A peculiar observation about infinity.

Question

Let ${\sqrt2^\sqrt2}^{\sqrt2^...}=y$.

Then $\sqrt 2^y=y$

$\implies \sqrt 2=y^{1/y}$ $\implies \sqrt 2 =1$ $\implies 2 =1$ !! but how come that be. Can anyone explain this and point out what is wrong in my reasoning?

How is $y^{\frac 1 y} = 1$? In fact, I'm almost certain that it's not. — Matthew Levy, Dec 27 '14 at 08:44
lol, okay must be a first year calculus student or something. I see that error all the time — Matthew Levy, Dec 27 '14 at 08:51

score 3 · Answer 1 · answered Dec 27 '14 at 08:46

3

First, the value of that power tower is 2. (Which converges for $e^{-e}\le x\le e^{1/e}$) Then we can see that $$\sqrt{2}=2^{1/2}=\sqrt2.$$ You are assuming $y=1$ and therefore simplifying $y^{1/y}$ to 1.

answered Dec 27 '14 at 08:46

Teoc

8,700

What is $x$? What notion of convergence is being used? – ajd Dec 27 '14 at 08:48
Sequence of finite towers.. obviously – Matthew Levy Dec 27 '14 at 08:49
2

x is the value that is being tetrated, in this case, $\sqrt 2$ – Teoc Dec 27 '14 at 08:50

score 1 · Answer 2 · answered Dec 27 '14 at 09:35

1

$\sqrt 2=y^{1/y}$ has solutions $y=2$ or $y=4$, though one of these is implausible for the power tower

answered Dec 27 '14 at 09:35

Henry

157,058

A peculiar observation about infinity.

2 Answers2