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Topic: Can a fraction have none noneending and nonerepeating decimal representation?
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David Bernier

Posts: 3,276
Registered: 12/13/04
Re: Can a fraction have none noneending and nonerepeating decimal
representation?

Posted: Aug 4, 2013 12:08 AM
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On 08/03/2013 10:43 PM, David Bernier wrote:
> On 08/03/2013 06:54 PM, David Bernier wrote:
>> On 08/02/2013 01:38 PM, David Bernier wrote:
>>> On 08/02/2013 06:49 AM, Richard Tobin wrote:
>>>> In article <85ffb478-968f-4de6-9284-b1e53fc19f0d@googlegroups.com>,
>>>> <jonas.thornvall@gmail.com> wrote:
>>>>

>>>>> But one 1/phi is not a rational ratio, but the 1/6 hexagon (center to
>>>>> vertex/sum of sides) is and so is all polygons derived from
>>>>> multiplying
>>>>> the vertices.

>>>>
>>>> You were claiming that pi is rational. It is not.
>>>>
>>>> -- Richard
>>>>

>>>
>>> This jogged my memory about Lagrange and irrational numbers.
>>> Lambert was the first to prove the irrationality of pi:
>>> in 1761,
>>>
>>> < http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational > .
>>>
>>> Lagrange showed that if x > 0 is rational, then sin(x)
>>> is irrational.
>>>
>>> Ref.: (from sci.math in 2011)
>>> ======================================================================
>>> In Jean-Guillaume Garnier's
>>> "Analyse algebrique, faisant suite a la premiere section de l'algebre",
>>> 1814,
>>>
>>> pages 538 and 539, he mentions Lagrange and C. Haros.
>>> He writes that Lagrange showed that, if x > 0 is a rational
>>> number, say x = p/q with gcd(p, q) = 1, then
>>> sin(x) is irrational.
>>>
>>> This uses the Taylor series:
>>> <snip>
>>> =======================================================================
>>> d.b. in sci.math, http://mathforum.org/kb/message.jspa?messageID=7364740
>>>
>>>
>>> sin(pi) = 0 is rational. pi>0.
>>>
>>> If we assume Lagrange's result:
>>> x>0, x rational ==> sin(x) is irrational,
>>> then we can conclude that pi must be irrational.

>>
>> [...]
>> Jean-Guillaume Garnier's argument is very sketchy.
>>
>> It relies on a modification of the "greedy algorithm", where
>> (a) The signs alternate in the representation of a rational
>> p/q with 0 < p/q < 1 as an alternating sum of
>> Egyptian fractions.
>>
>> (b) If m_k is the k'th summand (up to +/-), and
>> m_{k+1} is the k+1'st summand, then
>> m_k divides m_{k+1}.
>>
>> Garnier gives the algorithm as applied to p/q = 887/1103 .
>>
>> The algorithm is simple, and an explanation is given
>> for why it terminates for rational numbers p/q .
>> I haven't checked this through, however.
>>
>> Then, 887/1103 is represented as:
>>
>> ? 1 - 1/5 + 1/(5*47) - 1/(5*47*50)
>> + 1/(5*47*50*367) - 1/(5*47*50*367*551) + 1/(5*47*50*367*551*1103)
>>
>>
>> %67 = 887/1103 [ PARI/gp ].
>>
>> Garnier then agues in general about sin(x), so
>> in particular sin(1), assuming x is rational.

>
>
> If we look at the proof by contradiction by Fourier that
> e is irrational at Wikiepedia,
> < http://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Proof >
>
> It uses the Taylor series of exp(x) at x=1;
> similarly, I think we can get a proof by contradiction
> that sin(1) is irrational.
>
> But then, trying to do the same with sin(2), sin(3) etc.
> doesn't seem so easy.
>
> Dave L. Renfro wrote in sci.math:
>
> <<
> [A] Joseph Liouville, "Sur l'irrationnalité du nombre
> e = 2,718...", Journal de Mathématiques Pures et
> Appliquées (= Liouville's Journal) (1) 5 (May 1840), 192.
>
> Liouville's paper is on-line at
> http://gallica.bnf.fr/Catalogue/noticesInd/FRBNF34348784.htm
>
> This paper modifies Fourier's method of proving e is
> irrational (see below) to prove that e is not quadratically
> irrational. (This doesn't follow from the already known
> fact that e^r is irrational for every nonzero rational
> number, by the way.)
>
> [B] Joseph Liouville, "Addition a la note sur
> l'irrationnalité du nombre e", Journal de Mathématiques
> Pures et Appliquées (= Liouville's Journal) (1) 5
> (June 1840), 193-194.
>
> Liouville's paper is on-line at
> http://gallica.bnf.fr/Catalogue/noticesInd/FRBNF34348784.htm
>
> This paper extends the proof of Liouville [A] to prove
> that e^2 is not quadratically irrational.

