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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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Ed Hook

Posts: 131
Registered: 12/6/04
Re: How's this? Proof (hopefully) that pi is irrational.
Posted: Jun 13, 1996 11:54 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, since 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 argument to the difference
quotient for a monotone function at any point in its domain: since
f(x+h) - f(x) and (x+h) - x are never 0 *except* at h, then the derivative
can be calculated by plugging in 0 for h ?? Put 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 two functions agree
at precisely _one_ point. namely x = k ... if you want to evaluate the
limit, then you're going to have to resort to something like L'Hopital's
rule. But, then, your argument vanishes in a puff 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 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_ !!! ...







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