From: dacosta@prl.philips.nl (Paulo da Costa 42147)
Subject: Re: TIME HAS INERTIA
Keywords: CORRECTION        FUNDAMENTAL THM. OF ALGEBRA
Organization: none
Date: Sun, 22 Nov 1992 12:54:37 GMT

In article <1992Nov20.230233.18271@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
[...] Whether or not it can be, you are
>forgetting that any reductio ad absurdum *must* contain some absurdity
>in order to succeed.  Pointing out a falsehood during a reductio ad
>absurdum does not demonstrate a flaw in the proof. [...]

Yes, it does. The whole point of a proof by contradiction is the
following:

(1) You assume a proposition you want to disprove.

(2) By a sequence of _logical_ steps, and making only _valid_
    assumptions, you arrive at a contradiction.

(3) The fact that you arrived at a contradiction tells you that one
    of your assumptions is invalid, or one of the steps in the proof is
    illogical. SINCE ALL INTERMEDIATE ASSUMPTIONS AND STEPS ARE VALID,
    you can conclude that your original assumption was false, QED.

If you introduce a falsehood "during" a reductio ad absurdum -- i.e.,
during step (2) above -- you're bound to arrive at a contradiction. 
So what? You'll never know whether the contradiction was caused by
your original assumption or by the falsehood you introduced, and
so step (3) fails and your "proof" proves nothing.

[...] If his expansion is
>valid for a complex function with a derivative defined everywhere you
>cannot complain about such a valid use, regardless of how absurd it may
>seem to you.

But I _can_ complain about an invalid use, which is the case. I find it
appalling that people so ignorant of basic subjects like logic, algebra
and complex variables dare post to a sci.* newsgroup as if they know
what they are talking about. Sorry if this sounds too harsh, but the
people involved in this discussion should really reach for their
textbooks and brush up on the basics before posting.

>I haven't checked whether the division itself is valid, but this
>becomes a moot point if my objection is sustained.

"I haven't checked whether it makes sense or not, but I object to it." 
Sigh.

Just for fun, let's point out the hole in the original "proof":

>For the record here's Abian's proof (that 1+z+z^2 has a zero) again.
[...]
>>    1/ (1+z+z^2)  =   1 - z + z^3 - z^4 + z^6 - ...
>> 
>>    1/ (z^2+z+1)  =   1/(z^2) - 1/(z^3) + 1/(z^5) - 1/(z^6) + ... 
>>   
>>The two expressions on the right side of =  must be identical. [...]

No. Why should they? The series on the first line converges only for
|z|<=1; the series on the second line converges only for |z|>=1. You
can use either series in its own region of convergence, but you can
only equate them at the intersection of those regions (i.e., |z|=1),
where they do indeed converge to the same value (though not absolutely,
so you can't even change the order of the terms in either side of the
equation!). The next assertion,

[...] But
>>they are not.  Contradiction. [...]

is therefore utter bullshit, and you got nowhere. Sigh (again).

>Vaughan Pratt	  		A fallacy is worth a thousand steps.

-- 
-- Paulo da Costa           /\/\   Minha terra tem palmeiras   /\/\
-- dacosta@prl.philips.nl   \/\/   Onde canta o sabia'...      \/\/
Newsgroups: sci.physics,sci.astro