I would like to know how to conclude this from the Archimedean property of $\mathbb{Q}$: Let $x$, $y \in \mathbb{Q}$ and $x > 0$. Then there is an integer $n$ such that $nx < y \leq (n + 1)x$.
Suppose now I have proved the Archimedean property of $\mathbb{Q}$:
Let $x$, $y \in \mathbb{Q}$ and $x > 0$. Then there is an integer $m$ such that $mx > y$.
Then would it be okay to say that take the smallest such integer and denote it $n+1$, and then we are done?
Thanks in advance!
There exists an integer $m_0$ such that $m_0x>y$. Consider the set $S:=\{n\in\mathbb N:y>(m_0-n)x\}$.
By the archimedean property for $-y$ there exists an integer $m$ such that $mx>-y$, so $m_0+m\in S$, where $m+m_0\in\mathbb N$ since $-mx<y<m_0x$. Thus by the well-ordering principle $S$ has a minimal element $m_1$. Then, I claim $(m_0-m_1)x<y\le (m_0-m_1+1)x$, as desired. Indeed, $m_1\ne0$, since $0\notin S$ (since $y<m_0x$). Thus, $m_1-1\in\mathbb N$, and so if $y>(m_0-m_1+1)x$ then $m_1-1\in S$, contradicting the minimality of $m_1$.
Now, you can conclude by letting $n=m_0-m_1$.