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Topic: How's this? Proof (hopefully) that pi is irrational.
Replies: 5   Last Post: Sep 12, 2009 7:10 AM

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Re: How's this? Proof (hopefully) that pi is irrational.
Posted: Sep 12, 2009 7:02 AM
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> >
> >
> > In article <tkidd.834615695@hubcap>,
> > tkidd@hubcap.clemson.edu (Travis Kidd) writes:
> > |> hook@cscsun3.larc.nasa.gov (Ed Hook) writes:
> > |> >|>Assume pi were rational. Then there would

> be
> > some integer k such that
> > |> >|>sin kn = 0 for all integers n. Take the
> > function sin x^2 / sin kx.
> > |> >|>What is the limit of this function as x
> > approaches k? Well, Do not judge. Fair

> warning.Support a cure for childhood cancer: Alex's
> Lemonade
>
> ©2009 MeAmI.org "Search for the Psince sin x^2

> > |> >|>and sin x are both continuous, this limit
> could
> > be expressed by substitutiong
> > |> >|>k in for x, or sin k^2 / sin k^2 = 1.

> However,
> > since both sin x^2 and sin kx
> > |> > Nope -- not gonna work. Above, you chose 'k'

> as
> > a magical integer with
> > |> > the property that sin(kn) = 0 for _all_
> > integers n; in particular, you
> > |> > have to accept that sin(k^2) = 0, so you've

> got
> > an indeterminate form
> > |> > here.
> > |> Yes. But nowhere is either the numerator *or*

> the
> > denominator 0 until the
> > |> limit is reached. (I'll repeat what I just

> said
> > in reply to another post...
> >
> > Makes no difference -- you could apply the same
> > me argument to the difference
> > quotient for a monotone function at any point in
> > in its domain: since
> > f(x+h) - f(x) and (x+h) - x are never 0 *except*

> at
> > at h, then the derivative
> > can be calculated by plugging in 0 for h ?? Put
> > ut this way, it's blatantly
> > wrong ... but this is exactly what you're about.
> >
> > |> I neglected to mention that k is positive.)
> > Therefore (I hope...I agree this
> > |> is the weakness in the proof), since

> substituting
> > k for x yields equivalent
> > |> functions, the limit of the quotient would be

> 1.
> > |>
> > How do you define "equivalent" here ?? Near k, the
> > he two functions agree
> > at precisely _one_ point. namely x = k ... if you
> > ou want to evaluate the
> > limit, then you're going to have to resort to
> > to something like L'Hopital's
> > rule. But, then, your argument vanishes in a puff
> > ff of smoke.
> >
> > |> The fact that the limit is of an indeterminate
> > form *then* allows me to use
> > |> L'Hopital's rule to find a *different* limit,
> > which is the contradiction.
> > |>
> > |> Indeed, the fact that sin(k^2) = 0 is the
> > (hopeful) key to the proof!
> > |>
> > |> >|>
> >
> >
> >
> >

>
> QED(?)

> > |> > Ed Hook |
> > Coppula eam, se non posit
> > |> > Computer Sciences Corporation |
> > acceptera jocularum.
> > |> > NASA Langley Research Center | Me?

> Speak
> > for my employer?...<*snort*>
> > |> > Internet: hook@cscsun3.larc.nasa.gov |
> > ... Get a _clue_ !!! ...
> > |> -Travis
> >
> > --
> > Ed Hook |

> Coppula
> > a eam, se non posit
> > Computer Sciences Corporation |
> > | acceptera jocularum.
> > NASA Langley Research Center | Me? Speak

> for
> > r my employer?...<*snort*>
> > Internet: hook@cscsun3.larc.nasa.gov | ...
> > . Get a _clue_ !!! ...
> >
> >
> >
> >


<
//source>>=========================================
>M+artinMusatov.p=np.q.e.d.@==MuaROTOV
http://MeAmI.org? (TM)symbol=trademark==symbol;==?



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