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Say you are a high school student or a young undergrad. You are being taught about complex numbers and you are told that $i = \sqrt{-1}$.

You go home and you write this: \begin{equation} \begin{aligned} 1 & = 1 \\ 1 & = \sqrt{1} \\ 1 & = \sqrt{(-1)(-1)} \\ 1 & = \sqrt{-1}\sqrt{-1} \\ 1 & = i \times i \\ 1 & = i^2 \\ 1 & = -1 \end{aligned} \end{equation}

You are dismayed. The infamous imaginary numbers are inconsistent after all!

The best answer would be the clearest explanation of why the "proof" is faulty, keeping in mind the level of the student. An explanation is extra clear if t presents insights to the student, as well as being correct and concise.

Andrea
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  • Who says you can do the step $1=\sqrt{-1}\sqrt{-1} $ ? – imranfat Jan 25 '16 at 16:57
  • $$\sqrt{(-1)(-1)}\ne\sqrt{-1}\sqrt{-1}$$ – Jan Eerland Jan 25 '16 at 16:57
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    I am guessing the OP wants a clear explanation of why multiplication and taking the root doesn't commute for $-1$. – copper.hat Jan 25 '16 at 16:59
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    I think this is not a duplicate. Aparently, the OP knows at least one 'proof' of this. He's not asking "why" is the proof faulty, he's asking for a clear and easy explanaton of this fact for a particular public: school students. For example, the property $\sqrt{x^2} = |x|$ can be a little bit tough to school students, let alone non-intuitive. – Daniel Jan 25 '16 at 17:12
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    I think the real issue, which is unfortunately embedded deeply in our pedagogy, is the notation. Instead of a functional (though not unique) characterisation ($i^2 = -1$), we have the appealing $\sqrt{-1}$ which comes with many pitfalls. – copper.hat Jan 25 '16 at 17:36
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    @SolidSnake Thanks for defending the question: you understood the spirit. Unfortunately it is now closed. I would have liked to see what Physics SE users would have come up with. – Andrea Jan 25 '16 at 17:41
  • I must admit however, that the accepted answer in https://math.stackexchange.com/questions/49169/why-sqrt-1-times-1-neq-sqrt-12 would have been a perfect answer to this question too. – Andrea Jan 25 '16 at 17:42
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    Explain slowly and carefully that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ isn't true not even for real numbers* (as in real numbers the square root of negative numbers is undefined) and that every time you've done it in the past you've assumed both terms were positive. – fleablood Jan 25 '16 at 17:44
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    I think this should be reopened (possibly with 'high school' in the title :-)) as the issue is about explaining the fallacy at that level. – copper.hat Jan 25 '16 at 17:49

2 Answers2

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A priori, $\sqrt{-1}$ is just notation for some new object which, when squared, yields $-1$. It is not some operator $\sqrt{}$ applied to the number $-1$. There is no reason one should expect it to behave exactly like the traditional square root function, which isn't defined for negative numbers.

Alex Provost
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$$ \sqrt{(-1)(-1)} \neq \sqrt{-1}\sqrt{-1} $$

One way to see this is to use the property $\sqrt{x^2}=|x|$ for all $x \in \mathbb{R}$, so that

$$ \sqrt{(-1)(-1)}=\sqrt{(-1)^2}=|-1|=1. $$

Only for non-negative $x$ and $y$ is it true that $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

mzp
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  • But then the student would think $\sqrt{x^2} = |x|$ for complex. $i = \sqrt{-1} = \sqrt{I^2} = |i| = 1$ so $-1 = i^2 = 1^ = 1$. We're no better off. – fleablood Jan 25 '16 at 17:49
  • I mention in the answer that $\sqrt{x^2}=|x|$ holds for $x \in \mathbb{R}$, so I don't understand the confusion. – mzp Jan 25 '16 at 18:07
  • Think of it from a student's point of view. We are throwing all these definitions at him/her and suddenly saying they only work when we want them to. $\sqrt{x^2} = |x|$ is specifically brought up to point out that $\sqrt {x^2} \ne x$. And we seem to be bringing it up to show that $\sqrt{ab} \ne \sqrt{a}\sqrt{b}$ (I presume because ... well, I'm not sure why) And then we are saying that somehow because a rule works in the reals an entirely different rule won't work another way in the complex. It doesn't really held the student understand the fault in the proof. – fleablood Jan 25 '16 at 20:49
  • @fleablood A conceptual question requires a conceptual answer. If you have an answer that does not use the properties I used I think we would all be better off if you devoted your energy to submitting it as an alternative. I'm not doubting you, I just don't see the point of what you are saying if you don't have an alternative. – mzp Jan 26 '16 at 00:30