# Math Help - Rational Powers

1. ## Rational Powers

Course: Advanced Calculus aka Real Analysis

Prove that, for a positive number $x$ and integers $m$and $n$,

$(x^{\frac{1}{n}})^m = (x^m)^{\frac{1}{n}}$

How is this proof done?

I tried this:

$(x^{\frac{1}{n}})^m = (x^{\frac{1}{kn}})^{km}$

for $u > 0$

$u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{nk}$

then i tried to set $u = x^m$ and got completely lost.

Can anyone tell if I was doing it correctly or show some steps on how to do this ?

2. Originally Posted by harish21
Course: Advanced Calculus aka Real Analysis

Prove that, for a positive number $x$ and integers $m$and $n$,

$(x^{\frac{1}{n}})^m = (x^m)^{\frac{1}{n}}$

How is this proof done?

I tried this:

$(x^{\frac{1}{n}})^m = (x^{\frac{1}{kn}})^{km}$

for $u > 0$

$u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{nk}$

then i tried to set $u = x^m$ and got completely lost.

Can anyone tell if I was doing it correctly or show some steps on how to do this ?
You need to realize that there are different levels at which this function is introduced. Give us exactly how you're class has defined the function.

3. Originally Posted by Drexel28
You need to realize that there are different levels at which this function is introduced. Give us exactly how you're class has defined the function.
This is what the book has stated:

Definition: For $x>0$ and rational number $r= m/n$, where m and n are integers, with n positive, we define

$x^r \equiv (x^m)^{\frac{1}{n}}$

since rational numbers can be expressed in different ways as quotients of integers, we need to establish that if m is an integer and n and k are natural numbers then,
$(x^m)^{\frac{1}{n}} = (x^{km})^{\frac{1}{kn}}$ .......(1)

Since, for u>0,

$u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{kn}$,

setting $u = x^m$, we have

$x^{km} = (x^m)^k = [(x^m)^{\frac{1}{n}}]^{kn}$, from which (1) follows and so rational powers are properly defined.

Thats all the book has given about rational power and asked us to verfify the question I posted.