Can you solve this math problem? Please help .
dan.ms...@gmail.com
2 ^ k is a natural number
3 ^ k is a natural number .
Prove that k is always a natural number , or disprove it .
William Elliot
2/3 is alegbraic and /= 0,1. By theorem by Gelfand
If k is not rational, then (2/3)^k is transcendental.
However (2/3)^k is rational, thus k is rational.
Thus without undue effort k is positive integer.
Anon
For the OP's benefit, that should be Gelfond. Ref:
<http://mathworld.wolfram.com/GelfondsTheorem.html>
OwlHoot
On Mar 2, 10:01 am, William Elliot <ma...@hevanet.remove.com> wrote:
That does the job, but I wonder if there's a self-contained
elementary proof.
On the contrary assumption that k is irrational, none of
2k, 3k, .. can be integers but denoting the floor and
fractional part of m.k resp by:
n, e := [m.k], {m.k}
there must be values of m for which e > 0 is very small,
and theorems of Diophantine approximation quantify how
small it can and does get for given values of m. Here's
a good online reference:
http://www.math.jussieu.fr/~miw/articles/pdf/IntroductionDiophantineMethods.pdf
So denoting 2^k = p and 3^k = q this would imply:
2^e, 3^e = p^m / 2^n, q^m / 3^n
in which 2^e and 3^e are greater than 1 but very close
to it (and of course rational).
If for suitable e these were close enough to 1 then we
could conclude
p^m, q^m = 2^n + 1, 3^n + 1
giving a contradiction, because if 2^(n + e) = 2^n + 1
then 3^(n + e) = 3^n + 1 + "something > 0".
Cheers
John R Ramsden
jhnr...@yahoo.co.uk
>
> That does the job, but I wonder if there's a self-contained
> elementary proof.
>
> On the contrary assumption that k is irrational, none of
> 2k, 3k, .. can be integers but denoting the floor and
> fractional part of m.k resp by:
>
> n, e := [m.k], {m.k}
>
> there must be values of m for which e > 0 is very small,
> and theorems of Diophantine approximation quantify how
> small it can and does get for given values of m. Here's
> a good online reference:
>
>
> So denoting 2^k = p and 3^k = q this would imply:
>
> 2^e, 3^e = p^m / 2^n, q^m / 3^n
>
> in which 2^e and 3^e are greater than 1 but very close
> to it (and of course rational).
>
> If for suitable e these were close enough to 1 then we
> could conclude
>
> p^m, q^m = 2^n + 1, 3^n + 1
>
> giving a contradiction, because if 2^(n + e) = 2^n + 1
> then 3^(n + e) = 3^n + 1 + "something > 0".
Using Google Groups at work, I can see from the headers that Gerry
Myerson has posted in this thread, but when I click on the header
his post doesn't appear. Unbelievable! (Well actually all too
believable, given that we're talking about Google Groups.)
Hopefully it was clear the above was only meant as a sketch of a
possible proof approach rather than a sloppy attempt at a proof
itself.
But on further thought, it doesn't seem like a goer, because 2^n
will increase so much faster with n than 2^min({n.k}) typically
is likely to decrease towards 1.
However, the result follows if one can prove that assuming k is
irrational implies integers p, q such that 0 < |2^(pk) - 3^(qk)| < 1
Cheers
John Ramsden
no comment
> On Sat, 1 Mar 2008 dan.ms.ch...@gmail.com wrote:
>
> >kis a positive real number .
> > 2 ^kis a natural number
> > Prove thatkis always a natural number , or disprove it .
>
> 2/3is alegbraic and /= 0,1. By theorem by Gelfand
> Ifkis not rational, then (2/3)^kis transcendental.
>
> However (2/3)^kis rational, thuskis rational.
> Thus without undue effortkis positive integer.
Hello,
This will not quite work. The theorem of Gel'fond (and others) has
not been quoted correctly. The theorem implies that if k is ALGEBRAIC
and not rational, then (2/3)^k is transcendental. But it says nothing
about what happens if k is not algebraic. And indeed it is easy to
see by continuity that for example there is a k such that (2/3)^k =
1/2. The k happens to be transcendental.
One can prove the result using more delicate results due to Baker. I
remember also seeing an "elementary" solution to the 2^k = a, 3^k = b
problem, but do not recall the details.
