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

H. Peter Anvin (hpa@transmeta.com)
11 Mar 1997 16:36:42 GMT

Followup to: <Pine.LNX.3.91.970311081149.811B-100000@chaos.analogic.com>
By author: "Richard B. Johnson" <root@analogic.com>
In newsgroup: linux.dev.kernel
> >
> > no. Any non-zero number divided by zero results in an infinite answer.
> > 0/0, however, is strictly "undefined".
> >
> > 3/0 = X is the same as X*0 = 3, thus X=infinity
> >
> > 0/0 = X is the same as X*0 = 0, thus X can be ANY number (thus,
> > undefined).
>
> Since this continues....
>
> N/N = 1 For all N
> Therefore:
> 0/0 = 1
> Proof: (actually a conjecture)
> Numbers from -N to +N are LINER, there is no discontinuity through zero.
> Therefore:
> +N/+N = 1
> -N/-N = 1
> And everything in between...
> (-.000000000000001 N) / (-.000000000000001 N) = 1
> ...........
> ........... All real numbers in between (lots)
> ...........
> (+.000000000000001 N) / (+.000000000000001 N) = 1
>
> All the real numbers between - N and + N, when divided by themselves must
> equal 1. Therefore 0/0 = 1.
>

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

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