, consider squaring a/b (or cubing etc.) to attempt to result in
2,
or 11, etc. We'd have by exponent laws the same product series above
as
before, but with each exponent times two (or three, etc.) Below we
have
the same series as before with each exponent times two (or three,
etc.)
Yet again, we can't divide them out to get an integer because there
are
no matching primes to cancel out. (Remember, there is only one unique
representation in product series of powers of primes, of any given
integer. Therefore you can't say, "Suppose the series on top can
*also*
be represented as the series below multiplied by the number we want
to
get, so it cancels out for the number we want.") Therefore there is
not
any a/b that could be the square root of two, or the fifth root of
77,
etc. (This is a more powerful proof than simply for one integer or
exponent.) Has anyone seen this before?
Well, yes, this is the basic idea of the standard proof that an
algebraic
integer that is rational is an (ordinary) integer.
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Robert, thanks for the prompt reply. I checked around, and found the
following from Wikpedia article
http://en.wikipedia.org/wiki/Irrationals#General_roots:
"
General roots
The proof above for the square root of two can be generalized using the
fundamental theorem of arithmetic which was proved by Gauss in 1798.
This asserts that every integer has a unique factorization into primes.
Using it we can show that if a rational number is not an integer then no
integral power of it can be an integer, as in lowest terms there must be
a prime in the denominator which does not divide into the numerator
whatever power each is raised to. Therefore if an integer is not an
exact kth power of another integer then its kth root is irrational.
"
That is indeed the proof I independently discovered and described above.
(I'm still proud of doing that, it's no big deal but lets me know I have
savvy for such things however humbly.)
So I still wonder a few things:
Did someone, like Gauss, think of this proof about the irrationality of
roots soon after discovery of the FTOA?
[/quote:00cf085571]
I don't know about the history, but if it was not known before Gauss
then Gauss knew it about 1.5 seconds after he finished proving FTOA.
[quote:00cf085571]Who, when, etc. - note Wikipedia
says "can be ..." not who actually first said so. (I suppose someone
needs to say so and get credit thereby, however "obvious" of an
implication follows from something?) That leads into my second question,
why doesn't this (apparently, to me anyway) "get around" more in
discussions of the history of math? They talk a lot about the old proof,
but not IIRC the better one referred in Wikipedia.
[/quote:00cf085571]
I suspect the reason is that for most people in the intended audience
the distinction is going to be boring - we've shown sqrt(2) is
irrational, great, now what's the big deal about the cube root
of 17?
In fact it took me a minute just now to convince myself that the
general result really _is_ deeper than the fact that sqrt(2) is
irrational. To make the proof work we need to know this:
(*) If p is prime and p divides ab then p divides a or p divides b.
That seems obvious, but giving an actual proof is not quite
trivial if you haven't seen it before. And come to think of it,
(*) is precisely equivalent to FTOA (it's trivial that FTOA
implies (*), while if you know (*) then FTOA follows by
a simple induction) so if someone doesn't know FTOA then
he almost certainly doesn't know how to prove (*) and
hence he doesn't know how to make the proof of the
general fact about roots of integers work.
Of course (*) is trivial for p = 2 (if neither a nor b is divisible
by 2 then a = 2n+1 and b = 2m+1 so ab = 2(2nm+n+m)+1),
and for any _specific_ p one could prove (*) by enumerating
cases (for example, if neither a nor b is divisible by 3 then
a = 3n+1 or 3n+2 and b = 3m+1 or 3m+2, leaving four
possibilities for ab...), but that doesn't give a proof in general.
So it really _is_ possible for someone to know the proof that
sqrt(2) is irrational without knowing a proof of the general
result about roots of integers. Since Gauss is credited with
FTOA, although I don't know for sure I'd be willing to bet
he was the first to give a proof of (*) and hence a proof
of the result about roots of integers.
In fact suddenly I'm irritated by the fact that I'm not certain
I can give the proof of (*) that I saw many many years ago...
Ok, let's see. First, if I is a set of integers, we say that I is
an "ideal" if (i) x + y is in I for every x, y in I and
(ii) nx is in I for every x in I and every integer n.
