Quote:

Originally Posted by

**NuclearHellRaven**
What? An infinite root? You mean like (1/2)^(1/infinity)? Or (1/2)^0? Or 1?

Yeah. The "infinitieth root" of any number (except for zero of course) is 1.

I meant infiniteth as (1/2)^(1/infinity), I simply didn't find a way at that moment to really put it into words, and I didn't think of using the fractional exponent...

The problem is that if you say that the "infiniteth root" of any numbers is 1, that means that while 1 to any powers is 1, 1^infinity = every signle numbers... that's why I say it's an "integer approximation".

1 is actually a number that is closer to the real value than could actually be useful... if our goal was anything but the truth...

---- EDIT ----

Now, this course is long past, and I can't remember the exact terms, but bear with me if I screwed over any of those terms, I'm talking in term of concepts, not vocabulary.

Basically, every sets has 2 operators, * and +. And for every sets, you have to define their operators behaviors.

For starters, you will always have a null value. An N, where X+N=N, and XN=N. For regular numbers, these null values are 0 (for +) and 1 (for *).

Next, you have symmetrical values, where X+S=N=0, and XS=N=1. These are -X, and 1/X.

There are also a lot of other rules that defines the sets (actually, not a lot, more like 3-5, but they're overly complex and basically all includes more than 1 meaning), but it can be said that all operations a succession of those 2. So basically, a Power is simply multiple multiplications.

Also, all operations has its "opposite", so if X*Y=R, then R/Y=R*(1/Y)=X, because of the symmetrical values.

From there, you can safely say that if X^Y=R, then R^(1/Y)=X

So, if X^(infinite) = 0, then 0^(1/infinite) = X, no matter X's value... or basically, X = {Real Numbers}

Or, if X^(1/infinite) = 1, then 1^(infinite) = X, no matter X's value... same conclusion