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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


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_ !!! ...


Date

Subject

Author

6/13/96


Ed Hook

9/12/09


Guest

9/12/09


Guest

6/13/96


Ted Alper

9/12/09


Guest


