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I am comfuse about something.

I want to compute $(-8)^\frac{2}{3}$

Is it $(-2^3)^\frac{2}{3}$=$(-2)^{3\cdot\frac{2}{3}}$=$(-2)^2=4$ ?

Is there any problem here because the base is negative?

Thanks.

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  • $\begingroup$ but my Transitions are ok? i mean i could do: $(-16)^3$ which is negative but $(-4^2)^3=(-4)^6$ which is positive. $\endgroup$
    – ruth
    Oct 3, 2014 at 7:19
  • $\begingroup$ @ruth: $(-4^2)^3\not=(-4)^6$, however, it is $-4^6$. $\endgroup$
    – Paul
    Oct 3, 2014 at 7:23
  • $\begingroup$ but $(a^b)^c=a^{bc}$ $\endgroup$
    – ruth
    Oct 3, 2014 at 7:27
  • $\begingroup$ so i dont get it. because then my answer should be $-4$ $\endgroup$
    – ruth
    Oct 3, 2014 at 7:31
  • $\begingroup$ The rule $(a^b)^c$ actually means $[(a)^b]^c$, so $(-4)^6=[(-4)^2]^3$. Also note that, by convention, $-a^b$ is understood to mean $-(a^b)$ and not $(-a)^b$. $\endgroup$
    – JRN
    Oct 3, 2014 at 11:46

5 Answers 5

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$$(-8)^{\dfrac23}=\left[(-8)^2\right]^{\dfrac13}=(64)^{\dfrac13}=(4^3)^{\dfrac13}$$

$=4\cdot w$ where $w$ is a cube root of unity

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    $\begingroup$ @ruth, If $x=\sqrt[3]{64},x^3=64$ and mathworld.wolfram.com/FundamentalTheoremofAlgebra.html says "a polynomial of degree $n$ has $n$ root" $\endgroup$ Oct 3, 2014 at 6:42
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    $\begingroup$ @ruth, See also: en.wikipedia.org/wiki/Principal_value#Complex_argument and en.wikipedia.org/wiki/Principal_branch $\endgroup$ Oct 3, 2014 at 6:45
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    $\begingroup$ ok but here it's not $X^3=64$ but $x=\sqrt[3]{64}$ so there is just one value which is $4$ i cant see why it's $4\cdot w$. in your link they say that $w=\frac{-1+\sqrt3i}{2}$ $\endgroup$
    – ruth
    Oct 3, 2014 at 6:47
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    $\begingroup$ @labbhattacharjee. This is a very common situation. In fact, I heard that Mathematica contains now a cube root function. $\endgroup$ Oct 3, 2014 at 8:46
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    $\begingroup$ I don't agree that $64^{1/3}$ has three values. The cube root of a real number is understood to be a real number, unless there is a specific reason to think otherwise. (If not, where do you stop? What about quaternions, octonions,...?) $\endgroup$
    – TonyK
    Oct 4, 2014 at 21:53
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Your answer is correct.

$(-8)^{2/3}=\left((-8)^{1/3}\right)^2=(-2)^2=4$

$(-8)^{2/3}=\left((-8)^2\right)^{1/3}=(64)^{1/3}=4$

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    $\begingroup$ The answer of lab bhattarcharjee is better. $\endgroup$
    – JRN
    Oct 3, 2014 at 6:37
  • $\begingroup$ No, your answer is much better. $\endgroup$
    – TonyK
    Oct 6, 2014 at 20:33
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Please have a look at: How do you compute negative numbers to fractional powers?

With rational exponents for negative numbers you are in trouble. You can do $x^{1/3}$ but if you try the thing you want then there are problems. You find that it depends on the order how you perform the operations resp. which root you choose. Therefore if we consider only real numbers, such things as $(-8)^{2/3}$ are not well defined. See also: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents

here is the example from this site: $−27 = (−27)^{((2/3)⋅(3/2))} = ((−27)^{2/3})^{3/2} = 9^{3/2} = 27$

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  • $\begingroup$ In general, powering then root extraction is not commutative with root extraction then powering. It matters which operation is done first. However, positive real bases are a special case for which this is commutative. When the root being extracted is odd, it is commutative for all real numbers. I explain here. math.stackexchange.com/a/4864176/928654 $\endgroup$ Feb 16 at 19:07
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Yes. You are right. Or you can trace in this way:

$=\sqrt[3]{(-8)^2}=\sqrt[3]{64}=4$

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  • $\begingroup$ see my last comment $\endgroup$
    – ruth
    Oct 3, 2014 at 7:19
  • $\begingroup$ @ruth: $(-4^2)^3\not=(-4)^6$, however, it is $-4^6$. $\endgroup$
    – Paul
    Oct 3, 2014 at 7:24
  • $\begingroup$ then why my answer isn't $-4$ $\endgroup$
    – ruth
    Oct 3, 2014 at 7:34
  • $\begingroup$ Look at my answer. The writer of the question somehow assumes that you should take the negative square root, but this is not justified. $\endgroup$
    – Karl
    Oct 3, 2014 at 14:39
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As $2$ and $3$ are coprime integers, riaising to the ($\frac23$)th power (equivalently, taking the ($\frac32$)th root, is esdsentially squaring and cubing.

While generally, using the simple $p$th power then $q$th root to calculate the ($\frac pq$)th power generally works for only positive real bases, it works in the case of negative numbers for a $(\frac23$)th power.

$$\sqrt[3]{(-8)^2}=\sqrt[3]64=2$$

$$(\sqrt[3]{(-8)})^2=(-2)^2=2 $$

This does not seem to work for all exponents.

$$\sqrt{(-1)^3}=\sqrt{-1}=i$$

if $i^2=-1$

However,

$$(\sqrt{-1})^3=i^3=-i\neq-1$$

But why?

Let us look at the numerator and denominator.

We were taking a square root, which is an even root, in this example. In the $(-8)^{\frac23}$ example, we were taking a cube root, which is an odd root.

Anbd here we have it. For all nonzero integers $z$, none of the $2z$th roots of negative real numbers are real numbers.

But for a ($2z+1)$th root, every real number, including negative numbers, has a unique real root. It follows that for integers $z_1$ and $z_2$, and negative real numbers $-r$,

$$(-r)^{\frac{z_1}{2z_2+1}}=\sqrt[\frac{2z_2+1}{z_1}]{-r}=\sqrt[2z_2+1]{(-r)^{z_1}}=(\sqrt[2z_2+1]{(-r)})^{z_1}$$

has a unique real value, and powering then root extraction yields the same value as root extraction then powering.

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