# Find all rational numbers $r$ for which $\log_{2}(r)$ is rational as well.

Find all rational numbers $r$ for which $\log_{2}(r)$ is rational as well.

I found this problem in one of the Serbian competition books marked with a star. I looked at the solution, but it's confusing. Could someone help me by providing a detailed solution and how to approach problems similar to this one?

• Can you make your question more clear? If $r$ has no other restrictions, when $r$ is $2,4,8,16$ and so on, then $\log_2{r}$ is rational, so there are infinitely many rational numbers. – Toby Mak Jan 1 '18 at 11:42

This is the same thing as asking: for which rational numbers $r$ is it true that $2^r$ is also rational? The answer is: when $r\in\mathbb Z$. In fact, if $r\in\mathbb{Q}\setminus\mathbb Z$, $r=\frac ab$ with $b>1$ and $\gcd(a,b)=1$. The number $2^{\frac ab}$ is a root of the polynomial $x^b-2^a$. By the rational root theorem, all rational roots of this polynomial are integer numbers, but $2^{\frac ab}$ is not an integer.
• Alternatively, use unique factorization to see that $2^{a/b}$ is not an integer. – GEdgar Jan 1 '18 at 11:50
Assume $\log_2(r)=\frac{p}{q}$, with $q\in\mathbb{Z}^+,p\in\mathbb{Z}$ and $\gcd(p,q)=1$. We have $$2^\frac{p}{q} = r \in\mathbb{Q} \tag{A}$$ $$2^{k+\frac{p}{q}} = \frac{a}{b},\qquad a,b\in2\mathbb{Z}+1,\gcd(a,b)=1\tag{B}$$ $$b^q\cdot 2^{qk+p} = a^q \tag{C}$$ hence $qk+p=0$, since otherwise the LHS and RHS of $(C)$ would have a different value of $\nu_2$, where $\nu_2(m)=\max\{n\in\mathbb{N}: 2^n\mid m\}$. This implies $$r\in\mathbb{Q},\;\log_2(r)\in\mathbb{Q}\quad\Longrightarrow\quad \log_2(r)\in\mathbb{Z},\;r\in 2^{\mathbb{Z}}.\tag{D}$$