If $(m_1, n_1)$ and $(m_2, n_2)$ are two equal rational numbers, i.e. $m_1*n_1^{-1} = m_2 * n_2^{-1}$, so $m_1*n_2 = m_2*n_1$.
And if $m_1$ and $n_1$ are coprime, prove that $n_2 * n_1^{-1} = m_2 * m_1^{-1} = k \in \mathbb Z$.
The main problem is to show that k is an integer.
We consider the equation $m_1*n_2=m_2*n_1$
Since $m_1$ and $n_1$ are coprime, and $$m_2*n_1 \text{ mod } m_1 \equiv m_1*n_2 \text{ mod } m_1\equiv 0$$ Then, $$m_2 \equiv 0 \text{ mod } m_1$$
This implies $m_2 = m_1*a$ for some $a \in Z$. Thus,
$m_2 * m_1^{-1} = m_1*a * m_1^{-1}=a=k\in Z$, completing the proof.