Proof the the field of rational numbers has the Archimedean property

real-analysissolution-verification

I was tasked by the question to prove that the field of rational numbers has the Archimedean property

Proof:
Let $x$ and $y$ $\in$$ Q$ and $n$ be a positive integer. If $Q$ does not have the Archimedean property, then $nx\leq y$

For the case where $nx=y$, then $n+1\gt y$
however as $n+1$ is a positive integer then by contrapositive $Q$ must have the archimedean property

Is this valid and how can I phrase this better?

Best Answer

Perhaps the following is a bit cleaner:

Fix $x,y\in\mathbb{Q}$ with $x>0$. We need to produce a positive integer $n$ with $nx>y$.

By definition of $\mathbb{Q}$, there are integers $a,b,c,d$ with $x=\frac{a}{b}$, $y=\frac{c}{d}$ and $b>0$ and $d>0$.

As $x>0$, we have that $a>0$. Now, there are two cases:

  • If $y\le 0$, then taking $n=1$ does the trick, since then $y\le 0<x=1x$.
  • If $y>0$, then $c>0$. But then $$ y=\frac{c}{d}\le c \le ac=bcx<bcx+x=(bc+1)x $$ Since $bc+1$ is a positive integer, we are done.
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