Let $H = \{2^m\ |\ m \in \mathbb{Z} \}$. Define a relation $\sim$ on the set $\mathbb{Q}^+$ by $p \sim q$ if $\dfrac{p}{q} \in H$.

Question

For reflexive, I have: Let $p\in\mathbb{Q}$. Since $\frac{p}{p}$ is $\frac{2}{2}$ we can write this as $2^0 \in \mathbb{H}$. Since $0$ is an integer, then $pRp$, and $R$ is reflexive.

For symmetric, I have: Let $p,q \in \mathbb{Q}$ and assume $pRq$. This means $\frac{p}{q}=2^m$, for some integer $m$. Let $p=2$ and $q=1$, then $\frac{q}{p}=\frac{1}{2}$, which can be written as $2^{-1}$, but $\frac{p}{q}=\frac{2}{1} \neq 2^{-1}$. So $R$ is not symmetric, and therefore not an equivalence relation.

For transitive, I have: Let $p,q,z \in \mathbb{Q}$ and assume $pRq$ and $qRz$. This means $\frac{p}{q}=2^{m}$ and $\frac{q}{z}=2^{m}$. Let $p=2$, $q=4$, and $z=1$. Therefore $\frac{2}{1}\sim\frac{4}{1}$ which equals $\frac{2}{1}*\frac{1}{4}$ = $\frac{1}{2}$ or $2^{-1}$.

Answer 1

For symmetric, I have: Let $p,q∈Q$ and assume $pRq$. This means $\frac pq=2^m$, for some integer $m$.

Right. But it doesn't have to be same $m$ for different values if $p,q$.

In your example $\frac 12= 2^{-1}$ so $1R2$. But we also have $\frac 21 = 2^{1}$. So $2 R 1$. There is utterly no reason to think that if $\frac 21 = 2^m$ and $\frac 12 = 2^n$ then $m$ and $n$ must be the same integer.

Anyway.... to be symmetric we must have $aRb \implies bRa$.

So $aRb \implies \frac ab = 2^m$ for some integer $m\implies$

$\frac ba =\frac 1{\frac ba} = \frac 1{2^m} = 2^{-m}$ and as $-m$ is an integer

$\implies bRa$.

So yes it is symmetric.

And transitive. If $aRb$ and $bRc$ then $\frac ab = 2^m$ for some $m$ and $\frac bc = 2^n$ for some $n$. Then $\frac ac =\frac ab\frac bc = 2^m2^n =2^{m+n}$. And $m+n$ is an integer. So $aRc$.

So yes it is transitive.