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To evaluate $\sqrt{-1}$ $\times$ $\sqrt{-1}$ we cannot use $\sqrt{A}$ $\times$ $\sqrt{B}$ = $\sqrt{AB}$ as the result would be 1.

I know (?) that we must first respect that the initial numbers become imaginary, and of course $i^{2} $ is $-1$.

This came up in school (high school) yesterday, a teacher called me in as I walked by, and while I was confident of the answer, we are still looking to understand what rule, theorem, etc, makes this so.

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    $\begingroup$ Voting to close because this seems more like a math question than a how-to-teach question. This particular question has been asked and answered many times on SE Mathematics, and the answers showing up here so far are mostly restating those, e.g.: math.stackexchange.com/questions/49169/… $\endgroup$ – Daniel R. Collins Feb 14 at 2:25
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    $\begingroup$ I asked here because I know the correct answer and was in fact looking for the reasons behind it, for the very purpose of explaining to a class. $\endgroup$ – JTP - Apologise to Monica Feb 14 at 2:27
  • $\begingroup$ If the symbol $x = \sqrt{y}$ is defined at all, it means that $x^2=y$. Meaning that $\sqrt{-1} \times \sqrt{-1}$, which is what we abbreviate as $(\sqrt{-1})^2$, must be $-1$ by definition if it is defined at all. That is not due to some rule; it is due to the definition of that symbol (as in "what else, pray, could you mean when you write $\sqrt{-1}$?") $\endgroup$ – Torsten Schoeneberg Feb 15 at 6:34
  • $\begingroup$ I keep cringing when I think about "calculate i". There is no calculation. It is just a notation for what is really the same thing. "...before we write values using i" might be a better title. $\endgroup$ – Sue VanHattum Feb 15 at 21:46
  • $\begingroup$ Better? My phrasing wasn’t good, I agree. $\endgroup$ – JTP - Apologise to Monica Feb 15 at 21:49
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From Wikipedia:

The identities $(bc)^x=b^xc^x$ and $(b/c)^x=b^x/c^x$ are valid when $b$ and $c$ are positive real numbers and $x$ is a real number. But a calculation using principal branches shows that ${\displaystyle 1=(-1\cdot -1)^{\frac {1}{2}}\not =(-1)^{\frac {1}{2}}(-1)^{\frac{1}{2}}=-1}$
and
${\displaystyle i=(-1)^{\frac {1}{2}}=\left({\frac {1}{-1}}\right)^{\frac {1}{2}}\not ={\frac {1^{\frac {1}{2}}}{(-1)^{\frac {1}{2}}}}={\frac {1}{i}}=-i}$
On the other hand, when $x$ is an integer, the identities are valid for all nonzero complex numbers.
If exponentiation is considered as a multivalued function then the possible values of $(−1 \cdot −1)^{1/2}$ are $\{1,−1\}$. The identity holds, but saying $\{1\}=\{(−1\cdot −1)^{1/2}\}$ is wrong.

I suppose there is no name for the "rule" that says the identity $\sqrt{A}\sqrt{B}=\sqrt{AB}$ is wrong when $A,B<0$, just a rule saying that the identity is right if $A,B\ge 0$.

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  • $\begingroup$ Hmm. "On the other hand, when 𝑥 is an integer, the identities are valid for all nonzero complex numbers." Might be better to say "On the other hand, when 𝑥 is an integer, the identities are valid for all complex numbers (excluding 0 in denominators)." ie 0 is fine in (ab)^x = a^x*b^x. $\endgroup$ – Sue VanHattum Feb 13 at 18:24
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    $\begingroup$ This is really two answers, the first of which is not an answer that would be intelligible to high school kids, who don't know about stuff like "principal branches." Nor is it really true that the identity holds when $A$ and $B$ are positive. Focusing on the signs of $A$ and $B$ is the wrong idea. The focus should be on which square roots we're choosing. $\endgroup$ – Ben Crowell Feb 13 at 20:33
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    $\begingroup$ @BenCrowell Doesn’t $\sqrt$ conventionally denote the principal square root, though? (Unless your point is that needs to be explicitly made obvious, in which case your point is well-taken.) $\endgroup$ – BalinKingOfMoria Reinstate CMs Feb 14 at 2:24
  • $\begingroup$ @BalinKingOfMoriaReinstateCMs The point is that, while we can make a canonical choice of square root of a positive real numbers, there is no choice which will work for all complex numbers. If you assign a single square root to each complex number, the resulting function cannot be continuous on the entire complex plane. $\endgroup$ – Steven Gubkin Feb 15 at 13:02
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In modern pedagogy, there has been a tendency to eradicate the notion of a multiple-valued function, but it's really quite a useful notion, and it's the way people always used to talk about these things. In this language, both 0 and $\pi$ are inverse tangents of 0, and both 2 and $-2$ are square roots of 4. This is IMO the most natural way of dealing with the problem you describe. Then the identity $\sqrt{A}\sqrt{B}=\sqrt{AB}$ is not really an identity, because all three of the square root symbols represent multi-valued operators. Similarly, the identity $\tan^{-1}\tan x=x$ is not an identity; it's an identity modulo $\pi$.

The educational and practical problem with the eradication of multiple-valued functions is that it plays into students' belief that "the answer" is whatever their calculator says it is. Therefore, if their calculator says that the inverse tangent of 0 is 0, it must be right. This causes them to get wrong answers to real-life problems.

For students who have been carefully taught that all operations are single-valued, and who may not be able/willing to absorb this approach, you can also tackle this by asking them the following. If $x$ is a real number, and $x^2=4$, is it true that $x=2$? For students operating at this kind of concrete level, the message that they should be remembering is that when you take square roots of both sides of an equation, you have to write $\pm \sqrt{\ }$.

