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Topic: Can a fraction have none noneending and nonerepeating decimal representation?
Replies: 108   Last Post: Aug 16, 2013 5:22 PM

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 David Bernier Posts: 3,892 Registered: 12/13/04
Re: Can a fraction have none noneending and nonerepeating decimal
representation?

Posted: Aug 3, 2013 6:54 PM

On 08/02/2013 01:38 PM, David Bernier wrote:
> On 08/02/2013 06:49 AM, Richard Tobin wrote:
>> <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.

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:

So, in view of (I), it's sort of like some hand-waiving
is going on. But neverthelss, Garnier says that Haros
in cinquieme cahier du Journal de l'Ecole Polytechnique.

david

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

Date Subject Author
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