So {0} is an ideal, the "trivial" ideal. And if d is any
positive integer then I_d, the set of all multiples of d,
is an ideal.
Note that if I is a non-trivial ideal then I contains
some positive integer, since -x is in I for all x in I.
Lemma 1. Suppose that I is a non-trivial ideal. Let
d be the smallest positive element of I. Then
I = I_d.
Proof: Since d is in I it is clear from the definition that
I_d is a subset of I. Suppose on the other hand that
x is in I but not in I_d. Then we have
x = nd + r
where 1 <= r <= d-1.
Since d is in I it follows that r is in I, contradicting
the fact that d is the smallest positive element of I.
QED.
(Details for why it follows that r is in I: Since
d is in I, property (ii) shows that -nd is in I.
Also x is in I, and hence property (i) shows that
r = x -nd is in I.)
Lemma 2. Suppose x and y are positive integers with
no (positive) common factor (other than 1). Then
there exist integers n, m with nx + my = 1.
Proof: Let I be the set of all numbers of the form
nx + my, where n and m range over integers. It's
easy to see that I is an idea. Say d is the smallest
positive element of I.
Lemma 1 shows that I equals the set of all integer
multiples of d. Since x is in I, d is a factor of x.
Similarly d is a factor of y, so the fact that x and
y have no common factor greater than 1 shows that
d = 1. So 1 is in I, which is what we wanted to prove.
QED.
(Note the "Euclidean algorithm" gives a way to
actually find n and m, also giving a more constructive
proof of the lemma.)
Proposition. If p is prime, a and b are positive
integers, and p divides ab then p divides a or
p divides b.
Proof: Suppose that p does not divide a. Then
a and p have no common factor larger than 1
(since the only factor of p larger than 1 is
p itself, and p is not a factor of a).
So Lemma 2 shows that there exist integers
n, m with np + ma = 1. Now we should be
done...
Say
kp = ab.
Then
mkp = mab = (1 - np)b,
so
(mk + nb)p = b,
showing that p divides b.
QED.
There, I feel better now.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| David C. Ullrich... |
| Posted: Mon Mar 16, 2009 5:08 am |
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Guest
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On Mon, 16 Mar 2009 02:22:58 -0700 (PDT), achille
<achille_hui at (no spam) yahoo.com.hk> wrote:
[quote:58f0518a8a]On Mar 16, 4:40 pm, David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Sun, 15 Mar 2009 18:42:08 -0700, William Elliot
Anyway:
rb^n = a^n; b^n | a^n; b | a (by simple theorem)
Do you know a complete proof of this "simple
theorem"?
What do you mean?
[/quote:58f0518a8a]
A _question_ doesn't "mean" anything. I was curious
whether William knew how to prove the result he
called a "simple theorem". (Because he seemed to be
saying the result was entirely trivial; I suspect it's
less trivial than he thought, because it depends on
unique factorization...)
[quote:58f0518a8a]I though these are well known facts:
[/quote:58f0518a8a]
I _wasn't_ curious whether _you_ knew how to prove it...
[quote:58f0518a8a]If factorization into prime factors EXIST and is UNIQUE,
then matching the prime factors of b^n with those in a^n
immediately give b | a.
AND factorization is unique over Z because of the
existence of Euclid algorithm.
[/quote:58f0518a8a]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| stephen.p.harris... |
| Posted: Mon Mar 16, 2009 6:01 am |
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Guest
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David C. Ullrich wrote:
[quote:c0403acf50]
I suspect the reason is that for most people in the intended audience
the distinction is going to be boring - we've shown sqrt(2) is
irrational, great, now what's the big deal about the cube root
of 17?
[/quote:c0403acf50]
http://everything2.com/index.pl?node=The%20square%20root%20of%20any%20prime%20number%20is%20irrational
I think it is interesting that the square root of any prime has been
proven to be irrational because it provides a means of generating an
infinity of different initial finite segments. |
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| Frederick Williams... |
| Posted: Mon Mar 16, 2009 7:28 am |
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"David C. Ullrich" wrote:
[quote:3fcf76ca48]A _question_ doesn't "mean" anything.
[/quote:3fcf76ca48]
If so, what are we to make of it when people ask them?
