Let $x$ and $y$ be positive elements of a linearly ordered group $G$.
Then there exists a natural number $n\in \mathbb{N}$ such that $nx>y$.

Proof

Let $x,y\in \mathbb{R}$.
Without loss of generality, choose $x=1$.

Suppose for the sake of contradiction that there is no $n\in \mathbb{N}$ such that $1n>y$.
Then $n\le y$ for all $n\in \mathbb{N}$.
So $y$ a supremum in $\mathbb{N}$.

Since $(y-1)$ not a supremum in $\mathbb{N}$, there exists some $m>(y-1)$ in $\mathbb{N}$.
Since $m\in \mathbb{N}$, $(m+1)\in \mathbb{N}$.
However, $(m+1)>y$,
so $y$ not a supremum in $\mathbb{N}$ (a contradiction).

reductio ad absurdum, QED