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

[split]

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

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

Thanks.

EDIT: Finally GOT IT RIGHT.

[synergy]

b^(m/n)= n�ã(b^m) "the nth root of b to the m power"
you could also write (n�ãb)^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√m)^m?


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

[HallsofIvy]

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).

[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.