Digg ran a story that showed an fallacious proof that 1 = 2 using complex numbers. This is a rehash of the 0 = 1 fallacy. Which goes something like this.

- a = b
- a – b = 0 (subtract both sides by b)
- (a – b)/(a – b) = 0 / (a-b) (divide both sides by a – b)
- 1 = 0 (simplify)

Although there tends to be some steps in between to mask the a – b step. Since a and b are equal, we are clearly dividing by zero producing the nonsensical result. The divide by 0 is a common caveat that might be forgotten in the factor and divide rule, especially when it is masked by unknowns.

In more advanced math, more seeming oddities can sometimes arise. Such as the following proof, posted in the thread discussion.

- e^(pi * i) + 1 = 0
- e^(pi * i) = -1 (subtract 1)
- e^(pi * i) * e^(pi * i) = 1 (square both sides)
- ln e^(2 * pi * i) = ln 1 (take the ln of both sides)
- 2 * pi * i = 0 (simplify)

Every thing is fine up to step 5. Although, it looks fairly innocuous. Because ln is usually defined as ln y = x where y = e^x, so it would seem that ln e^x would be x, by definition. Although, the way we evaluate e^(y) changes slightly when y is complex. Namely, e^(a + b * i) = e^a * cos(b) + e^a * sin(b) * i. Therefore, the ln must take this into account, the proper definition for complex numbers is ln(e^a * cos(b) + e^a * sin(b) * i) = a + b * i. In short, we can’t leave the i in the exponent when dealing with complex numbers and taking the ln.

Using the correct definition, we evaluate e^(2 * pi * i) = 1 * cos(2*pi) + 1 * sin(2 * pi) * i = 1 + 0i = 1. ln 1 = 0, where we reach the correct solution.

This particular fallacy is possibly because of the nature of the sin and cos evaluation, it is fairly trivial to prove e^(a + b * i) = e^(a + (2 * pi * k + b) * i) where k is an integer, which this particular fallacious proof was exploiting in a hidden way. Since, everyone would recognize e^(2 * pi * i) = e^0, ln e^0 = 0.

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