# Resolving index laws for fractional exponents

• B
Gold Member
2019 Award

## Main Question or Discussion Point

I was just thinking about this earlier and couldn't come up with a good enough resolution. I'm guessing it's a matter of convention more than anything. If we have $x^{2} = a$, taking the principle root of both sides gives $\sqrt{x^{2}} = \sqrt{a} \implies |x| = \sqrt{a}$.

Yet evidently if the rule ${(a^{b})}^{c} = a^{bc}$ is taken to be true, then we end up with ${(x^{2})}^{\frac{1}{2}} = a^{\frac{1}{2}} \implies x = \sqrt{a}$, which disregards the potential negative root.

If we use the definition $a^{b} = e^{b\ln{a}}$, then negative bases make no sense since the domain of $\ln{x}$ is greater than zero. And, if I write $a$ in complex form, it turns out I can use the normal power rules to get both answers as expected:

$x^{2} = ae^{2n\pi i} \implies x = \sqrt{a} e^{n\pi i} = \pm \sqrt{a}$

Is the failure of this particular index law in the second example then just something we need to be aware of, or is there a resolution of some sort?

Mark44
Mentor
I was just thinking about this earlier and couldn't come up with a good enough resolution. I'm guessing it's a matter of convention more than anything. If we have $x^{2} = a$, taking the principle root of both sides gives $\sqrt{x^{2}} = \sqrt{a} \implies |x| = \sqrt{a}$.

Yet evidently if the rule ${(a^{b})}^{c} = a^{bc}$ is taken to be true, then we end up with ${(x^{2})}^{\frac{1}{2}} = a^{\frac{1}{2}} \implies x = \sqrt{a}$, which disregards the potential negative root.
For rational exponents, the rule above applies only when the base is positive.
For example $((-4)^2)^{1/2} = 4$, but $((-4)^{1/2})^2$ is not a real number.
etotheipi said:
If we use the definition $a^{b} = e^{b\ln{a}}$, then negative bases make no sense since the domain of $\ln{x}$ is greater than zero. And, if I write $a$ in complex form, it turns out I can use the normal power rules to get both answers as expected:

$x^{2} = ae^{2n\pi i} \implies x = \sqrt{a} e^{n\pi i} = \pm \sqrt{a}$

Is the failure of this particular index law in the second example then just something we need to be aware of, or is there a resolution of some sort?

• etotheipi
Stephen Tashi
I was just thinking about this earlier and couldn't come up with a good enough resolution. I'm guessing it's a matter of convention more than anything.
An underlying problem is that, in texts treating only the real numbers, $a^r$ is not defined for rational numbers $r$. Instead, $a^r$ is defined in a way that depends on how the rational number $r$ is denoted.

For example, consider $r = 5/3 = 10/6$. Typical secondary school texts in the USA give the definition:

For integers $m,n$ if $a^{\frac{1}{n}}$ is a real number then $a^{\frac{m}{n}}$ is defined to be $(a^{\frac{1}{n}})^m$.
Using this definition, students may be expected to compute $(-1)^{\frac{5}{3}}$ but they are expected to say that $(-1)^{\frac{10}{6}}$ does not exist - or perhaps the text avoids confusing the students by never asking for such a computation.

• etotheipi