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I know these seems to be splitting hairs, but according to two different calculators, there's a real difference.

For e.g, $(-8)^{1/3}$ computes to NaN, where $-8^{1/3}$ computes to $-2$.

Though am having a hard time seeing what the brackets actually mean, I mean, in this context at least. They seem arbitrary, though is there any actual use for the former notation?

By the way, this came about when trying to compute the more complex expression, $(3/-96)^{1/5}$, where of course, you have to bracket the $3/-96$, to tell the calc you want it to raise both numbers to said power.

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3 Answers 3

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In English, "$(-x)^{1/n}$" is the "$n$-th root of negative $x$", whereas "$-x^{1/n}$" is the "negative $n$-th root of $x$".

The difference is in the order of operations. In "$(-x)^{1/n}$", one negates $x$, then takes the $n$-th root, whereas in "$-x^{1/n}$", one takes the $n$-th root of $x$, then negates the result.

Clear?

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  • $\begingroup$ Thanks very much, yes, I understand :) $\endgroup$ Sep 6, 2015 at 21:20
  • $\begingroup$ And whether $(-8)^{1/3}$ is defined or not is a matter of conventions. $\endgroup$
    – egreg
    Sep 6, 2015 at 21:31
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Well, there's an order of operations you have to take into account. Performing $-8^{1/3}$ should translate to doing $$-\left(8^{1/3}\right)=-(2)=-2,$$

while $(-8)^{1/3}$ translates to taking a cuberoot of a negative number first. Now, taking the cuberoot of a negative number is certainly possible, but for some reason your calculator doesn't like it.

Note that there are technically multiple solutions to both $x^3=8$ and $x^3=-8$, but there is only one real solution in each case, and generally the real solution of $x=2$ and $x=-2$ respectively are thought of as the cuberoots.

Now, as far as your actual question, note that $$\left(-\frac{3}{96}\right)^{1/5}=\left(-\frac{1}{32}\right)^{1/5}$$ This can be evaluated without a calculator.

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  • $\begingroup$ Ah, I see, so basically, they're two different ways of expressing the same notation, but I suppose that it's a technicality that leads to the varying results, rather than the result of mathematical law. $\endgroup$ Sep 6, 2015 at 21:21
  • $\begingroup$ @user108262 Well, sort of. We have to have some rules. It's not so obvious when it's just $-8^{1/3}$ by itself, but it becomes more recognizable when we say look at $1-8^{1/3}$. The order of operations says that we do exponentiation before we do subtraction. $\endgroup$ Sep 6, 2015 at 21:24
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    $\begingroup$ Ahh, I see what you mean, if it was say, 2-8^1/3, then the bracketed notation makes the difference between 2 * -8^1/3, or 2 - 8^1/3. Well, rather, it would make the difference between the former and the latter, if not for the freak out when -8 is bracketed, thus being 2*(-8)^1/3, but hey, all's well in the end :) Thanks Peter! $\endgroup$ Sep 6, 2015 at 21:31
  • $\begingroup$ @user108262 No. In reply to the first comment, the notations $-x^n$ and $(-x)^n$ mean two entirely different things. Reread Peter's post.The brackets change the order of operations. Consider $-3^2 = -(3\cdot3) = -9$ but $(-3)^2 = -3 \cdot -3 = 9$. $\endgroup$
    – zahbaz
    Sep 6, 2015 at 21:38
  • $\begingroup$ Ah, zah, you raise a very good point. Though prior to this I had considered the difference between a value squared and cubed, where the brackets matter particularly with the former, but with the latter, an extra layer of complexity is introduced given my scenario, as I was also looking to find the cube root of a value. Anyway, we'll leave it at that because I feel that this convo could get a bit out of hand :D $\endgroup$ Sep 6, 2015 at 21:48
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Normally rational exponents are reserved toh positive numbers because with negative numbers, this notation can lead to inconsistencies: $$-2=(-8)^{\tfrac13}=(-8)^{\tfrac26}=\begin{cases}\bigl((-8)^2\bigr)^{\tfrac16}=(64)^{\tfrac16}=2,\\\Bigl((-8)^{\tfrac16}\Bigr)^2\enspace\text{which is not defined}.\end{cases}$$ Of courseone may speak of the cube root of a negative number, but this is to be denoted $\sqrt[3]{x}$, not $x^{\tfrac13}$.

One more argument against the use of rational exponents for negative numbers: the definition is: $$x^{\tfrac1n}\stackrel{\text{def}}{=}\mathrm e^{n\ln x},$$ which definitely excludes non-positive numbers.

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