Irrationality of $x$ if $x > 1$ and $x^x = 2$

Question

Show that if $x>1$ and $x^x=2$, then $x$ must be irrational.

I know you have to show that it cannot be reduced into a form $\frac pq$, but get stuck with quite ugly algebra.

Answer 1

Suppose $x$ is rational. Then $x=p/q$ in lowest terms with $p,q$ integers, and $p>0$. $x>1$ so $p>q$, and $$\left(\frac{p}{q}\right)^{p/q}=2 \implies p^{p}=2^{q}q^{p}$$ So $2$ divides $p$; write $p=2r$. $$2^{p}r^{p}=2^{q}q^{p} \implies 2^{p-q}r^{p}=q^{p}$$ So $q$ is divisible by $2$; contradiction.

Answer 2

Suppose $x=p/q$ with $(p,q)=1$, then $$\left(\frac{p}{q}\right)^{\frac{p}{q}}=2\Longrightarrow p^p =2^q q^p$$ But $(p,q)=1$ means $(p^p, q^p) = 1$. Therefore $(p^p,q^p)=(2^qq^p, q^p)= q^p=1$, or $q=1$. So we need an integer $p$ such that $p^p=2$. Clearly no such integer exists.