Messages in this thread | | | Date | Tue, 11 Mar 1997 14:37:56 -0500 (EST) | From | Nathan Bryant <> | Subject | Re: 2.0.30 - its all a numbers game, I tell ya! |
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On Tue, 11 Mar 1997, Donald R Willhoit wrote:
> Excerpts from internet.computing.linux-kernel: 11-Mar-97 Re: 2.0.30 - > its all a numb.. by H. Peter Anvin@transmeta > > Proof by obfuscation doesn't count. You can "prove" that 0/0 = a for > > any finite number a by observing that: > > > > a x > > lim --- = a > > x -> 0 x > > > > Q.E.D. > > > > (a doesn't even have to be real for the proof to hold!) > > > > -hpa
Well, since you mention it:
Given: a = b, a^2 = ab a^2 - b^2 = ab - b^2 (a + b)(a - b) = b(a - b) a + b = b a = 0
Thus, all numbers equal zero. Thus, all numbers are equal. So the point is moot. Now can we end this offtopic thread? ;)
> ON THE CONTRARY NOT QED. this may not be obfuscation itself but it > relies on the obfuscation of the definition of a limit. > > Okay my apologizes to continue this TERRIBLY OFF TOPIC thread but > something has to be set straight here. The above is a limit - it > doesn't prove that 0/0=a. In a slightly better form, since people are > always abusing things likes limits and infiinites heres the definition.: > > DEFINITION > Let f be a function defined on some open interval that contains a number > a, except possibly a itself. Then we say that the limit of f(x) as x > approaches a is L, and we write > lim f(x) = L > x->a > if for every number E > 0 there is a corresponding number d >0 such that > |f(x) - L| < E whenever 0 < |x-a| < d > > >From this you should see that your "proof" fails. > > D. Ryan Willhoit >
+-----------------------+---------------------------------------+ | Nathan Bryant | Unsolicited commercial e-mail WILL be | | nathan@burgessinc.com | charged an $80/hr proofreading fee. | +-----------------------+---------------------------------------+
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