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Article #353346

Re: Lightspeed constant... <--off topic on an off topic thread.

#353346
From: mum...
Date: Wed, 20 Nov 1996 00:00
70 lines
4207 bytes
In article <32927A...@mail.sgi.net>,  <tyr...@mail.sgi.net> wrote:
>> > > It has been proven.  It was a theory when Einstein first came up with
>> > > it.  It was then proven.  If you wish to find out what some of the
>> > > proofs are, then just post something else stupid, and I'll respond
>> > > again.
>> >
>> > Oh?  Can you prove it?  I can prove 2 = 3 via
>> >
>> > assume x = y,
>> > then 2(x-y)=3(x-y),
>> > divide through by (x-y),
>> > thus 2=3
>> >
>> > We KNOW that's false because, by definition, dividing by 0 (or x-y in
>> > the example) is undefined.  You really can't prove anything, since any
>> > proof can contain faulty reasoning or may be based on flawed
>> > assumptions, not to mention exceptions to the rule, like classical
>> > physics in subatomic situations.
>> >
>> > -JAmes
>> >
>> If there is an exeption, it CANNOT be called a proof. A proof cannot
>> contain any faulty reasoning as you said. For what you said to be a
>> proof it would have to also have 'for x-y not equal to 0'.

Umm, you just cancelled your own point... for x-y not equal to 0 (better
to say x != y) is an exception to algebra theories pertaining to the
above formulas.  The above formula cannot be called a proof because it
is based on the fauly premise that you can divide by zero, and also
for any x,y such that x != y, 2(x-y)=3(x-y) is not true, so that was
a faulty step as well:  if you already know x = y, there's no reason
for the variables.. just say
   0 = 0 (start)
 2*0 = 3*0 (multiplying both sides by different amounts?)
   2 = 3 (incorrect assumption from incorrect multiplication &
          undefinied division)

In mathematics, even this is an incorrect proof
   0 = 0 (start)
 3*0 = 3*0 (multiply both sides by the same number, still correct and legal)
   3 = 3 (dividing by zero)

because division by zero is not defined.. even when you get the correct
answer, your steps are incorrect and therefore your proof is faulty.

>
>What if an exception isn't known, as in Einstien's theories of
>relativity?  Can we ASSUME that there aren't no exceptions?  REAL proofs
>don't exist.

In any theory (as in life, really), the basic idea is to assume there are
no exceptions until there are shown to be any.  Then they are dealt with.
There's no point in trying to deal with something that you don't know exists
yet.

This would have been roughly comparable to someone from Newton's day trying
to figure out the mass increase a falling apple would have gained because of
its acceleration.  They didn't know such a phenomenon happens, and would not
have thought to consider it...  The same is true for us and modern theory.

You can go around saying 'what if' 'what if' 'what if' all you want, but
that accomplishes nothing.  You must *prove* that there is an exception
to an accepted rule, or you'll just be ignored (politely).

-- 
Glenn Lamb - mum...@netcom.com -- Finger mum...@netcom.com for PGP Key.
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From: mum...@no-spam.netcom.com
Subject: Re: Lightspeed constant... <--off topic on an off topic thread.
Date: 1996/11/20
Message-ID: <mumfordE16own.A8G@netcom.com>
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