By Archimedean property I mean:

For any positive rational numbers $x = \frac{a}b$ and $y = \frac{c}d$, there is an integer $n$ such that $nx > y$, namely, $nx \equiv (x+x+ \ldots+x)$ with $n$ copies of $x$, it is larger than $y$.

I'd like to find a formula involving $a,b,c,d$ of obtaining such $n$ that is large enough.

So here's what I tried. I first assumed there exists an $n$ that works. Then determined that IF such an $n$ exists, it would have to be greater than a formula I determined. And then say that there exist numbers bigger than that formula, so my assumption is possible in the first place.

It feels like circular logic and I'm assuming what I'm trying to prove.

I proved that $n > \frac{bc}{ad}, but how do I even know such an integer exists?

  • $\begingroup$ crossmultiply and work with integers. Rewrite $nx > y$ as $n(ad)>(bc)$. $\endgroup$ – Mirko Nov 5 '18 at 3:26

Simply, there always exists exactly one integer larger than z in the interval of (z,z+1].


It's okay to show if what we want to prove were true something else would need to be true and then prove that, provided the implications work "in the right directions".

$nx > y$ for $n \in \mathbb Z; x > 0; y > 0$. If and only if $n > \frac yx$. So it is sufficient to prove that an $n$ greater than $\frac yx$ exists.

Now $x$ and $y$ are rational and positive so $\frac yx$ is rational and positive. So there are positive integers $a,b$ so that $\frac yx = \frac ab$.

So it is sufficient to prove there is an integer $n > \frac ab$. Well $b \ge 1$ so $\frac ab \le a$. So if $n = a + 1 > a \ge \frac ab = \frac yx$ we are done.

Or if you want to do it more directly......

Let $x = \frac ab$ where $a,b$ are positive integers. Let $y = \frac cd$ where $c,d$ are positive integers. Then let $n = cb + 1$.

$n\frac ab = (cb +1)\frac ab > cb \frac ab \ge \frac {cb}{da}*\frac ab = \frac cd = y$


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