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Hi Chris & All,
But notice that one doesn't quite get to divide by zero because the
previous line (below) says that a*zero = b*zero and this condition alone
voids any conclusions that one might attempt to make about a and b.
Yt, DW
Factor :
a ( a - b -c) = b ( a - b -c)
c.majka@ns.sympatico.ca wrote:
> Hi Dave,
>
> Nice riddle: made me scratch my head for a few minutes, However, as
> Paul points out (a - b - c) = 0 and dividing by 0 (at the last step)
> allows one to prove whatever one wishes. ;->
>
> Cheers,
>
> Chris
>
> On 20-Dec-06, at 9:17 AM, David & Alison Webster wrote:
>
>> Dear All, Dec 20, 2006
>> I recently came across an old article about Raymond Smullyan, a
>> highschool dropout who e.g. satisfied the course requirements for a PhD
>> in Math and Philosophy by teaching the required courses. He also
>> created
>> many logic puzzles. One thing leads to another, so I blew the dust off
>> of an old copy of Martin Gardner's 1959 Mathematical Puzzles and
>> Diversions.
>>
>> One of his fallacies follows. I hope you have not seen this
>> previously.
>> Proof that unequal numbers are equal
>>
>> Given two numbers a & b such that b is smaller than a by an amount
>> c; thus
>>
>> a = b + c
>>
>> prove that a = b
>>
>> multiply both sides by (a - b) to obtain--
>>
>> a^2 - ab = ab +ac - b^2 - bc
>>
>> subtract ac from both sides to obtain
>>
>> a^2 - ab - ac = ab - b^2 - bc
>>
>> Factor :
>> a ( a - b -c) = b ( a - b -c)
>>
>> Divide both sides by ( a - b - c) to obtain
>>
>> a = b
>>
>> QED
>>
>> "The road to correct conclusions is full of pitfalls" (DHW, Dec
>> 20, 2006).
>>
>> Merry Christmas, Dave Webster, Kentville
>>
>>
>
>
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