# Are hyperoperations < 3 to a reciprocal of a positive integer equivalent to the 'root' inverse to that integer?

Can the logic of $\sqrt[n]x = x^{1/n}$ be applied to tetration and other natural numbered hyperoperations greater than exponentiation, or, do reciprocals of positive integers as the second argument of a hyperoperation greater than exponentiation always equate to the root analog of that operation? Is there a general proof of this?

An example of what I mean is x tetrated to the one half and the super-square root of x being the same value, then extended to all positive integer denominators and all hyperoperations above exponentiation.

• Your phrase "the logic of" is a bit vague. Can you please clarify? – Adrian Keister May 17 at 14:31
• Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos May 17 at 14:40
• I don't see why not. Just define it that way. However, I'm not certain it's all that useful. – Arthur May 17 at 14:56

No. The reason why $\sqrt[n]x = x^{1/n}$ is because $(x^a)^b = x^{ab}$, so assuming $x \in (0,\infty)$ we would have $(\sqrt[n]x)^n = x = x^{n/n} = (x^{1/n})^n$, which implies $\sqrt[n]x = x^{1/n}$. There is no similar identity for hyperoperations.

For example, let's look at tetration. There is no standard definition of ${^b}a = a^{a^{{...}^a}}$ with height $b$ for noninteger $b$, so the expression ${^{1/b}}a$ is not well-defined without a continuous interpolation of some kind. One could imagine defining ${^{1/b} a}$ to be the solution to ${^b}x = a$. However, there's no reasoning to justify this definition, because unlike exponentiation, tetration does not satisfy ${^{bc}}a = {^b}({^c}a)$. Why not? Well let's just look at an example with $a = b = 2$ and $c = 3$. Then ${^{2\cdot 3}2 = 2^{2^{2^{2^{2^2}}}}} \ne {^2}({^3}2) = {^2}(2^{2^2}) = 16^{16}$. Thus the reasoning behind $\sqrt[n]x = x^{1/n}$ doesn't apply to tetration. You might define ${^{1/b}}x$ to be the inverse of ${^b}x$, but this definition will break down once you try extending to noninteger $b$. To take your logic a bit farther, suppose we define $^{1/n}x$ as the inverse function of ${^n}x$ and then define $^{m/n}x = {^{1/n}}({^m}x)$. This actually not even well defined: By our working definition, ${^{1/2} 4} = 2$, because ${^2}2 = 2^2 = 4$. But also by our definition this should be the same as ${^{2/4} 4} = {^{1/4}}({^2}4) = {^{1/4}}(4^4) = {^{1/4}}256$, i.e. the solution to $x^{x^{x^x}} = 256$, which numerically comes out to be about $x = 1.8973...$, clearly not equal to $2$.

We can define $n-$fold tetration as $f(x,n)=x^{x^{x^{\ldots x}}}$ with $n$ layers in the stack. As this function is monotonic in $x$ (for large enough $x$) we can define an inverse in the first argument by $g(y,n)=x$ when $f(x,n)=y$. According to Wikipedia, there is no accepted way to extend $f(x,n)$ to allow the second argument to be real, so we do not have continuity in the second argument to allow an inverse there. Given $x$ and a target $y\ge 1$ you can generally find a nonnegative integer $n$ such that $f(x,n) \le y \lt f(x,n+1)$

If you want to use the $1/n$ symbolism, we could just define $f(x,1/n)=g(x,n)$ using the above $g$. As we hadn't defined $f$ on fractional arguments before this is not a problem. In this case we will have $f(x,1/2)=y$ where $y^y=x$ We need to require that $x \gt (1/e)^{(1/e)}\approx 0.6922$ to get an invertible function.