> >>
> cf.:
> < http://mathforum.org/kb/message.jspa?messageID=3853756 > .
>
> So in 1840, Liouville showed that e isn't the root
> of a degree two integer polynomial.
>
> So, e^2 is irrational (and more).
>
> So, maybe proving sin(2) is irrational isn't too hard ...
>
>
>
>

>> The Taylor series for sin(x) gives:
>>
>> sin(1) =
>> 1 - 1/3! + 1/5! - 1/7! + 1/9! - ... (***)
>>
>> He writes:
>> "The series for sin(1) [ more generally: sin(p/q) ]
>> is also an alternating modified greedy algorithm expansion
>> series , as we had for 887/1103. But we know that the
>> alternating modified greedy algorithm expansion terminates
>> for rationals. Since (***) is non-terminating,
>> sin(1) is irrational [ more generally sin(p/q) ]. "
>>
>> I accept the part:
>> "the alternating modified greedy algorithm expansion terminates
>> for rationals".
>>
>> But,
>>
>> (I) The "alternating modified greedy algorithm expansion"
>> is not explained rigorously for irrationals.
>>
>> It's Chapter xxvii, "Transformations des fractions",
>> pp. 528-540,
>> Analyse algébrique, faisant suite a la première section de l'algèbre,
>> 1814. On the web:
>>
>> http://books.google.ca/books?id=iS4PAAAAQAAJ
>>
>> So, in view of (I), it's sort of like some hand-waiving
>> is going on. But neverthelss, Garnier says that Haros
>> told him about this and that Lagrange had a Memoire
>> in cinquieme cahier du Journal de l'Ecole Polytechnique.



Journal de l'Ecole Polytechnique, 5ieme cahier:

http://gallica.bnf.fr/ark:/12148/bpt6k433664s.image

On pages 92-114, Lagrange explains his algorithm(s),
which allow for choice of signs between terms.

He claims that if u := 1/m, where m is a strictly
positive integer, then
exp(1/m) is irrational,
sin(1/m) is irrational,
cos(1/m) is irrational .

Continued fractions and their convergents also appear
explicitly.

Intuititevely, with u:= 1/m , m>=1 an integer, it may be
that the partial sums of the Taylor series
for exp(1/m), sin(1/m), cos(1/m) are all
convergents in the c.f.s. expansion of
exp(1/m), sin(1/m), cos(1/m) [respectively].
But there are infinitely many convergents, so
exp(1/m), sin(1/m), cos(1/m) are all irrational.

===

The method of Fourier's proof should work also for
exp(1/m), sin(1/m), cos(1/m):
http://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Proof

===

In the concluding section, he mentions that the article looks at
contined fractions anew (e.g. his algorithm(s)) and refers to
"the late Mr Lambert".

The most redeeming ("salvaging") thing that emerges from this
Odyssey is that
Jean-Guillaume Garnier wrote that Haros (of Farey series fame)
told him of a proof of the irrationality of pi/2, based
on the memoire of M. Lagrange in
Journal de l'Ecole Polytechnique, 5ieme cahier.

Haros (and Garnier) appeal to:
x rational and x>0 ==> sin(x) is irrational.

As I noted above, Lagrange claimed this for x = 1/m, m
a positive integer, but not arbitrary rational x > 0 ...


http://books.google.ca/books?id=iS4PAAAAQAAJ
Garnier, Chapter XXVI.


Anyway, a first step would be to do sin(2), sin(3), sin(4), sin(5) ...
irrational. But all those seem quite a bit harder than
sin(1) or exp(1) ...

dave



--
new:
http://sci.math.narkive.com/


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