# Re: 2.0.30 - its all a numbers game, I tell ya!

Nathan Bryant (nathan@burgessinc.com)
Tue, 11 Mar 1997 14:37:56 -0500 (EST)

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
>

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