This is my first post on this forum so I apologise for not knowing how to ask this question in the desired format however I will do my best to make it clear.
First consider: $$ y = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}} $$ Now, we can rewrite this as simply: $$ y = (\sqrt{2})^y $$ Then further rearrange this to: $$ y^{1/y} = \sqrt{2} $$ Thus leaving y to equal both $2$ and $4$, which obviously can't be correct. However, which is correct and why?
Thanks!
On
This is really a problem in logic. When you solve $x^{x^{x...}}=4$ as above, you are not really finding that $x=\sqrt {2} $. You find that if such a value of $x $ exists, then $x=\sqrt {2} $. Similarly if$x^{x^{x...}}=2$ has a solution then that solution must be $x=\sqrt {2} $. Clearly both equations cannot be satisfied by the same value of $x $, so the "fallacy" is actually an indirect proof that at least one of them in fact has no solution.
Having narrowed the possibilities, you now check whether $x=\sqrt {2} $ is in fact a solution to either one by putting that value into the infinite power tower and seeing where, if anywhere, it converges. Thus $x^{x^{x...}}=2$ has the proposed solution $x=\sqrt {2} $ leaving none for $x^{x^{x...}}=4$ .
It converges to $2$ because for the sequence defined by $a_1=\sqrt{2}$ and $$a_{n+1}=\sqrt{2}^{a_n}$$ We have $a_n\leq 2$ which can easily be proved by induction if $a_n\leq 2$ then $\sqrt{2}^{a_n}=a_{n+1}\leq \sqrt{2}^2=2$.This is because $a_1\leq 2$ so once the recurrence gets near $2$ it can't get past it since $\sqrt{2}^2=2$.Formally since the sequence is bounded and increasing it is convergent.
If you had $4\geq a_1>2$ instead then the sequence would converge to $4$.