$x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent How do we prove that the three statements below about the real number $x$ are equivalent?
(i) $\displaystyle x$ is rational
(ii) $\displaystyle \frac{x}{2}$ is rational
(iii) $\displaystyle 3x-1$ is rational

It is enough to prove that $$(i) \implies (ii) \implies (iii) \implies (i)$$
$1$. $(i) \implies (ii)$. Let $x = \dfrac{p}q$, where $p,q \in\mathbb{Z}$. We then have $\dfrac{x}2 = \dfrac{p}{2q}$ and we have $p,2q \in \mathbb{Z}$. Hence, $$(i) \implies (ii)$$
$2$. $(ii) \implies (iii)$. Let $\dfrac{x}2 = \dfrac{p}q$, where $p,q \in \mathbb{Z}$. This gives $x = \dfrac{2p}q$, which in-turn gives $$3x-1 = \dfrac{6p}q - 1 = \dfrac{6p-q}q$$ Since $p,q \in \mathbb{Z}$, we have $q,6p-q \in \mathbb{Z}$. Hence, $$(ii) \implies (iii)$$
$3$. $(iii) \implies (i)$. Let $3x-1 = \dfrac{p}q$, where $p,q \in \mathbb{Z}$. This gives $$3x = \dfrac{p}q + 1 \implies x = \dfrac{p+q}{3q}$$
Since $p,q \in \mathbb{Z}$, we have $3q,p+q \in \mathbb{Z}$. Hence, $$(iii) \implies (i)$$
