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

I don't know that it does. 3^1/2=(2[squ]1)¹=2?

HallsofIvy

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

I take it you mean: "Why is b^m/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": b^mbⁿ= b^m+n and (b^m)ⁿ= b^nm.

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

Finally, if (b^m)ⁿ= b^mn is to be true for all numbers, we must have (b^1/m)^m= b¹= b. Since n &radic (b) is define as "the positive number whose nth power is b, we must define b^1/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.