About using Archimedean property to prove the existence of least upper bound

I am studying the proof of existence of least upper bound, but I can not understand how the autor applies the Archimedean property.

Archimedean property. Let $x$ and $\epsilon$ be any positive real numbers. Then there exists a positive integer $M$ such that $M \epsilon > x$.

So, in the proof of existence of least upper bound

Let $E$ be a non-empty subset of $\mathbb R$ with an upper bound $M$, we have to show that $E$ has at least one least upper bound. Since $E$ is non-empty, we can choose some element $x$ in $E$. Let $n \ge 1$. By the Archimedean property, we can find an integer $K$ such that $K/n \ge M$, and hence $K/n$ is also an upper bound for $E$. By the Archimedean property again, there exists another integer $L$ such that $L/n < x$. Since $x$ lies in $E$, we see that $L/n$ is not an upper bound for $E$. Since $K/n$ is an upper bound but $L/n$ is not, we see that $K \ge L$.

I understand the rest of the proof but the part "there exists another integer $L$ such that $L/n < x$" it is no clear for me.

• Here $L$ is only required to be an integer, not necessarily a positive integer. If $|x|< M/n$ for some positive integer $M$, then $(-M)/n < x < M/n$. – Xiao Oct 8 '14 at 4:19
• I don't think you can prove the existence of least upper bound using the Archimedean property alone, because $\mathbb Q$ is an Archimedean ordered field that does not have the least upper bound property. – lhf Oct 8 '14 at 16:12
• @lhf: the proof follows from a Cauchy sequence whose formal limit is the least upper bound. – Cristhian Gz Oct 8 '14 at 16:35

Can help to reinterpret the proof. Let $L$ be a integer. By Archimedean property, exists a $n \ge 1$ such that $nx > L$. Now using that $n$, by Archimedean property again, exists a integer $K$ such that $K/n \ge M$.