If x is a positive rational number, but not an integer, then can xxxx be a rational number ?

We can prove that if x is a positive rational number but not an integer, then xx can not be rational:

Denote x=ba,(a,b)=1,xx=dc,(c,d)=1,

(ba)ba=dc⇒(ba)b=(dc)a⇒bbca=daab

Since (a,b)=1,(c,d)=1, we have ca∣ab and ab∣ca, hence ab=ca.

Since (a,b)=1, ab must be an ab-th power of an integer, assume that ab=tab, then a=ta, where t is a positive integer, this is impossible if t>1, so we get t=1,a=1, hence x is an integer.Then from Gelfond–Schneider theorem , we can prove that if x is a positive rational number but not an integer, then xxx can not be rational. In fact, it can not be an algebraic number, because both x and xx are algebraic numbers and xx is not a rational number.

- Can we prove that xxxx is irrational?
- Can xx…(n−th)…x (n>1) be rational?

**Answer**

Let x1=x and by induction xn+1=xxn: so x1=x is rational by hypothesis, x2=xx is algebraic irrational, x3=xxx is transcendental by the Gelfond-Schneider theorem, and the question is to prove that x4,x5,… are transcendental (or at least, irrational).

I will assume Schanuel’s conjecture and use it to prove by induction on n≥2 that x3,x4,…,xn are algebraically independent (and, in particular, transcendental). For n=2 there is nothing to prove: so let me assume the statement true for n and prove it for n+1.

Since x2 is irrational, x1 and x2 are linearly independent over Q. The induction hypothesis implies that 1,x3,…,xn are linearly independent over Qalg (the algebraics), so in particular x1,…,xn are linearly independent over Q, and, of course, this implies that x1⋅log(x),…,xn⋅log(x) are also such.

Now Schanuel’s conjecture then implies that among the 2n quantities x1log(x),…,xnlog(x),x2,…,xn+1 at least n are algebraically independent. Of course, we can remove x2 from that list since it is algebraic, we can similarly replace both x1log(x) and x2log(x) by simply log(x): so among log(x),x3,…,xn+1,x3log(x),…,xnlog(x) at least n are algebraically independent. But (for any i) this independent set cannot contain all three of xi, log(x) and xilog(x), and if it contains two of them then we can choose any two (namely, xi and log(x)): so that, in fact, the n quantities log(x),x3,…,xn,xn+1 are algebraically independent, which concludes the induction step (and moreover shows that log(x) is also independent with the rest).

**Attribution***Source : Link , Question Author : lsr314 , Answer Author : Gro-Tsen*