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Re: Help with confirmation on a small bit of assembly?
David Empson wrote:
> John B. Matthews <nospam@nospam.com> wrote:
>
>> In article <pvgcr2ljj2nilnh3em53656c16eiihj3to@4ax.com>,
>> "Jeff Blakeney" <jeff.blakeney@a2central.com.remove-xme-this> wrote:
>>
>> > To: mdj
>> > On Tue, 23 Jan 2007 05:32:11 -0800, "mdj" <mdj.mdj@gmail.com> wrote:
>> >
>> > >David Empson wrote:
>> > >
>> > >> I'm sure someone can come up with a proof that 16 = 15, based on the
>> > >> 1 = 0 concept. :-)
>> > >
>> > >Damn you Empson! I'd never seen that before, and now that I've
>> > >Googled, I'll be awake for days trying to decide if it's funny or not
>> >
>> > A college classmate showed me one of those on a chalk board one day
>> > many years ago but the fact that he used a divide by zero in the
>> > process made the result meaningless. :-)
>>
>> It is well known that 16 = 15 for sufficiently large values of 15:-)
>
> Tee hee.
>
>> 1. -240 = -240 ; identity
>> 2. 256 - 496 = 225 - 465 ; still equal
>> 3. 16^2 - 16(31) = 15^2 - 15(31) ; factor each side
>> 4. 16^2 - 16(31) + (31/2)^2 = 225 - 15(31) + (31/2)^2 ; complete square
>> 5. (16 - 31/2)^2 = (15 - 31/2)^2 ; factor each side again
>> 6. 16 - 31/2 = 15 - 31/2 ; take square root & add 31/2, each side
>> 7. 16 = 15 ; qed?
>>
>> No division by 0; can you spot the bogus step?
>
> Without looking at the answer, it must be step 6 (square root).
>
> If A^2 = B^2, it does not follow that A = B. There are two possible
> solutions: either A = B, or A = -B.
>
> In this case, 16 - 31/2 is 0.5, and 15 = 31/2 is -0.5. Squaring both of
> these produces the same value (+0.25) and the square roots are therefore
> also equal (+0.5), but the square root of (15 - 31/2)^2 is not (15 -
> 31/2), it is (16 - 31/2).
>
> So there. :-)
>
> Enough sillyness for one night.
I don't see the problem with using division by 0 in things. I recall needing
to do it in subjects such as linear algebra and calculus at university. I
always got a kick out of dealing with "big infinity" and "little infinity
though".
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