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Why $x^{\frac{8}{3}}$ is defined only on $\mathbb R^+$ ? Indeed, if $x>0$, then $$(-x)^{\frac{8}{3}}=(-x^{1/3})^8=x^{8/3}$$ and all operation are well defined...

I agree with irrational power, but in this case, it looks strange.

Indeed, $x^8$ is defined over all $\mathbb R$ and $x^{\frac{1}{3}}$ also. Therefore, the composition should also be defined over all $\mathbb R$. enter image description here

user330587
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  • What makes you think it's not defined on $\mathbb{R}_{<0}$? –  Mar 12 '17 at 12:56
  • @vrugtehagel: The program grapher (see my picture) ! – user330587 Mar 12 '17 at 12:57
  • Try Desmos instead – Rob Mar 12 '17 at 12:58
  • A graph doesn't show anything. The fact that the program only graphed it on the positive numbers just shows the program isn't very intelligent –  Mar 12 '17 at 12:58
  • @vrugtehagel: Indeed, but I saw this question : here, and I also don't understand why in one case we they get $y=Cx$ and in the other case, they don't take off the absolute value (i.e.e they let $C|x|^{8/3}$) – user330587 Mar 12 '17 at 13:03
  • Most likely, the program you used to draw the graph converted $8/3$ to floating point first. At that point, any information on whether the denominator is even or odd is lost, so it can't do the powers of negative numbers. (Actually, any non-integral floating point number is an exact representation of a fraction whose denominator is a power of $2$.) – Harald Hanche-Olsen Mar 12 '17 at 13:12
  • This question has some discussion. – TonyK Mar 12 '17 at 13:13
  • It's a question of conventions; you can define $x^{8/3}$ also for negative $x$, but you have to be careful about doing algebra with the exponents, because $x^{16/6}$ would, by the same reasons, be defined only for $x\ge0$. – egreg Mar 12 '17 at 13:27
  • @egreg: I don't agree with that. $8/3$ is the same as $16/6$, so $x^{16/6}$ must equal $x^{8/3}$. See my answer to the question I linked to in the previous comment. – TonyK Mar 12 '17 at 14:35
  • @TonyK If you consider them as numbers, they are obviously equal. Otherwise you couldn't justify $\frac{1}{6}+\frac{8}{3}=\frac{1}{6}+\frac{16}{6}=\frac{17}{6}$. Therefore, why should $x^{8/3}$ be different from $x^{16/6}$? But, as I said, it's a question of conventions. Provided you don't do algebra on these expressions, you can define whatever you prefer. But when your student will do $-1=(-1)^{1/3}=(-1)^{2/6}=1^{2/6}=1$, you'll know why I warn about carelessly using fractional exponents. – egreg Mar 12 '17 at 14:54
  • @egreg: There is no convention that defines $x^{8/3}$ to be different from $x^{16/6}$, because such a system would be unworkable. So either you insist that the exponent be interpreted as if it was in lowest terms, or you disallow fractional powers of negative numbers altogether. That is the choice. – TonyK Mar 12 '17 at 18:00
  • @TonyK Exactly what I say: choose a convention and follow it, but be aware that others could be using a different one. I'm more concerned with $x^{1/3}=x^{2/6}$, of course. – egreg Mar 12 '17 at 18:57

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One of the problems here is that you assumed $\left(-x\right)^\frac 8 3=\left(-x^\frac 1 3\right)^8$, when it equals $\left(\left(-x\right)^\frac 1 3\right)^8$. This has multiple solutions. For positive non-zero $x$,

$$\left(-x\right)^\frac 8 3 = x^\frac 8 3\left(\left(-1\right)^\frac 1 3\right)^8$$

Where $\left(-1\right)^\frac 1 3$ has 3 solutions:

$$-1\\​\\0.5+\frac{i\sqrt3}2\\​\\0.5-\frac{i\sqrt3}2\\​\\$$

If you define $n^\frac 1 3$ as the real root, then $\left(-x\right)^\frac 8 3 = x^\frac 8 3\left(-1\right)^8=x^\frac 8 3$, which is defined over all of $\mathbb R$.

If you define $n^\frac 1 3$ as the principal root,

$$\begin{align} \left(-x\right)^\frac 8 3 &= x^\frac 8 3\left(0.5+\frac{i\sqrt3}2\right)^8 \\&=x^\frac 8 3\left(-0.5+\frac{i\sqrt3}2\right) \\&=-0.5x^\frac 8 3+\frac{ix^\frac 8 3\sqrt 3}2 \end{align}$$

which is a complex number, and the principal root is usually taken.

Artyer
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