If $0<x<1$ and $x$ is rational, does $2^x$ have to be irrational? Why? Also, if $2^x$ is rational, and $0<x<1$ and $x$ is rational, does $x$ have to be irrational? (i.e. contrapositive of the first statement)
I believe a better (and easier) way would be to prove the latter statement and then the former one would be proven as well (due to contra-positivity), however I was unable to prove that anyway.
On
$2^\frac{p}{q}$ is the solution of the equation $x^q-2^p=0$.
As per Rational root theorem, the only way for a rational number $x=\frac{m}{n}$ (in its reduced form) to be a root of that equation is when $n\mid 1$ (and $m\mid 2^p$), which implies that $x$ itself is an integer (dividing $2^p$).
(Note: in Algebra we say that $\mathbb Z$ is the ring of integers over $\mathbb Q$, as $\mathbb Z$ is precisely the set of rational zeros of monic polynomials, i.e., polynomials with leading coefficient $1$. The proof is the same as above, using the Rational Root Theorem.)
Thus, no such solution can be strictly between $1=2^0$ and $2=2^1$, which would be the case for $0<\frac{p}{q}<1$.
On
Let $x=\frac ab\in \mathbb Q$, where $\gcd(a,b)=1$ and $0<a<b$
Prove by contradictions, assume $2^x\in \mathbb Q$, then $$2^x=2^{\frac a b}=\frac p q,~~~\gcd(p,q)=1\Longrightarrow 2^aq^b=p^b\Longrightarrow p\in \text{even}$$
Since $\gcd(p,q)=1$ and $p$ is even, then $q$ must be odd.
Let $p=2^kp_o$, where $\color{red}{k\ge1}$ and $p_o$ is odd. We get
$$2^aq^b=2^{bk}p_o^b$$
Since both $q^b$ and $p_o^b$ are odd, it implies $$2^a=2^{bk}\Longrightarrow a=bk$$
since we know $0<a<b$, hence
$$bk<b\Longrightarrow k<1$$
which contradicts with the fact $\color{red}{k\ge1}$
Assume that $a,b,p,q\in\Bbb N$ and $\gcd(p,q)=1$. If $2^{p/q}=\frac ab$ then $2^pb^q=a^q$.
Let's call, for any natural $n$, $v(n)$ the greatest natural $r$ such that $2^r$ divides $n$.
So $v(2^pb^q)=v(a^q)$, and $p+qv(b)=qv(a)$. Then $p$ is a multiple of $q$. Since $p$ and $q$ are coprime, $q=1$.