Gerry Myerson
<d4bcea2d-c754-4885...@e6g2000prf.googlegroups.com>,
no comment <adler...@gmail.com> wrote:
-> On Mar 2, 2:01 am, William Elliot <ma...@hevanet.remove.com> wrote:
-> > On Sat, 1 Mar 2008 dan.ms.ch...@gmail.com wrote:
-> >
-> > >kis a positive real number .
-> > > 2 ^kis a natural number
-> > >3^kis a natural number .
-> > > Prove thatkis always a natural number , or disprove it .
-> >
-> > 2/3is alegbraic and /= 0,1. By theorem by Gelfand
-> > Ifkis not rational, then (2/3)^kis transcendental.
-> >
-> > However (2/3)^kis rational, thuskis rational.
-> > Thus without undue effortkis positive integer.
->
-> Hello,
->
-> This will not quite work. The theorem of Gel'fond (and others) has
-> not been quoted correctly. The theorem implies that if k is ALGEBRAIC
-> and not rational, then (2/3)^k is transcendental. But it says nothing
-> about what happens if k is not algebraic. And indeed it is easy to
-> see by continuity that for example there is a k such that (2/3)^k =
-> 1/2. The k happens to be transcendental.
->
-> One can prove the result using more delicate results due to Baker. I
-> remember also seeing an "elementary" solution to the 2^k = a, 3^k = b
-> problem, but do not recall the details.
Get yourself to a hypnotherapist immediately,
so you can recover those details - nobody else has seen a solution,
elementary or otherwise.
A little personal note. Problem A6 on the 1971 Putnam exam went,
Let c be a real number such that n^c is an integer for every positive
integer n. Show that c is a non-negative integer.
I took that exam, and when I got to that question I said to myself,
surely if 2^c and 3^c are integers then c is a non-negative integer,
and I spent - wasted - a fair bit of time trying to solve A6 by
proving the stronger statement.
After the exam, I asked Serge Lang about the problem. He told me
that the 2^c, 3^c problem was still unsolved, and that it had only
been proved in the last few years that if 2^c, 3^c, and 5^c are all
integers then c is a non-negative integer. No wonder I hadn't been
able to make any progress on the problem!
Robert Israel will remember that 1971 Putnam exam - he finished
in the top 5 that year.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
jhnr...@yahoo.co.uk
>
> On Mar 2, 2:01 am, William Elliot <ma...@hevanet.remove.com> wrote:
> >
> > However (2/3)^kis rational, thuskis rational.
> > Thus without undue effortkis positive integer.
>
> not been quoted correctly. The theorem implies that if k is ALGEBRAIC
> and not rational, then (2/3)^k is transcendental. But it says nothing
> about what happens if k is not algebraic. And indeed it is easy to
> see by continuity that for example there is a k such that (2/3)^k =
> 1/2. The k happens to be transcendental.
We can obviously rule out 2^k being an integer power of 2, or 3^k
an integer power of 3, because in either of these cases the result
follows by definition.
Also, generally if 2^p || 2^k then p <= k, and similarly 3^q || 3^k
means q <= k. But otherwise, on the assumption that k is irrational,
you can conclude nothing about GCDs or largest powers dividing this
and that, and so on.
Gelfond & Schneider's result also means we can rule out the
following for irrational k and integers p, q, r, s all >= 0:
2^k, 3^k = 2^p.3^q, 2^r.3^s
because dividing each by 2^p, 2^r resp, and taking logs gives:
(k - p).log(2), (k - s).log(3) = q.log(3), r.log(2)
and multiplying these and dividing throughout by log(2).log(3),
which is non-zero gives:
(k - p)(k - s) - q.r = 0
This means k is either rational (at worst half-integral), in
which case William Elliot is right and the result follows, or
at most a quadratic irrational, and hence not transcendental
and consequently the G-S Theorem applies.
So if k is irrational we must have:
2^k, 3^k = 2^p.3^q.(6.u +/- 1), 2^r.3^s.(6v +/- 1)
with at least one of u, v > 1, and for independent sign options
(requiring the +ve sign only if the corresponding multiple of 6
is zero).
So somehow just knowing that one of these factors must be > 1
should be relevant to a proof of the general result.
