Question 221921
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Let *[tex \Large x] represent the chosen number.


Square the number: *[tex \Large x^2]


Subtract 1: *[tex \Large x^2 - 1]


Divide by one less than the original number: *[tex \Large \frac{x^2-1}{x-1}]


Before we continue, simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2-1}{x-1}\ =\ \frac{(x+1)(x-1)}{x-1}\ =\ x + 1]


Subtract the original number: *[tex \Large x + 1 - x = 1]


Works for all *[tex \Large x \neq 1].  (Extra credit: why can't x = 1?)


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New Number Game:  Prove that 2 equals 1.


Let *[tex \Large a = b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a = b]


Multiply both sides by *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2 = ab]


Subtract *[tex \Large b^2] from both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2 - b^2 = ab - b^2]


Factor both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a+b)(a-b) = b(a - b)]


Divide both sides by *[tex \Large a - b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a+b = b]


But *[tex \Large a = b], so substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a+a = a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a = a]


Divide both sides by *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2 = 1]


The game is "What's wrong with this proof?  What did I do wrong that allowed me to achieve a result that is obviously absurd?"



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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