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    $\begingroup$ Maybe it’s too pedantic, but (at least from my ugrad math-class perspective) a “multi-valued function” is a contradiction in terms since functions must be injective [sic?]. $\endgroup$ – BalinKingOfMoria Reinstate CMs Feb 14 at 2:26
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    $\begingroup$ @BalinKingOfMoriaReinstateCMs ncatlab.org/nlab/show/red+herring+principle $\endgroup$ – hobbs Feb 14 at 3:52
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    $\begingroup$ @Balin Maybe you're thinking about the idea that in conventional terminology functions must be single-valued. The word injective doesn't mean that. But in any case, there's no problem with introducing a separate notion of a multivalued function, it would just be a different type of object than a conventional function. $\endgroup$ – YiFan Feb 14 at 4:22
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    $\begingroup$ @hobbs To confirm that I understand (for my own knowledge): Does "function" sometimes have a broader meaning in modern mathematical usage? If so, is there a specific agreed-upon definition somewhere, or is it more of "context makes it clear what I mean"? $\endgroup$ – BalinKingOfMoria Reinstate CMs Feb 14 at 20:56
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    $\begingroup$ @BalinKingOfMoriaReinstateCMs it's more that "sometimes an adjective broadens rather than restricts, and that's okay". In this case, as in many others, history factors in. Multi-valued functions were generally considered functions up to the 20th century. $\endgroup$ – hobbs Feb 15 at 1:48
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The first radical is negative i. The second one is positive i. ;-) Almost not trolling. Similar point as with Ben.

This question may not be 100% identical, but reminds me of a lot of the other questions we get that sort of key around the issue/confusion of radicals meaning "positive root" or meaning "either root". FWIW, i is neither positive nor negative.

I think I have seen i defined as the number squared that gives you -1. (Not using a radical in the definition.) Of course this makes it clear that either i or negative i satisfy.

From a practical teaching standpoint, I would avoid getting into this unless the kids bring it up. Sounds like you are confused on it. (If they do, just explain that the identify only holds for positive numbers and leave it at that.) I guess you could similarly wonder if y=3x/x is 3 at x=0 or undefined.

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    $\begingroup$ I was less confused regarding the answer, more about articulating the rule it resulted from. As far as your last example goes, I would state there is a removable discontinuity at x=0. I’d ready myself for a lower level teacher to ask me to not do that, but I’d expect calc/or pre-calc (and the model calculator we recommend, the TI84), to agree. $\endgroup$ – JTP - Apologise to Monica Feb 14 at 16:15
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    $\begingroup$ That discontinuity is not removable. $\endgroup$ – Sue VanHattum Feb 15 at 21:50
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The underlying idea is that "the" square root of a number is only well defined when that number is positive, as we then have a consistent way to discriminate both square roots by taking the positive one. So $\sqrt{x}$ is the positive number that, when squared, gives the positive number $x$. This works because the product of two positive numbers is a positive number.

The general statement is that the product of a root of $a$ and of a root of $b$ is a root of $ab$. This still works for $i$: $i$ is a root of $-1$, so $i^2=-1$ is a root of 1. It simply isn't "the" root of $1$, as it is not positive.

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I think trying to invoke a "rule" handed down from above would not be too beneficial for students. It just reinforces the idea that math is a bunch of rules to memorize and not think about (I know that's not your intention of course). The reason the statement is false for imaginary numbers is pretty simple: there are two possible ways to define $\sqrt{-1}$, and neither one makes the statement true (at least using the standard definition of square root for nonnegative reals; see below). There's nothing more or less to it.

I think what would be helpful is for the students to try different combinations of definitions (say, $\sqrt{x}$ is nonpositive for nonnegative reals and $\sqrt{-1}=i$) and see whether the rule holds for reals and/or imaginary numbers then. This reinforces the idea that the "rules" of math depend on our definitions and objects of study, and may be true in some cases and false in others.

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  • $\begingroup$ I am working, in general, with students who are not really original thinkers, and are more likely to embrace a set of rules. $\sqrt{A}\sqrt{B}=\sqrt{AB}$ is one of the (equalities?) we introduce what talking about roots. I was actually good with Joel's answer, that these variables need to be >0 for that to hold. And a big 'yes' to your last line. The square of 4 can be just 2, or can be +/-2, depending on the context. $\endgroup$ – JTP - Apologise to Monica Feb 14 at 17:02
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    $\begingroup$ @JTP-ApologisetoMonica I'm a little taken aback by your first sentence. Almost every high school math student (in the US at least) is more likely to embrace a set of rules because that's how they've been taught since they were little kids. Then they get labeled (perhaps some demographics more than others) as "not original thinkers" and are given more of the same type of terrible instruction. What a shame. $\endgroup$ – Thierry Feb 14 at 17:41
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    $\begingroup$ I appreciate your concern. When I find myself trying to engage with a HS junior, not knowing that student well, and just trying to introduce the problem they face - I ask 'ok, so what's 3 x 3?' and get "one second"(as she reaches for a calculator), and a minute later, "nine". At what level do you suggest I engage with her? I am not a classroom teacher. I am an in-house tutor. The honors students don't need my help. The shame is whatever happened in grade and middle school. By the time they visit me, walking out understanding PEMDAS is progress. $\endgroup$ – JTP - Apologise to Monica Feb 14 at 18:21
  • $\begingroup$ You engage by gently telling her that you know she can do that without her calculator. You help her find ways to do mental math. You help her see how to think with math. $\endgroup$ – Sue VanHattum Feb 15 at 21:52
  • $\begingroup$ "understanding PEMDAS" Does that mean understanding how to use the "rule"? Or does it mean understanding both how to use this convention, and why we have this convention? $\endgroup$ – Sue VanHattum Feb 15 at 21:53

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