> One can prove the result using more delicate results due to Baker. I
> remember also seeing an "elementary" solution to the 2^k = a, 3^k = b
> problem, but do not recall the details.
It's easy to express conditions as forms with integer coefficients
and log unknowns. But the snag is none of those forms are linear
in the logs, which I believe the results of Alan Baker & colleagues
(and people since?) all require. So has there been any progress,
or even a start made, on "log-quadratic" forms I wonder?
Cheers
John R Ramsden
jhnr...@yahoo.co.uk
>
> [..]
>
> So if k is irrational we must have:
>
> 2^k, 3^k = 2^p.3^q.(6.u +/- 1), 2^r.3^s.(6v +/- 1)
>
> with at least one of u, v > 1,
annoying niggle: that should be "with at least one of u, v > 0"
jhnr...@yahoo.co.uk
> following for irrational k and integers p, q, r, s all >= 0:
>
> 2^k, 3^k = 2^p.3^q, 2^r.3^s
>
> because dividing each by 2^p, 2^r resp, and taking logs gives:
>
> (k - p).log(2), (k - s).log(3) = q.log(3), r.log(2)
>
> and multiplying these and dividing throughout by log(2).log(3),
> which is non-zero gives:
>
> (k - p)(k - s) - q.r = 0
I'd bet a pound to a pinch of pig sh*t the correct approach
to prove the general result is a generalization of the above,
i.e. by showing that for any irrational k there is a set of
n >= 2 distinct primes {p_i} (including 2 and 3) and a set
of distinct positive integers {n_i} such that for integers
m_i >= 0 :
a^{k.n_i} = prod{j = 1..n} (p_i^m_i} (a = 2 or 3)
Taking logs gives a set of n homogenous linear equations in
the n distinct non-zero unknowns log(p_i} and with every
coefficient of the form either p.k + q with p, q integers
(with possibly p = 0 and q <= 0)
This in turn requires the vanishing of a determinant which
is a polynomial in k with integer coefficients, and leading
coefficient prod(n_i) which is non-zero. So k is algebraic.
Cheers
John R Ramsden
Robert Israel
wrote:
> A little personal note. Problem A6 on the 1971 Putnam exam went,
>
> Let c be a real number such that n^c is an integer for every positive
> integer n. Show that c is a non-negative integer.
>
> I took that exam, and when I got to that question I said to myself,
> surely if 2^c and 3^c are integers then c is a non-negative integer,
> and I spent - wasted - a fair bit of time trying to solve A6 by
> proving the stronger statement.
>
> After the exam, I asked Serge Lang about the problem. He told me
> that the 2^c, 3^c problem was still unsolved, and that it had only
> been proved in the last few years that if 2^c, 3^c, and 5^c are all
> integers then c is a non-negative integer. No wonder I hadn't been
> able to make any progress on the problem!
>
> Robert Israel will remember that 1971 Putnam exam - he finished
> in the top 5 that year.
Yes. That was the problem that really stumped me. Fortunately, it
also stumped everybody else: nobody got 8, 9 or 10 on it,
according to the official results in
The American Mathematical Monthly, Vol. 80, No. 2. (Feb., 1973), pp.
170-179
<http://links.jstor.org/sici?
sici=0002-9890%28197302%2980%3A2%3C170%3ATWLPMC%3E2.0.CO%3B2-S>
IIRC there was speculation that whoever made up that question
was thinking of a proof that turned out to be wrong.
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Gerry Myerson
<dfda0994-a43e-45e2...@h11g2000prf.googlegroups.com>,
Robert Israel <isr...@math.ubc.ca> wrote:
I can well understand why there would be such speculation,
but fortunately I am in a position to refute it. I also spoke to
Andrew Gleason after the exam, and he told me he had made up
the question. He wasn't responsible for it appearing on the exam,
though - he told the problem to a friend (I don't remember
whether he told me who the friend was; if he had, the name probably
wouldn't have meant anything to me, and wouldn't have stayed
with me), and through that friend the problem made its way onto
the exam. Gleason showed me a way to do the problem - it was
a hard problem, but his proof was, of course, correct.
Thinking about it a little more, it's possible that Gleason's friend
thought he had a simpler proof, which proof turned out to be wrong.
On the other hand, it was the last problem in the session, and those
are supposed to be hard.