$$(e^{2πi})^{1/2}=1^{1/2}$$$$(e^{πi})=1$$ $$-1=1$$ I think it is due to not taking the principle value but please can someone explain why this is wrong in detial, thanks.
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3Since this appears in your question, what is your definition of $x^{1/2}$ when $x$ is not a nonnegative real number? – Did Jun 21 '14 at 12:50
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Note that "proof" is a thing: a noun. "Prove" is the corresponding verb: We aim to prove that X holds by writing a proof that X holds. – amWhy Jun 21 '14 at 12:59
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1sorry i am dyslexic and hence why there is often a lot of grammar + spelling mistakes in my questions! – Jun 21 '14 at 13:00
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possible duplicate of $-1$ is not $ 1$, so where is the mistake? – Jyrki Lahtonen Jun 21 '14 at 13:26
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You used two different branches of the function $x^{\frac{1}{2}}$.
Note that even in exponential form $(e^{x})^\frac{1}{2}$ has two different branches: $e^{\frac{x}{2}}$ and $e^{\frac{x}{2}+\frac{2 \pi i}{2}}$.
N. S.
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You basically claimed
$$ \sqrt{(-1)^2} = \sqrt{(1)^2} \quad \Rightarrow \quad -1 = 1 $$
but the square root is not a map, i.e. $\sqrt{a}$ can have multiple solutions. In particular, it is not the inverse of $x^2$.
Adam
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