# Why does b^(m/n) = sqrt(n,m)?

split
Why does b^(m/n) = (nsqrt(b))^m?

Hi, as the subject says, why does b^(m/n) = (n&radic;b)^m?

I don't understand how you can multiply a number by itself less than one times.

Thanks.

EDIT: Finally GOT IT RIGHT.

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synergy
b^(m/n)= nã(b^m) "the nth root of b to the m power"
you could also write (nãb)^m
Aaron

split
I meant that but I was just thinking about too many things. It's been fixed. I'm asking for an explanation of why that is true.

StephenPrivitera

Originally posted by split
Hi, as the subject says, why does b^(m/n) = (n&radic;m)^m?

I don't know that it does. 31/2=(2[squ]1)1=2?

Homework Helper
The stupid errors just keep piling up don't they!

I take it you mean: "Why is bm/n= n &radic (b)m?"

Let's start with "I don't understand how you can multiply a number by itself less than one times."

You can't. bn is defined as "multiply b by itself n times" only if n is a positive integer (counting number).

However, in that simple situation, we quickly derive the very useful "laws of exponents": bmbn= bm+n and (bm)n= bnm.

We then define bx for other number so that those laws remain true.

For example, IF the laws of exponents are to be true for x= 0, then we must have bn= bn+0= bnb0. As long as b is not 0 we can divide both sides of the equation by bn to bet b0= 1. That is, we MUST define b0= 1 or the laws of exponents will no longer hold.

Now we can see that bn+(-n)= b0= 1. If the laws of exponents are to hold for negative exponents as well, we must have bnb-n= bn+(-n)= 1 or, again dividing both sides of the equation by bn, b-n= 1/bn.

Finally, if (bm)n= bmn is to be true for all numbers, we must have (b1/m)m= b1= b. Since n &radic (b) is define as "the positive number whose nth power is b, we must define b1/m= m &radic (b).

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split
Thanks HallsOfIvy, your explanation was very clear.

And yes, the errors kept piling up! I have fixed everything but the subject (I don't believe it can be changed. Am I wrong?) so if anyone wants to read it in the future it should make sense.