Cube root of -8

When you take the root $a^{\frac{p}{q}}$ you require $\dfrac{p}{q}$ to be in its lowest terms (specifically, you require the gcd of p and q to be 1).

When you take the root $a^{\frac{p}{q}}$ you require $\dfrac{p}{q}$ to be in its lowest terms (specifically, you require the gcd of p and q to be 1).

Heh, saw you reply and knew you'd own my post...is that by definition? Never knew that...

It's sort of by definition, but not really. I mean, clearly we must have $(-8)^{2/6} = (-8)^{1/3}$ otherwise it wouldn't make sense. However, the rule $a^{p/q} = \sqrt[q]{a^p} = (\sqrt[q]a)^p$ only necessarily holds when $\frac{p}{q}$ is in its lowest terms. It follows from a similar reasoning that gives the result that if you say that $x^2=1 \Rightarrow x=1$ then you lose a root of the equation (namely $x=-1$).

It's sort of by definition, but not really. I mean, clearly we must have $(-8)^{2/6} = (-8)^{1/3}$ otherwise it wouldn't make sense. However, the rule $a^{p/q} = \sqrt[q]{a^p} = (\sqrt[q]a)^p$ only necessarily holds when $\frac{p}{q}$ is in its lowest terms. It follows from a similar reasoning that gives the result that if you say that $x^2=1 \Rightarrow x=1$ then you lose a root of the equation (namely $x=-1$).

Now... who negged me?

You're a sub aren't you? Surely you can see who negged you anyway?

I've not been a sub for aaaaages!

I've not been a sub for aaaaages!

Anyway OP, here's a more thorough explanation.

Suppose that we have $x=a^{p/q}$, then $x^q=a^p$, so far so good. Suppose that $q$ is even, so that $x^q = |x|^q$, so that $|x|^q=a^p$. We can't then deduce that $|x|=a^{p/q}$, since that would imply that $x=|x|$ (which isn't true when $x<0$).

So, although we can say that $x=(-8)^{1/3} \Rightarrow x^3=-8 \Rightarrow x = \sqrt[3]{-8}$, we can't say $x=(-8)^{2/6} \Rightarrow x^6 = (-8)^2=64 \Rightarrow x = \sqrt[6]{64}$ and then take the positive root.

When you take the root $a^{\frac{p}{q}}$ you require $\dfrac{p}{q}$ to be in its lowest terms (specifically, you require the gcd of p and q to be 1).