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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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 JT Posts: 1,388 Registered: 4/7/12
Re: Can a fraction have none noneending and nonerepeating decimal representation?
Posted: Aug 2, 2013 1:52 PM

Den fredagen den 2:e augusti 2013 kl. 19:38:23 UTC+2 skrev David Bernier:
> 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.
>
>
>
> david
>
>
>
>
>
> --
>
> On Hypnos,
>
> http://messagenetcommresearch.com/myths/bios/hypnos.html

Honestly i think i may have solved it alot earlier than that, problem when young is to realise who is a halfwit and who isn't.

It may have ended up with some retarded Nasa employed, i just don't know.

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