Register to reply

Simple Proof

by bomba923
Tags: proof, simple
Share this thread:
bomba923
#19
Oct16-05, 04:02 PM
P: 736
Also in which you mean [itex] p \ne 1 [/itex],

because where [tex] p = 1 [/tex],

[tex] \left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \Rightarrow [/tex]

[tex] a ^ {1/q} = a ^ {1/q} [/tex]

Again, why is it incorrect to say:

[tex] \left( { - 1} \right)^{\frac{1}{3}} = - 1\;{\text{because }}\left( { - 1} \right)^3 = - 1 [/tex]

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

Quote 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....
Zurtex
#20
Oct16-05, 07:03 PM
Sci Advisor
HW Helper
P: 1,123
Actually I meant:

[tex]\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\}[/tex]

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.


Register to reply

Related Discussions
My simple proof of x^0 = 1 General Math 30
Simple Proof Calculus & Beyond Homework 1
Very simple proof of P(A) > P(B) Precalculus Mathematics Homework 3
A simple proof... General Math 2
Simple proof General Math 9