[NatureNS] Christmas brain teazer

Date: Wed, 20 Dec 2006 19:22:30 -0400
From: David & Alison Webster <dwebster@glinx.com>
User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:0.9.2) Gecko/20010726 Netscape6/6.1 (CPQCA3C01)
To: naturens@chebucto.ns.ca
References: <458937F0.5010101@glinx.com> <AFCB71B8-9101-4189-96E1-0F1BCEB1BE9B@ns.sympatico.ca>
Precedence: bulk
Return-Path: <naturens-mml-owner@chebucto.ns.ca>
Original-Recipient: rfc822;"| (cd /csuite/info/Environment/FNSN/MList; /csuite/lib/arch2html)"

next message in archive
next message in thread
previous message in archive
previous message in thread
Index of Subjects

Index of Subjects
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
>>
>>
>
>



next message in archive
next message in thread
previous message in archive
previous message in thread
Index of Subjects