Re: OT: White Goddess
From: | Andreas Johansson <and_yo@...> |
Date: | Friday, April 13, 2001, 12:59 |
Henrik Theiling wrote:
>Hi!
>
>Andreas Johansson <and_yo@...> writes:
> > >But it isn't a polygon with an infinite number of sides, because the
> > >length of each of those sides would have, perforce, to be zero, which
> > >means that the circumference, being a sum of zeros, would be zero.
> >
> > According to your reasoning, if I took a line 1m long and divided it
>into an
> > infinite amount of pieces the total length of the fragments would be
>zero.
> > 1m has disappeared without anyone removing any length, eh?
>
>You cannot calculate like that. What works for finite numbers does
>not work for infinity. See why:
> a) with infinitely many pieces, you cannot add them by iterating
> (you will not stop working)
Who said anything about adding? Common sense would seem to dictate that if I
start out with 1m of length and don't remove or add any length, I'll finish
still having 1m of length. Now, common sense doesn't always apply in math,
but here it does.
Lemme explain in another way: My 1m-length consists of an infinite number of
"sub-lengths" of infinitely short length. This does not mean that the length
off my 1m-length is zero.
> b) with infinitely many pieces, you cannot multiply a finite
> length > 0 with infinity to get to a finite result
Not a "finite" length, sure. But the length we're multiplying with here
isn't finite - it's INfinitely short.
> c) assuming each piece has length 0, you still may not multiply
> 0 times infinity. It is like dividing by zero.
What? You're telling me I can't multiply infinity with zero?! What'd stop
me?
If we're dealing with lengths of zero (boring lengths if you ask me), then
the total length of ANY number of such lengths is zero. But of course those
fragments of my 1m-length weren't of zero length - the sum of their lengths
are 1m.
Andreas
_________________________________________________________________________
Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com.