0

In my school book it says that if $f(x) =$ $x^{\frac{m}{n}}$. The base ($x$) should be greater than $0$ or in other words $Af=(0,+\infty)$. Why is that? If the base is negative I run to the contradiction that while $(-1)^3 = -1$, if you write it as $(-1)^{\frac{6}{2}}$ then it is the square root of $x^6 $which is equal to $1$ and not $-1$. Why does this happen and what constraints do I have to get to use fractional exponents on negative numbers? Thanks

Xristos Ntallas
  • 1
  • 2
    If you use fractional exponents on negative numbers, be prepared for things like complex numbers, lack of continuity, and cases of $\left(a^b\right)^c \not = a^{(bc)}$ – Henry Sep 23 '22 at 11:14
  • 1
    $(-1)^{\frac mn}$ is not a real number, where $\frac mn \not \in \mathbb Z$. It is undefined in only real number set. – nonstudent Sep 23 '22 at 11:15
  • 1
    Shall this be the formula for $x^{\frac{m}{n}}$ ? We then have however $x^{\frac{m}{n}}=exp(\ln(x)\cdot \frac{m}{n})$ for $x>0$. And $\ln(x)$ is (as long we only deal with real numbers) only defined for $x>0$. Even if we allow complex numbers, the ln of negative numbers is problematic since it is not uniquely defined and $\ln(0)$ is still undefined. – Peter Sep 23 '22 at 11:16
  • So , if the base is negative, the exponent should be an integer in which case we can apply the usual power rules ($a^{-m}=\frac{1}{a^m}$ , $a^0=1$ for $a\ne 0$ , and $a^m=a\cdots a$ where we have $m$ $a$'s for positive $m$) – Peter Sep 23 '22 at 11:26

0 Answers0