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Suppose $J \in \mathbb{R}$.

Is $J^{-\frac{2}{3}} \in \mathbb{R}$ for all $J$ ?

I think yes, because: $J^{-\frac{2}{3}} = \frac{1}{J^{\frac{2}{3}}} = \frac{1}{\sqrt[3]{J^2}} \in \mathbb{R}$. Only for $J=0$, the expression is undefined.

Is that true?

Simon
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1 Answers1

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Long story short, $\frac{1}{\sqrt[3]{J^2}}$ is unambiguously defined for all $J\in\mathbb R\setminus\{0\}$, while $J^{-\frac 23}$ is an ambiguous expression for negative $J$.

Regarding the problem of raising a negative number to a non-integer exponent, consult Non-integer powers of negative numbers or How do you compute negative numbers to fractional powers? or What is $(-1)^{\frac{2}{3}}$? etc.

Andreas Tsevas
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  • "...while $J^{−{\frac{2}{3}}}$ is an ambiguous expression for negative J." I see, but for negative J, the expression is also unambigious if J is restricted to be a real number. Right? – Simon Feb 14 '23 at 09:45
  • @Simon No, I do not understand what you mean by "restricting a negative $J$ to be a real number". Anyways, $(-1)^{-\frac 23}$ is ambiguous, for example. – Andreas Tsevas Feb 14 '23 at 09:49
  • Sorry, I was not clear enough. The expression $J^{-\frac{2}{3}}$ can be a real or a complex number for $J<0$. But there is only real number, right? – Simon Feb 14 '23 at 10:01
  • @Simon No, for example $(-1)^{-\frac 23}$ can either be real or complex, depending on which root you pick. See this answer. – Andreas Tsevas Feb 14 '23 at 10:42
  • "can either be real or complex" But if it is real, 1 is the only "solution", right? – Simon Feb 14 '23 at 14:12
  • @Simon In this case, yes. – Andreas Tsevas Feb 14 '23 at 16:42