$n^n$ being rational for irrational $n$ within $x_{n+1} = x_n^{-x_n}$

Question

Previously, in this question, I tried to prove the rationality of $n^n$ when $n$ is irrational, and I proved it using a recurrence relation rather than the answer I accepted, which was proving that $n$ is irrational if $n^n=2$. My question was closed because I was looking for a general proof verification and counter examples (there were duplicates of this). However now, I am looking for something different, specifically related to my proof. Here is the proof:

I decided to employ the use of a recurrence relation in order to prove it's rationality. The sequence I defined is: $$x_{n+1} = x_n^{-x_n},$$ where the first term, $x_0$ is $> 0$ to make sure the argument stays well defined. This sequence is not monotonic as it alternates between two values that seem to converge to $1$. But this sequence can be split into two, for every other number, and those sub-sequences are monotonic. These are $x_{2n+1}$ (decreasing) and $x_{2n+2}$ (increasing). Now returning to the original relation, we know that it stays above $0$ as $x_0 > 0$, so the base $x_n$ is always positive. By then placing an upper bound, we can conclude that the sequence is convergent and solve using the limit. The upper bound is simply the maximum of $x^{-x}$, as it would be the most the sequence could reach. This can be calculated by finding when the derivative is equal to $0$. $$\frac{d}{dx}[x^{-x}]=-\frac{1}{(x^x)^2}\times\frac{d}{dx}[x^x]$$ $$\frac{d}{dx}[x^{-x}]=-\frac{x^x(1+\ln(x))}{(x^x)^2}$$ $$\frac{d}{dx}[x^{-x}]=-\frac{1+\ln(x)}{x^x}$$ Now solving for when it is equal to $0$: $$-\frac{1+\ln(x)}{x^x}=0$$ $$1+\ln(x)=0$$ $$x=e^{-1}$$ Plugging into $x^{-x}$: $$= (e^{-1})^{(e^{-1})}$$ $$= e^{\frac{1}{e}}$$ $$\approx 1.445$$ Now that it's convergence is verified, we can now solve using the limit ($L$ represents the limit): $$\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n^{-x_n}$$ $$L=L^{-L}$$ $$L^{-(1+L)}=1$$ $$L=1,-1$$ We obtain two answers for $L$, however since the sequence is strictly positive then $1$ is the only answer. After solving this sequence, we get $1$, a rational number. This contradicts my initial belief that $n^n$ is irrational if $n$ is irrational. The reason for this is the way the recurrence relation is structured. $x_2$ will always be irrational as shown: $$x_2=(x_0^{-x_0})^{-(x_0^{-x_0})}$$ $$x_2=x_0^{x_0^{1-x_0}}$$ Since $x_0>0$, then $1-x_0$ will be negative, resulting in $x_2=x_0^{(\frac{1}{x_0^{x_0-1}})}$. We know that the $n$th root of $n$ is irrational, and since $x_0^{x_0-1}>x_0$, then $x_2$ must be irrational. Every term onwards follows $n^{-n}$. Since we only care about the rationality, this can be adapted to $n^n$, as the reciprocal of an irrational number is irrational. If we assumed the initial statement to be true, that $n^n$ is irrational if $n$ is irrational, then every term in the sequence after $x_2$ would be irrational, meaning that it would converge to an irrational number. This contradicts the solution found beforehand that it converges to $1$, a rational number, therefore $n^n$ is not necessarily irrational if $n$ is irrational.

EDIT: A weak point in my proof is the irrationality of $x_2$, as there are counter examples such as $x_0=1$. However I still think that the sequence becomes irrational at some point, but I still need to prove that.

In my proof, I concluded that $x_{n+1} = x_n^{-x_n}$ must have contained a counter example that led to the rational answer of $1$. My question now is, how can I find these counter examples that are specifically from the sequence I defined, not including $x_0$ and the fixed point?

Please tell me if the question does not meet guidelines and I will be happy to change it.

Answer 1

There are several incorrect points in your proof

Since $x_0>0$, then $1−x_0$ will be negative, resulting in $x_2=x_0^{(\frac{1}{x_0^{x_0-1}})}$

The former is wrong for $0 < x_0 < 1$. The latter doesn't need the former. $x_2=x_0^{\frac{1}{\left(x_0^{x_0-1}\right)}} = x_0^{x_0^{1-x_0}}$ is true regardless of whether $x \geq 1$ or not

We know that the $n^{th}$ root of $n$ is irrational, and since $x_0^{x_0−1}>x_0$, then $x_2$ must be irrational.

I don't see how all of these are connected

If we assumed the initial statement to be true, that $n^n$ is irrational if n is irrational, then every term in the sequence after $x_2$ would be rational, meaning that it would converge to an irrational number.

An infinite sequence of irrational numbers does not always converge to an irrational number, and vice versa. For example, $\frac{1}{2^n\sqrt{2}}$ converges to $0$ and $x_{n+1} = (x_n + \frac{2}{x_n})/2$ converges to $\sqrt{2}$ for $x_0 > 0$ and $-\sqrt{2}$ for $x_0 < 0$. So this point is invalid.

In short, you didn't show why it is wrong to believe that $n^n$ would be irrational if $n$ is irrational. I suggest moving to a different method. For example, the belief implies that $x^x = r$ must have a rational root for $r > \frac{1}{e}$ and rational, otherwise $r$ would be a counterexample. Then $x$ can be expressed as a reduced fraction $p/q$. We then have $(p/q)^p = r^q$ or $p^p = r^qq^p$. However, this equation does not have solution for $r = 2$ or any other non-power integer values (you should find this familiar if you have already proven roots of $2$ are irrational) therefore the initial assumption is disproven.