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Index of Subjects 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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