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Possible Duplicate:
-1 is not 1, so where is the mistake?

I'm trying to understand the exact point of failure in the following reasoning:

\begin{equation*} 1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1. \end{equation*}

I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.

Edit:

Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.

My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.

My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):

\begin{equation*} \sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i. \end{equation*}

Instead, had I used the correct definition of i:

\begin{equation*} {(xi)}^2 = x^2i^2 = -x^2 = -9, \\ x^2 = 9, \\ x = +- 3. \end{equation*}

Now, analyzing the equations in the original problem, I can see at least the following two errors:

1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.

2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.

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Sami
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  • See also the discussion at http://math.stackexchange.com/questions/1211/non-integer-powers-of-negative-numbers . This kind of stuff seems to come up a lot here... – Qiaochu Yuan Aug 24 '10 at 16:46
  • Yes, they come up a lot, because people here are bad at explaining things at a level below their own. I'm still left wondering what exactly it wrong with the above. – Sami Aug 24 '10 at 17:05
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    Sami, in general one has to put in a lot of thought and work themselves to even see what it is they do not understand as well as extracting understanding from peoples comments. It's something to start practicing as early as possible. –  Aug 24 '10 at 17:10
  • If you consider $\sqrt{x} = y$ as a short hand notation for $x = y^2$ and start rewrite the derivation in this explicit way it should become clear where the error creeps in. –  Aug 24 '10 at 17:12
  • Muad, I agree that everyone should put in a lot of thought and effort to themselves figure out what exactly they do not understand. However, it seems to also be the standard excuse here to give vague answers, while giving a clear answer would not really take any more effort to give. – Sami Aug 24 '10 at 17:30
  • @Sami: If you feel that something in the answers to the previous instance of this question was unclear, please leave a comment on those answers, or mention exactly in your question what part you are having trouble with. Both Isaac and Casebash go beyond simply saying "sqrt is not a function," and they do point out the exact step where the "proof" fails. – Larry Wang Aug 24 '10 at 17:38
  • @Kaestur, I'll do that. Yes, they point of the exact step where the "proof" fails. And as Casebash said, it's easy to spot the step that goes wrong. However, the real explanation on what exactly goes wrong in that step is explained as: "because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers". Which to me sounds a bit like: "There are some formulas related to it. Just rewrite $$sqrt{-n}$$ as $$i\sqrt{n}$$ first, put your head in the sand and calculate." – Sami Aug 24 '10 at 18:02
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    @Sami: Actually, I think the point of failure in your reasoning above is different than in the linked question. Across your 5th =, you rewrite $\sqrt{z^2}$ as $z$ (for $z=\sqrt{-1}\sqrt{-1}$), but $\sqrt{z^2}\neq z$. For $z\in\mathbb{R}$, $\sqrt{z^2}=|z|$; for $z\notin\mathbb{R}$, it's a bit more complicated, but again becomes an issue of defining principal square roots and/or there being more than one root of a complex number. – Isaac Aug 24 '10 at 18:59

1 Answers1

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You need to pay attention to branches of multivalued functions, e.g. see the Wikipedia explanation here. Similar less-trivial questions often arise when symbolic mathematical sotfware systems exhibit bugs due to failure to stay on principal branches, e.g. see this thread where John McKay asks what your favorite system returns for $(-1)^{5/9} - (-1)^{2/9} - (-1)^{8/9}$. You may find such discussions instructive.

For the reader who may be interested in algorithms see for example

Thomas Breuer. [email protected]
Integral Bases for Subfields of Cyclotomic Fields. AAECC 8, 1997, 279-289 https://doi.org/10.1007/s002000050065

Abstract. Integral bases of cyclotomic fields are constructed that allow to determine easily the smallest cyclotomic field in which a given sum of roots of unity lies. For subfields of cyclotomic fields integral bases are constructed that consist of orbit sums of Galois groups on roots of unity. These bases are closely related to the bases of the enveloping cyclotomic fields mentioned above. In both situations bases over the rationals and over cyclotomic fields are treated.

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Bill Dubuque
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  • In other words: "Remember that thing they told you about in school where a number also has a negative (for real numbers) square root, but you didn't worry about it so much then? Now it comes back to bite 'ya!" – The_Sympathizer Sep 29 '14 at 09:10