# How to prove that $x\uparrow \uparrow 1/2 = \sqrt x_s$ [closed]

This may be a stupid question but when we work with exponentiation we can see that $x^{\frac 12}=\sqrt x$ because:

$x^{\frac 12}\times x^{\frac 12}=x^{\frac 12+\frac 12}=x^1=x$

and

$\sqrt x \times \sqrt x={\sqrt x}^2=x$

Now it seems obvious working with tetration that $x\uparrow \uparrow \frac 12 = \sqrt x_s$ (where $\sqrt x_s$ is the super square root so that $\sqrt x_s^{\sqrt x_s}=x$) but I'm not sure so how do I actually prove/disprove this ? Can it be generalized for higher degree operations like $x\uparrow \uparrow \uparrow \frac 12=\sqrt x_{ss}$ (where $\sqrt x_{ss}\uparrow \uparrow \sqrt x_{ss}=x$) and so on ?

## closed as off-topic by Did, Jonas Meyer, C. Falcon, Leucippus, ShaileshJul 3 '16 at 1:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

## 2 Answers

This would only be obvious if there's a rule that $(a\uparrow\uparrow b)\uparrow\uparrow c = a\uparrow\uparrow bc$, because in that case $a\uparrow\uparrow (1/b)$ ought to be some number $x$ such that $x\uparrow\uparrow b=a\uparrow\uparrow 1$. However such a rule does not hold in general. For example with $a=b=c=2$, $$(2\uparrow\uparrow 2)\uparrow\uparrow 2 = 4^4 = 256$$ but $$2\uparrow\uparrow (2\cdot 2) = 2^{2^{2^2}} = 65536$$

Currently, there is no accepted way for extending tetration to real numbers, so it might not be true.