The question of whether a rational number ( r \in \mathbb{Q} \setminus \mathbb{Z} ) can satisfy ( r^{r^{r^r}} \in \mathbb{Q} ) is not definitively resolved by current mathematical knowledge. Here's the breakdown:

1. First Exponentiation (( r^r )): For ( r \in \mathbb{Q} \setminus \mathbb{Z} ), ( r^r ) is irrational. This follows from properties of rational numbers and logarithms, as...

Short answer: this is open.

Details:

• Let r be a non-integer rational (typically we also restrict to r>0 to avoid branch issues with real exponentiation). Writing r=p/q in lowest terms, one checks that r^r=(p/q)^{p/q} is algebraic but irrational. Hence, by the Gelfond–Schneider theorem (which says that a^b is transcendental if a is algebraic ≠0,1 and b is algebraic irrational), we get that r^{r^...

Write r = p / q ∈ ℚ\ℤ in lowest terms and assume r > 0 (the question only makes sense for a positive base; the cases r = 0, –1 or r < 0 with real-valued branches of powers are irrelevant here).
Put

 x₁ = r ^ r  (algebraic, irrational),
 x₂ = r ^ x₁ (transcendental by Gelfond–Schneider),
 x₃ = r ^ x₂ = r ^{r ^{r ^ r }}.

The question is: “Can x₃ be rational?”

What we already know....

<think> The problem asks: If \( r \) is a rational number but not an integer, is it possible that \( r^{r^{r^r}} \) is rational? The expression is a power tower, specifically \( r^{r^{r^r}} \), which I believe means \( r^{(r^{(r^r)})} \), evaluated from the top down.

First, I recall that for a rational ( r ) not integer, ( r^r ) is irrational. The user mentions that it's straightforward to pr...

This is an excellent question that quickly goes from standard number theory results to the frontier of open problems in transcendental number theory. Your analysis of the first two levels of exponentiation is perfectly correct.

Let's break it down step-by-step.

Let be a rational number that is not an integer, with and .

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Step 1: Is $r^r \in \mathbb{Q}...

Let

  r = p/q ∈ ℚ \ ℤ ,   q > 1,  r > 0 (1)

and build the tower

  T3 (r) := r^{,r^{,r^{;}} } = r^{r^{r}} . (2)

Step 0 (r^r)
Because r is rational, r = p/q with (p,q)=1.
Write r^r = (p/q)^{p/q} = (p^{,p})^{1/q}/(q^{,p})^{1/q}.
Taking an n-th root of an algebraic number is still al...

This is indeed a very interesting question about iterated exponentiation (tetration) and transcendence theory.

Current State of Knowledge

You're correct that this appears to be an open problem. The question of whether for goes beyond what current transcendence theory can handle.

What We Know

As you noted:

1. For $r ...

The Rationality of for Non-Integer Rationals

This is an interesting question in transcendental number theory. Let me analyze it step by step.

What We Know

First, let's clarify what's already established:

1. For (rational but not an integer), we know that is irrational.

2. Using the Gelfond-Schneider theorem, we can prove that $r...

For convenience write a tower of height n as
Tₙ(r) = r^{r^{.^{.^{r}}}} (n copies of r, the usual right-associate convention).

1. It is classical (indeed an immediate corollary of the Gelfond–Schneider theorem) that if r∈ℚ∖ℤ and r≠0,±1 then
   • T₁(r)=r is rational (by hypothesis),
   • T₂(r)=r^r is transcendental (hence irrational),
   • T₃(r)=r^{r^r} is transcendental.

Th...