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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 >> >> > >
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