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- Newsgroups: sci.physics
- Path: sparky!uunet!mcsun!sun4nl!relay.philips.nl!prle!dacosta
- From: dacosta@prl.philips.nl (Paulo da Costa 42147)
- Subject: Re: TIME HAS INERTIA
- Message-ID: <1992Nov22.125437.16062@prl.philips.nl>
- Keywords: CORRECTION FUNDAMENTAL THM. OF ALGEBRA
- Sender: news@prl.philips.nl (USENET News System)
- Organization: none
- References: <1992Nov20.100104.17600@prl.philips.nl> <1992Nov20.190744.6915@meteor.wisc.edu> <1992Nov20.230233.18271@CSD-NewsHost.Stanford.EDU>
- Date: Sun, 22 Nov 1992 12:54:37 GMT
- Lines: 71
-
- 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'... \/\/
-