I have read through this thread. It mentions that there a several conventions for rational exponents.The first condition for the first convention for exponent laws to apply is that the base be greater than zero. I can understand that, if I assume exponent laws work the way they do only for real numbers and not complex ones. The second condition is that the exponent be in lowest form.

The second convention allows the base to be less than zero as well, but the exponent needs to have an odd number in the denominator.

Furthermore the first answer states:

The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd

None of these three statements make sense to me in their entirety and they are not being motivated in the thread linked (at least not in a way that makes sense to me). So, having established that these are the conditions for exponent laws to apply, why are these the conditions and where can I find a proof on them? Searching for "exponent laws" only gave me the typical school level exponent laws that make no mention of these extended conditions.

  • $\begingroup$ "The second condition is that the exponent be in lowest form" This isn't nescessary. If $b > 0$ and we define if $\frac pq = \frac mn$ then it is easy to show that $\sqrt[q] {b}^p = \sqrt[n]{b}^m$ so for $r = m/n = p/q$ defining $b^r = \sqrt[q]{b}^p$ is consistant. $\endgroup$ – fleablood Jan 25 '17 at 0:47

[THIS is only an intuitive argument!]

The definitions will make sense if you look at it this way...

The 1st condition: "The laws of exponents still apply to fractional exponents."

This condition is true, since a fractional exponent is still an exponent. This can also be proved in calculus, and the fact that $a^r = e^{r \ln a}$.

The 2nd condition: "Fractional exponent must be in simplest form."

Let's say you have a negative number (e.g. $-100$). Now, raise $-100$ to the $\frac 12$ power using the definition for a fractional exponent. You get: $$(-100)^{\frac 12} = \sqrt{-100}$$ which is imaginary. But, let's raise $-100$ to the $\frac 24$ power. You get: $$(-100)^{\frac 24} = \sqrt[4]{(-100)^2} = \sqrt[4]{10000} = 10$$ which is real. But, $\frac 12 = \frac 24$, so how could the two answers be different? So it is the mathematical convention to simplify the exponent first before doing anything else.

The 3rd condition: "The symbol $\sqrt[n]a$ is defined for negative values when $n$ is odd.

Let's say we have a negative number, $-64$. It is possible to have a cube root of $-64$, since every real number has a unique cube root ($\sqrt[3]{-64} = -4$). The index $n$ for the cube root is $3$, which is odd. $-64$ also has a unique 5th root ($-2.29739670999$) 7th root ($-1.81144732853$), etc. This shows that $\sqrt[n]a$ is defined for odd values of $n$.

Now, let us calculate the square root of $-64$. This number is $8i$, which is imaginary. So, $\sqrt[n]a$ will not be real if $n$ is even (the index of the square root is $2$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.