What is the Exponential Inverse for Identity Element 1?

[JKaufinger]

What is the exponential inverse? And first let me explain what I mean when I am talking about inverse.

I am not talking about anything where f(f^-1(x)) = x

What I mean is in one operation, when that operation is applied onto the inverse of that operation it equals the identity element of that operation.
For example, 5 + -5 = 0 because 0 is the identity element in addition (x+0=x), and negative is the inverse.
Also for example, 5 * (1/5) = 1 because 1 is the identity element in multiplication (x*1=x), and reciprocal or 1/x is the inverse.
In matrices, [matrix] * [inverse of that matrix] = [identity matrix]

So, what is this for exponentiation?

Let me define what I want to use for identity element: anything, that when that operation is applied to it, it equals the original number. So in exponentiation, the identity element is 1 because x^1 = x;

So now, we want x^what = identity element; And since the identity element of x is 1 then this can become: x^what = 1?

I have considered 0 as the answer, but that doesn't make any sense. Even though x^0 = 1, and thus the identity element, how is 0 the exponential inverse of x? All of the other inverses for addition and multiplication (-x, 1/x) all include x. So why, here, is 0 the inverse; what makes exponentiation and 0 so special?

In addition, root and logarithm do not work. Root doesn't work because x^(x^(1/anything)) does not equal 1 (unless x = 1). and neither does x^log_x(anything)

So x^what = 1, where "what" can't be zero, a root, or a logarithm.

 

[JKaufinger]

Oh and let me also add that taking a higher operation, and applying the negative, will result in the lower operation's inverse.

x * -1 = the additive inverse = -x
x ^ -1 = the multiplicative inverse = 1/x

So to find the exponential inverse, then that would be the negative tetration (the operation higher than exponentiation).

 

[confinement]

In the system you have defined y = 0 is the only solution to x^y = 1 so that is all you have.

 

[HallsofIvy]

How are you thinking of exponentiation? xⁿ as a function of x for fixed n? Or a^x for fixed a? If the former then the inverse function is the nth root. If the latter then it is the logarithm base a.

 

[JKaufinger]

Let me explain why I think log and root don't work for this, but first let me reiterate some things:

In addition and multiplication, the inverses have this property: When they are added (or multiplied) to the original, non-inverted number, they equal the identity element (0 and 1 respectively).

So if we try to apply this using exponentiation, then how can we do that?

[x] [operation] [inverse x] = [identity element]
x + (-x) = 0
x * (1/x) = 1
x ^ [exponential inverse x] = 1

what is exponential inverse of x that will fit here? (except 0)

While it is true, that to get the original x, root and logarithm can apply. sqrt(x^2)=x and logarithm works too. Allow me to explain my reasoning below.

The reason the inverse function works in addition and multiplication is because of the original property I defined above, where doing that operation on the inverse will equal the identity element. So for x+5, to get x you would make it x+(5+(-5))=x. In multiplication, x*5, you would do x*(5*(1/5))=x. Following this, then for x^2, to get x you would need to do sqrt(x^2); seems pretty obvious. But notice how in the examples for multiplication and addition, the inverse we are doing never touches x. The 5 and 1/5 cancel out, and the 5 and -5 cancel out. x is left alone. But here, the square root DOES affect x, and so this doesn't follow the pattern. What I mean is that the square root is ((x^2)^1/2). Here, you can just multiply 2 and 1/2 and they become 1, thus making it x^1=x; Great. It all works out fine. So what is the problem? Well, we are not exponenting 2^1/2, but we are exponenting the entirety of x^2, which we will then root. But we want x^(2^(whatever)). Essentially, in multiplication and addition, x is not affected when the inverse is done; In exponentiation, to get the original x, you need to affect x. What I want, is a number [defined as "whatever"] which only affects the 2. The only solution to this is 0, because x^2^0 = x^1 = x; Ok, that's great. We found it, it is 0. Now where's the problem? We want something other than 0, since 0 is not related to x; it is the same for all x; Whereas -x and 1/x include x, 0 does not include x;

Do you kind of understand my point a bit better?

 

[Werg22]

You should realize that in \mathbb{R} the exponential function f(t) = a^{t} is injective, so there is only one t for which a^{t} = 1, which is 0. So in the 2-variable function g(x,y) = x^y, if and only if y = 0 does g(x,y) = 1, whatever the choice of x. Even in \mathbb{C} it doesn't make a difference because it will only work for y \in \{2{\pi}k : k \in \mathbb{Z} \} which does not 'depend' on x either.

 

[qntty]

The only solution to this is 0, because x^2^0 = x^1 = x

Not true, it would be (x^2)^0 = 1. You have to apply it to the entire function. This doesn't matter with addition/multiplication because they're communicative but it does here.

So for x+5, to get x you would make it x+(5+(-5))=x ... But notice how in the examples for multiplication and addition, the inverse we are doing never touches x

This is because addition commutes. What you're actually doing is
x+5 = k
(x+5) + (-5) = k + (-5)
x+(5+(-5)) = k + (-5)
x + 0 = k + (-5)
x = k + (-5)No number fits your description. For real X and n, the only way to have X^n = 1 is to have n=0 (and X \not= 0).