Why do fractional exponents result in the square root operator?

[hackedagainanda]

Homework Statement

What does 9^(1/2) = to?

Relevant Equations

9^`1 = 9
9^0 = 1
(x^a)(x^b)= x^a+b)

So I got the answer through a little addition i.e 9^(1/2) multiplied by 9^(1/2) = 9^1 or 9

3 x 3 = 9 so 3 is the answer to what is 9^(1/2)

I've tested this out with a few other numbers and have made this generalization, x^(1/2) = √x

It seems to make the equations orderly and consistent but is it by definition or is there a reason why raising it to a fractional exponent gives you this answer. I'm trying to not just memorize this rule, so I can get a better intuitive grasp on how the algebra works.

 

[Mark44]

Homework Statement:: What does 9^(1/2) = to?
Relevant Equations:: 9^`1 = 9
9^0 = 1
(x^a)(x^b)= x^a+b)

So I got the answer through a little addition i.e 9^(1/2) multiplied by 9^(1/2) = 9^1 or 9

3 x 3 = 9 so 3 is the answer to what is 9^(1/2)

I've tested this out with a few other numbers and have made this generalization, x^(1/2) = √x

It seems to make the equations orderly and consistent but is it by definition or is there a reason why raising it to a fractional exponent gives you this answer. I'm trying to not just memorize this rule, so I can get a better intuitive grasp on how the algebra works.

$9^{1/2} = \sqrt 9 = 3$

 

[Office_Shredder]

This is because (at least for positive real numbers)

$$(x^2)^{1/2} = x^{2/2} = x$$

Because in general $(x^a)^b = x^{ab}$.
So raising to the 1/2 power performs the inverse operation of squaring, and hence it is the square root operator as you discovered.