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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 ?

AlienRem
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    I don't want to be boring, but I think that this question is a duplicate, have you tried to search about this on MSE? http://math.stackexchange.com/questions/402050/tetration-and-its-inverse-to-various-exponents http://math.stackexchange.com/questions/1274478/the-physical-meaning-of-tetration-with-fractional-power-tower http://math.stackexchange.com/questions/1099279/complex-and-decimal-tetration

    Just few examples, and not to mention the Tetration Forum... http://math.eretrandre.org/tetrationforum/

     – MphLee May 30 '15 at 14:49
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    Ops sorry @MphLee I made a quick search but probably it was too quick... I'll change my question in a few hours with something new (I hope). – AlienRem May 30 '15 at 15:39

2 Answers2

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

wythagoras
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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 $$ 2\uparrow\uparrow (2\cdot 2) = 2^{2^{2^2}} = 65536 $$

hmakholm left over Monica
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