# Simple Proof

by bomba923
Tags: proof, simple
P: 736
Also in which you mean $p \ne 1$,

because where $$p = 1$$,

$$\left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \Rightarrow$$

$$a ^ {1/q} = a ^ {1/q}$$

Again, why is it incorrect to say:

$$\left( { - 1} \right)^{\frac{1}{3}} = - 1\;{\text{because }}\left( { - 1} \right)^3 = - 1$$

I even searched for this problem in Dr. Math, at
http://mathforum.org/library/drmath/view/55604.html. It says:

 Quote by Dr.Math When the base is negative and the exponent is rational with an odd denominator, like (-64)^(1/3), there is a negative real number -4 which can be chosen to be the principal value. When the base is negative and the exponent is rational with an even denominator, there is no real root. For (-64)^(1/2), you have the two complex roots 8*i and -8*i. It is not clear which of these, if either, you can or should choose for the principal value. When the base is negative and the exponent is irrational, you will also not have any real root, and no clear choice for the principal value. Furthermore, there are problems making these choices in such a way that the function resulting, mapping the reals into the complex numbers, is continuous and differentiable. As a result of these considerations, it is very clear that it is a good idea to restrict one's attention to exponential functions with positive bases, and avoid the difficulties encountered with negative ones.
***Then again, I would have to look up what exactly a
***principle value is. I'll try MathWorld() or google()

Until then....
 Sci Advisor HW Helper P: 1,123 Actually I meant: $$\left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \quad \forall p \in \mathbb{Z} \quad \text{and} \quad \forall q \in \left. \mathbb{Z} \right\backslash \left\{ 0 \right\}$$ It's really worth considering and understaing why negative powers raised to rationals is defined on the princaple value. If this wasn't the case then there would be a lot of problems. Seriously try and prove where this is valid for a.

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