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Why isn't $\sqrt[12]{\left(-1\right)^6}$ equal to $\sqrt{-1}$? Clearly, the square root isn't defined. We have divided the index and the exponent by $6$. The theorem says that $$\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}$$ where $a\ge0$. How does it work when $a<0$, or it doesn't?

PP. I see that we can write $\sqrt[12]{(-1)^6}=\sqrt[12]{1^6}=1,$ but my question still holds.

nonuser
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user965851
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    That is a common question. There is nothing in the Universe to make it work for $a<0$. It only works for $a>0$ because there is a mathematical proof that it does. This proof does not work for $a<0$. –  Nov 10 '21 at 16:38
  • @StinkingBishop, thank you, but we're talking about the case when the index of the root is even, right? I think that when it is odd, it holds for all $a$. – user965851 Nov 10 '21 at 16:40
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    A different problem is why you have been taught (or you have assumed) that this works for both $a>0$ and $a<0$. Often this can be attributed to the approach in teaching, where students are incentivised to do calculations (and produce "results") rather than gain deep knowledge about what's actually happening as they do those calculations. This might bite later if you pursue mathematics as a topic. –  Nov 10 '21 at 16:40
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    Sure, $\sqrt[12]{(-1)^6}=\sqrt[12]{1}=1$. The issue here is that you are trying to use a formula that works only for positive $a$ - with $a<0$. I am not even saying that it works only for positive $a$ - there are special cases when it works with negative $a$ and with some conditions on $k, m, n$ attached. The complete breakdown when exactly it works is quite messy. –  Nov 10 '21 at 16:42
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    For odd indices in the root it works for real numbers and then breaks for complex numbers ;) (In fact, the whole notion of "radical/root" gets messy.) –  Nov 10 '21 at 16:45
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    One way of unpicking this is to ask yourself "which twelfth root do I mean?" There are contexts (eg roots of positive real number), where there is an option (the positive real root) which makes particular sense, but in general disambiguating is an arbitrary choice, and it is very easy to miss that two parallel computations are making different choices. – Mark Bennet Nov 10 '21 at 17:03
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    Although a different question, the answer at Why $e^{i(π/3)} \ne -1$? also addresses your question. – user21820 Nov 11 '21 at 23:37
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    @user21820, I have no idea what similarities you have found in the questions. We haven't even studied complex numbers. So you are stating that my question isn't good, I haven't provided examples of my work and etc? So with what exactly does your comment help me with understanding mathematics better? Completely nothing. – user965851 Nov 12 '21 at 07:49
  • Given your very poor attitude, this shall be my last comment to you unless you apologize for your ridiculous insults. I did not say the question is similar, but it is standard practice to close as duplicate if the answer addresses the query, even at the abstract level. Read the answer carefully. Secondly, you did not use Approach0 as mentioned in the linked meta post, which will easily find many related questions. It is not about work here. I have helped countless people here who are sincere in wanting to learn. If you are, I can help you. But never if you insult me. – user21820 Nov 12 '21 at 08:27

5 Answers5

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Because it is a compositon of two functions $f(x)= \sqrt[12]{x}$ and $g(x)=x^6$. You need $$(f\circ g)(-1) = f(g(-1)) = f(1) = 1$$ So, you act on $-1$ with $g$ and then on result with $f$.

nonuser
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If you call $a=\sqrt[12]{(-1)^6}$, this means by definition that $a^{12}=(-1)^6$. So $a^{12}=1$. There are two real roots to this equation, namely $a=1$ and $a=-1$ (there are ten other complex roots, including $i$ and $-i$).

Notwithstanding the above, the usual convention for the symbol $\sqrt{\ \ \ }$ is to mean the positive root.

Martin Argerami
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A related property of complex square roots that sometimes shows up in these 1=-1 proofs is that the $\sqrt{ ab}=\sqrt{a}\sqrt{b}$ property of positive real numers $a,b$ does not hold for complex numbers, i.e. $\sqrt{ ab}=-\sqrt{a}\sqrt{b}$, if $a,b<0$. That’s the exact point where these proofs are false.

Peter Sanctus
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If $\ x\in\mathbb{R}\ $ and $\ x>0,\ $ then $\ x^y\ $ is well-defined and unique for any $\ y\in\mathbb{R}.\ $ This is, for example, an exercise left to the reader in Rudin's PMA chapter $1.$

If $\ x\in\mathbb{R}\ $ and $\ x<0,\ $ then $\ x^y\ $ is well-defined if and only if $\ y\ $ is an integer. Basically, attempts to define $\ x^y\ $ if $\ x<0\ $ always runs into problems, one of which you encountered in your question.

However, in $\ \mathbb{C}\ $ we do not get analogous issues: if $\ x,y\in\mathbb{C}\ $ then $\ x^y\in\mathbb{C}.$

Adam Rubinson
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$\sqrt[n]{a^m}$ says do calculations in this order:

  1. Find $a^m$.
  2. Find the greatest real number you can raise to the $n$th power to get that previous result.

$\sqrt[nk]{a^{mk}}$ says do calculations in this order:

  1. Find $mk$ and separately $nk$.
  2. Find $a$ raised to the first result from 1.
  3. Find the greatest real number you can raise to (1)'s second result,such that you get the result from 2.

My point is that the sequence of things that you do in the two expressions are very different according to purely the order of operations. So in some sense it should be surprising that they would ever end up giving you the same result.

And that's right: they don't give the same result. Except in certain conditions, it can be proved that they will give the same result. One of those conditions is when $a>0$, in which case it can be proved these two things lead to the same end. That proof relies on $a$ being positive. In another situation where $a<0$, maybe the two things will end up the same, maybe not.

2'5 9'2
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