--
Science is a differential equation.
Religion is a boundary condition.
--Alan Turing |
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| William Elliot... |
| Posted: Mon Mar 16, 2009 9:01 am |
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On Mon, 16 Mar 2009, David C. Ullrich wrote:
[quote:216bcb01be]On Sun, 15 Mar 2009, Neil B. wrote:
roots of integers have to be either whole or irrational.
Assume r in N, r^(1/n) in Q. Thus some a,b in N with r^(1/n) = a/b.
I really wish you'd either write in complete sentences or
stop complaining to others about their grammar.
I'm not giving proofs, only outlines or sketches of proofs[/quote:216bcb01be]
with details left to the reader. Details like the reasons.
I strive to be complete and detailed. Complex logic is
better expressed in pseudo-symbolic logic than English.
You have no complaint about my writing.
[quote:216bcb01be]Anyway:
rb^n = a^n; b^n | a^n; b | a (by simple theorem)
Do you know a complete proof of this "simple theorem"?
No. As I've said before, only an outline or sketch.[/quote:216bcb01be]
Will you be able to fill in the details?
BTW, simple is not trivial. For example
dx^2 / dx = 2x
is simple and not trivial.
The theorem depends upon the equally not trivial simple theorem
(a^n,b^n) = (a,b)^n
which depends upon
(r^n,s^n) = 1
where
r = a/(a,b), s = b/(a,b)
and
(ku,kv) = k(u,v)
which of course isn't scalar vector multiplication
but multiplication of a positive integer by a gcf.
>> some k in N with a = kb; r = k^n, QED. |
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| Patrick Hamlyn... |
| Posted: Mon Mar 16, 2009 5:48 pm |
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Frederick Williams <frederick.williams2 at (no spam) tesco.net> wrote:
[quote:bb428041cf]"David C. Ullrich" wrote:
A _question_ doesn't "mean" anything.
If so, what are we to make of it when people ask them?
[/quote:bb428041cf]
Now you're just taking his comment out of context. To put it back in context,
you should ask
"If so, what are we to make of it when David C. Ullrich asks them?"
which apparently has a whole 'nother answer.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms-subscribe at (no spam) egroups.com) |
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| Bill Dubuque... |
| Posted: Mon Mar 16, 2009 9:53 pm |
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Robert Israel <israel at (no spam) math.MyUniversitysInitials.ca> wrote:
[quote:7790a0db84]"Neil B." <neil_delver at (no spam) caloricmail.com> writes:
Here is my proof that roots of integers have to be either whole or
irrational. I worked it out many years ago (pi day yesterday reminded
me of this exercise.) I figure someone thought of this or similar before
but I haven't seen it elsewhere, simple as it is. It's a reductio like
the very old, other proof you see around. It's interesting in any case,
due to its great power but simplicity.
If not an integer, assume that m^(1/n) is a fraction a/b. Consider that
all integers can be represented as a unique product series of powers of
primes (e.g., 637197957 = 3^2*11*23^5, etc.) We can represent a/b as
maximally reduced, so the fraction would have to be representable in the
form (5^3*13*29^7*...)/(7*17^4*61^2*...) That is, no prime on the top
can also be on the bottom.
Then, consider squaring a/b (or cubing etc.) to attempt to result in 2,
or 11, etc. We'd have by exponent laws the same product series above as
before, but with each exponent times two (or three, etc.) Below we have
the same series as before with each exponent times two (or three, etc.)
Yet again, we can't divide them out to get an integer because there are
no matching primes to cancel out. (Remember, there is only one unique
representation in product series of powers of primes, of any given
integer. Therefore you can't say, "Suppose the series on top can *also*
be represented as the series below multiplied by the number we want to
get, so it cancels out for the number we want.") Therefore there is not
any a/b that could be the square root of two, or the fifth root of 77,
etc. (This is a more powerful proof than simply for one integer or
exponent.) Has anyone seen this before?
Well, yes, this is the basic idea of the standard proof that an
algebraic integer that is rational is an (ordinary) integer.
[/quote:7790a0db84]
Really? Precisely what do you think is that "basic idea"?
--Bill Dubuque |
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