the tree Posted July 14, 2006 Posted July 14, 2006 This is something my teacher did accidently whilst showing something else. [math]-i = -\sqrt{-1} = \sqrt{(-1)^{2} \times -1} = \sqrt{1\times -1} = \sqrt{-1} = i[/math] He said "that can't be right", rubbed it off the board and started again. Thing is, I can't see where the mistake was. Can you point it out?
Severian Posted July 14, 2006 Posted July 14, 2006 It is this equality where the error is: [math]-\sqrt{-1} = \sqrt{(-1)^{2} \times -1} [/math] It should read [math]-\sqrt{-1} = -\sqrt{(-1)^{2} \times -1} [/math] (In other words [math]-1 \neq \sqrt{(-1)^2}[/math])
woelen Posted July 14, 2006 Posted July 14, 2006 Here catch is that -1 does not equal sqrt((-1)²). This is OK for positive numbers, e.g. 2 = sqrt(2²), but for negative numbers you cannot do this. The function sqrt(x) is defined to give a positive number for all positive x. You could introduce a new function sqrtn(x), which returns the negative value, e.g. sqrtn(4) = -2. But what is done in that "proof" above, is sometimes using the positive definition of the sqrt() function and sometimes using the negative definition of the sqrt() function. In the step from -sqrt(-1) to sqrt((-1)²*(-1)) the negative definition, equivalent to sqrtn() is used, while in the step from sqrt(1*(-1)) to sqrt(-1) the positive definition is used. EDIT: While I was typing, Severian was just before me Luckily we say the same.
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