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Is the following statement true? $$ (-1.2)^{1.6} = \sqrt[10]{(-1.2)^{16}}. $$

Is it because $a^{0.5} = a^{1/2} = \sqrt{a}$?

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It seems so...What's the problem? –  Zeta.Investigator Sep 1 '12 at 17:10
    
well because the first gives an unreal number... –  user1561559 Sep 1 '12 at 17:15
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now I'm confused! Can you show me why it is an "unreal"(I think you mean imaginary) number? –  Zeta.Investigator Sep 1 '12 at 17:25
    
non integer exponents on negative numbers are sometimes considered undefined. –  Karolis Juodelė Sep 1 '12 at 17:33
    
so are they both equal? –  user1561559 Sep 1 '12 at 17:37

2 Answers 2

up vote 1 down vote accepted

I'll take $\sqrt{}$ as example because for odd numbers, you get a bijection from $\mathbb{R}$ into itself (with positive number mapped to positive numbers and negative numbers mapped to negative numbers) so the problem only arises with even numbers and once you understand for $\sqrt{}$, you'll understand for $\sqrt[4]{}$, $\sqrt[6]{}$ etc. as well.

You problem is that you didn't understand the function $\sqrt{} : \mathbb{R_+} \to \mathbb{R_+}$ is just a convention.

When we write $\sqrt{4}$, we mean $2$ because we chose the positive real number as result for the function $\sqrt{}$ because if we didn't specify that, we wouldn't get a function since a function is something that has a unique value for each "input".

But if you try to extend that function to $\mathbb{R}$ we get $\sqrt{} : \mathbb{R} \to ?$ where does it go? Because now, for $\sqrt{-1}$, you have to decide between $i$ and $-i$...

You would pick $i$ because it looks positive but it is not... There has been a question about a way to differentiate $i$ from $-i$ recently on SO and the answers were, more or less, that you can't.

Of course, for $\mathbb{R}$, you could decide to always take the one that is a positive real number multiplied by $i$. But then you get $\sqrt{} : \mathbb{R} \to \mathbb{C}$ and you only use $i$ in $\mathbb{C}$ so you don't get a bijection.

And then if you try to extend it to $\mathbb{C}$ with $\sqrt{} : \mathbb{C} \to ?$ you can't decide which one to take unless you define some kind of sign over $\mathbb{C}$ to partition it. Which you usually don't want to do.

Plus when you work in $\mathbb{C}$, you usually just want one of the answers, of $a^2=b$ and you don't care which one you get.

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First, take a look at this Wikipedia article.

The problem here is that the base number $-1.2$ is negative. Some people think of negative numbers being raised to non-integer powers as being undefined. However, one might reasonably think of for example $(-1)^{\frac{1}{3}}$ as the cube root of $-1$, which is $-1$. It might be a bit confusing, but I think if you look at the Wikipedia article that I linked to above, you might find a good answer.

You have a rational exponent $1.6 = \frac{16}{10} = \frac{8}{5}$ (reduced fraction). $5$ is odd you have something well-defined and you get

$$ (-1.2)^{\frac{8}{5}} = \sqrt[5]{(-1.2)^8}. $$

This is equal to what you have written.

Now, you might have a calculator that doesn't evaluate the left hand side. I just tried with my calculator (TI-84) and had no problem.

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because when I enter the first in the calculator, it gives me an error –  user1561559 Sep 1 '12 at 17:37
    
@user1561559: I edited –  Thomas Sep 1 '12 at 17:42
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Also, relying on $(-1)^{\frac13} = -1$ can lead the unwary to conclude $-1 = (-1)^{\frac13} = (-1)^{\frac26}=((-1)^2)^{\frac16}=1^{\frac16}=1$. Be careful with this stuff! –  Hagen von Eitzen Sep 1 '12 at 17:55
    
@HagenvonEitzen: Indeed! Of course in your example $\frac{2}{6}$ is not reduced. –  Thomas Sep 1 '12 at 18:00

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