rational exponents

[richardt]

Consider the following:

(-1)^2/2 = ((-1)^1/2​)² = (+/- i)²= -1 = (-1)¹ as expected.

However, ((-1)²)^1/2 = (1)^1/2 = +/- 1. (Here I am considering all roots, not necessarily principal)

Hence it is not necessarily true that a^m/n = ((a)^1/n)^m = (a^m)^1/n. This, of course, flies in the face of a^mn = (a^m)ⁿ.

So, subsequent to some investigation, I believe I have an understanding, which I ask you to corroborate for me.

First, I believe that we typically define a^m/n as, a^m/n = ((a)^1/n)^m ; a real, m and n integers. In using this definition, I inevitably get the same result whether m and n are relatively prime or not. That is (-1)^4/6 = ((-1)^1/6)⁴ = (-1)^2/3 = ((-1)^1/3)² = 1, -1/2 +/- sqrt(3)/3.

And, in keeping with that definition, it seems that, in general, we have a^m/n = (a^m)^1/n provided m and n are relatively prime.

Does this sound reasonable?

Thanks,

Rich

 

[lookagain]

Consider the following:

(-1)^2/2 = ((-1)^1/2​)² = (+/- i)²= -1 = (-1)¹ as expected.


However, ((-1)²)^1/2 = (1)^1/2 = +/- 1. (Here I am considering all roots, not necessarily principal).

Let me split up art of what you posted:


\(\displaystyle (-1)^{2/2} \ = \ ((-1)^{1/2})^2 \ \ \ \ \ \ \)You think this to be true, but it could be the case that it doesn't apply as the base is negative.


\(\displaystyle ((-1)^{1/2})^2 \ = \ (+/- i)^2 \ \ \ \ \ \ \)No, (-1)^(1/2) = \(\displaystyle \ \sqrt{-1} \ = \ i, \ \ not \ \ \pm i.\)


\(\displaystyle (+/- i)^2 \ = \ -1 \ \ \ \ \ \ \ \)It would be (i)^2 = -1.


- - - - - - -- - - - - - - - - - - - - - -- - - - - - --- - - - - -- -- - - - - - - -


\(\displaystyle ((-1)^2)^{1/2} ** \ = \ (1)^{1/2} \ \ \ \ \ \ \ \ \)Yes.


\(\displaystyle (1)^{1/2} \ = \ +/- 1 \ \ \ \ \ \ \ \ \)No, (1)^(1/2) = \(\displaystyle \ \sqrt{1} \ = 1.\)



** This is how an absolute value can be defined.


Let x equal a real number.


(x^2)^{1/2} = \(\displaystyle \ (x^2)^{1/2} \ = \ \sqrt{ \ x^2 \ } \ = \ |x|\)


When x = -1, the value necessarily comes out to +1.