Is $ (((\sqrt{2})^ \sqrt{2})^ \sqrt{2})^{\cdots} $ an irrational number?

Question

It is well known that $ \sqrt{2} $ is an irrational number. Is there someone who can show me if this number: $$ \left(\left(\left(\sqrt{2}\right)^ \sqrt{2}\right)^ \sqrt{2}\right)^{\cdots} $$

is irrational?

Thank you for any help.

See also: How can I prove $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$ — Martin Sleziak, Jun 22 '15 at 14:33
The parentheses make this an entirely different problem than the incorrectly identified "duplicate". — rajb245, Jun 22 '15 at 14:39
@martin, the question is not duplicate see the parentheses in my formula — Zeraoulia rafik, Jun 22 '15 at 14:42
As written, this is the limit of the recurrence $a_n = (a_{n-1})^\sqrt{2}$ with $a_0=\sqrt{2}$, which can be proven to grow without bound: http://www.wolframalpha.com/input/?i=a_n+%3D+a_%28n-1%29%5Esqrt%282%29+and+a%280%29+%3D+sqrt%282%29 — rajb245, Jun 22 '15 at 14:45
Well, I did not vote to close a duplicate. At the moment in says that it is closed was marked as as a duplicate of Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct? by Community♦, see revision history. (I am not sure what it means. I thought that this indicates that somebody voted to close and the OP agrees.) It is true that I did not read the question carefully enough and did not notice the difference between the two expressions. — Martin Sleziak, Jun 22 '15 at 14:49
$x^{x^{x^{x...}}}$ Is infinite when $x>1$ it is undefined when $x=0,1$. — Aleksandar, Jun 22 '15 at 14:57
I don't know why the down vote to my question and it made other problem. — Zeraoulia rafik, Jun 22 '15 at 15:00
Other users than the downvoter themself can only speculate what was the reason for the downvote. But if you wish to discuss the reason for the downvote, there is a chatroom explicitly for this purpose. (Although there is no guarantee that somebody will be able to give you a reasonable answer there.) — Martin Sleziak, Jun 22 '15 at 15:04

score 6 · Answer 1 · answered Jun 22 '15 at 14:58

Your number is not finite. Let $x_0=\sqrt{2}$ and $x_{n+1}=x_n^{\sqrt{2}}$ for all $n\in\Bbb Z_{\ge 0}$.

Then $x_n\ge \sqrt{2}$. Proof: $x_0=\sqrt{2}$ and $x_k\ge \sqrt{2}$ means $x_k^{\sqrt{2}}>x_k\ge \sqrt{2}$.

Assume for contradiction your number is finite, i.e. $\displaystyle{\lim_{n\to\infty} x_n}:=L$ is finite.

$$\left(\left((\sqrt{2})^{\sqrt{2}}\right)^{\sqrt{2}}\right)^{\ldots}=\underbrace{\lim_{n\to\infty}x_n}_{L}=\lim_{n\to\infty} \left(\left(x_n^{\sqrt{2}}\right)^{\sqrt{2}}\right)$$

$$=\lim_{n\to\infty}\left(x_n\cdot x_n\right)=L\cdot L$$

$$\implies L=L^2\iff L\in\{0,1\}$$

Impossible, since $x_n\ge \sqrt{2}$.

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

You can write $x_1=\sqrt{2}$ and $x_{n+1}=(x_n)^{\sqrt2}$. Then the expression you are interested in is $\lim\limits_{n\to\infty} x_n$.

You have $x_2=(\sqrt 2)^{\sqrt2}$ and $x_3=((\sqrt 2)^{\sqrt2})^{\sqrt2}=(\sqrt2)^2=2$.

You also have that $x_{n+2}=(x_n^{\sqrt2})^{\sqrt2}=(x_n)^2$.

This implies that $x_{3+2k}=2^{2^k}$, and we see that this sequence is unbounded, $\lim\limits_{n\to\infty} x_n=+\infty$.

Sorry for misreading your question and suggesting in the comments that you are actually asking about this number: How can I prove $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$. (That was also the reason why I added the tetration tag.)

Ashkay · Answer 3 · 2015-06-22T14:39:44.633

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Note: my bad, I misinterpreted the parenthesization of OP's problem. As all exponents are greater than $1$, the given sequence will increase without bound.

However, here is my original proof that $\sqrt{2}^\left(\sqrt{2}^\left(\dots\right)\right) = 2$.

Let $x = \sqrt{2}^{\sqrt{2}^{\dots}}$. Then we have $\sqrt{2}^x = x$. Note that $x = 2 $ and $x = 4$ are the only solutions to this. Here, we prove that the actual solution is $2$ by showing that the sequence is bounded by $2$.

Let $x_1 = \sqrt{2}$, and let $x_n = \sqrt{2}^x_{n-1}$.

For our base case, we have $x_1 = \sqrt{2} < 2$. For our base case, assuming that $x_n \le 2$. Then, we have $x_{n+1} \le \sqrt{2}^2 = 2$. Therefore, of the two solutions, we must have $x = 2$.

edited Jun 22 '15 at 14:39

answered Jun 22 '15 at 14:27

Ashkay

1,821

3

it's not the same number is it? – hHhh Jun 22 '15 at 14:29
9

Not so! $x=4$ is another solution. In fact these are the only two solutions, but you still must check that the limit converges at all, and that it does indeed converge to 2. – Alex G. Jun 22 '15 at 14:29
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Aren't the parentheses nested the wrong way for this? It seems to be the limit of a sequence $\sqrt{2}$, $(\sqrt{2})^\sqrt{2}$, $((\sqrt{2})^\sqrt{2})^\sqrt{2}$, .... – hardmath Jun 22 '15 at 14:38
when you said ,you should prove that , then what does it meant your answer, just you proved the water by water – Zeraoulia rafik Jun 22 '15 at 14:38
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First of all, the equation $\sqrt{2}^x=x$ is not satisfied by the number the OP seeks, because of the way the original expression was parenthesized. Second, the sequence implicit in the OP's definition increases without bound, so there is no such number. – Jack Lee Jun 22 '15 at 14:38
My bad, I misread the problem! I edited the solution to reflect this. – Ashkay Jun 22 '15 at 14:43

score 0 · Answer 4 · answered Jun 22 '15 at 16:20

Claim. If $f$ is continuous, increasing and $f(x)>x$ on $(a,\infty)$, then for all $x>a$ we have

$$(\underbrace{f\circ f\circ\cdots\circ f}_n)(x)\to\infty \quad \textrm{as}\quad n\to\infty.$$

Proof. Pick $x_0>a$ and recursively define $x_{n+1}=f(x_n)$. Since $f$ is increasing, the sequence $(x_n)$ is increasing. Suppose it was bounded above. Then it must have a limit $L$. Since $f$ is continuous, we must have $f(L)=f(\lim x_n)=\lim f(x_n)=\lim x_{n+1}=L$. But $f(L)>L$ since $f(x)>x$ for all values $x>a$ (which applies to $L$ since $L>x_0>a$). We cannot have $f(L)=L$ and $f(L)>L$, so the assumption that $(x_n)$ was bounded was wrong. Any unbounded increasing sequence diverges to infinity, so in particular $x_n\to\infty$.

Application. The function $f(x)=x^{\sqrt{2}}$ is continuous, increasing and $>x$ on the interval $(1,\infty)$, and in particular we have the seed value $\sqrt{2}\in(1,\infty)$.

Is $ (((\sqrt{2})^ \sqrt{2})^ \sqrt{2})^{\cdots} $ an irrational number?

4 Answers4