
#19
Oct1605, 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: ***principle value is. I'll try MathWorld() or google() Until then.... 



#20
Oct1605, 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 