Date: Jun 13, 1996 4:07 PM
Author: Ted Alper
Subject: Re: How's this? Proof (hopefully) that pi is irrational.



In <tkidd.834615695@hubcap> tkidd@hubcap.clemson.edu (Travis Kidd) writes:

>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...
>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.
>


But this is definitely not true. You seem to want to think that
if f and g are continuous functions,
then
lim f(x)/g(x) = f(k)/g(k)
x->k
but it isn't true unless f(x)/g(x) is itself continuous at k,
i.e. unless g(k) not equal to 0.

it's not clear what you mean when you say
"substituing k for x yields equivalent functions" --
once you substitute k for x, you no longer have functions, you have
specific numbers.

as someone else pointed out, choosing any k for which sin(k^2) = 0
would work in your argument, whether or not k was an integer.

Ted Alper