# Is sqrt(2)^(sqrt(2)) rational ? And sqrt(2)^(sqrt(2)^sqrt(2))?. And sqrt(2)^(sqrt(2)^(sqrt(2)^cdots))?

Sep 16, 2016

${\sqrt{2}}^{\sqrt{2}}$ is irrational.
""^oosqrt(2) is rational.

#### Explanation:

While I haven't found a proof that ${\sqrt{2}}^{{\sqrt{2}}^{\sqrt{2}}}$ is irrational, here are answers to the first and third parts.

${\sqrt{2}}^{\sqrt{2}}$ is irrational:

The Gelfond-Schneider theorem states that given algebraic numbers $a , b$ where $a \ne 0 , 1$ and $b$ is irrational, ${a}^{b}$ is transcendental.

As $\sqrt{2}$ is a root of ${x}^{2} - 2$, it is algebraic. It is well known that $\sqrt{2}$ is irrational, although this fact may be proven by supposing that $\sqrt{2} = \frac{p}{q}$ with $p , q \in \mathbb{Z}$, squaring both sides, multiplying by ${q}^{2}$, and then noting that the left hand side will have an odd exponent for $2$ in its prime factorization, while the right hand side will have an even exponent, a contradiction.

By the above, ${\sqrt{2}}^{\sqrt{2}}$ fulfills the conditions for the theorem, and thus is transcendental (and therefore irrational).

sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) = ""^oosqrt(2) is rational:

We will prove the stronger result that ""^oosqrt(2)=2.

First, to show that ""^oosqrt(2) converges, we will show that the sequence a_n = ""^nsqrt(2) is monotone increasing and bounded above.

We proceed by induction. As our base case, note that
${a}_{1} = \sqrt{2} < {\sqrt{2}}^{\sqrt{2}} = {a}_{2} < {\sqrt{2}}^{2} = 2$. Now, suppose that for some positive integer $k$, the sequence ${a}_{1} , {a}_{2} , \ldots , {a}_{k}$ is increasing and ${a}_{k} < 2$. Then
${a}_{k} = {\sqrt{2}}^{{a}_{k - 1}} < {\sqrt{2}}^{{a}_{k}} = {a}_{k + 1} < {\sqrt{2}}^{2} = 2$
Thus, by induction, ${a}_{n}$ is increasing and bounded above by $2$.

Now that we know ""^oosqrt(2) converges, let a= ""^oosqrt(2). Then ${\sqrt{2}}^{a} = a$, meaning $a$ is a root of $f \left(x\right) = {\sqrt{2}}^{x} - x$. Finding the critical points of $f \left(x\right)$, we have
$f ' \left(x\right) = \frac{1}{2} \ln \left(2\right) {\sqrt{2}}^{x} - 1$
$\implies f ' \left(x\right) = 0 \iff x = {\log}_{\sqrt{2}} \left(\frac{2}{\ln} \left(2\right)\right)$

As $f \left(x\right)$ has only one critical point, it can have at most two roots. By observation, $2$ and $4$ are both roots of $f \left(x\right)$, meaning $a \in \left\{2 , 4\right\}$. But, by the above work, $a \le 2$, meaning $a = 2$.

Sep 16, 2016

Naive considerations.

#### Explanation:

Analysing

$x = {\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{\cdots}}}$ we can propose

$x = {\left(\sqrt{2}\right)}^{x}$

Now suppose that

$f \left(x\right) = f \left(a\right)$ Of course that a possible solution is $x = a$. Now posing

$x = {\left(\sqrt{2}\right)}^{x}$ we have

${x}^{\frac{1}{x}} = {2}^{\frac{1}{2}}$ or $g \left(x\right) = g \left(2\right)$ then a solution is

$x = 2$ rational.

Now if

$y = {\left({\left(\sqrt{2}\right)}^{\sqrt{2}}\right)}^{x}$

What is ${\lim}_{x \to \sqrt{2}} y$ ? or
What is ${\lim}_{x \to \sqrt{2}} \log y$ ? for

$\log y = x \log \left({\left(\sqrt{2}\right)}^{\sqrt{2}}\right) = x \sqrt{2} \log \left(\sqrt{2}\right) = x \frac{\sqrt{2}}{2} \log 2$ so

${\lim}_{x \to \sqrt{2}} \log y = {\lim}_{x \to \sqrt{2}} x \frac{\sqrt{2}}{2} \log 2 = \log 2$

so $y = 2$ rational. This is a long way to get at

${\left({\sqrt{2}}^{a}\right)}^{b} = {\left(\sqrt{2}\right)}^{a b}$. Here considering $a = b = \sqrt{2}$ we get at the same result.

Sep 16, 2016

${\left({\sqrt{2}}^{a}\right)}^{b} = {\sqrt{2}}^{a b}$, so ${\left({\sqrt{2}}^{\sqrt{2}}\right)}^{\sqrt{2}} = {\sqrt{2}}^{\sqrt{2} \sqrt{2}} = {\sqrt{2}}^{2} = 2$ is rational.