Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: How's this? Proof (hopefully) that pi is irrational.
Posted:
Sep 12, 2009 7:01 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, 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_ !!! ... > > > >


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


