Without loss of generality let us assume \(x > 0\).Hence we have \((y-x)>0\).

Now using the Archimedean principle we must have \((y-x) > \frac{1}{n}\)

Hence \((ny - nx)> 1\).

\(ny> nx+1\).

Now let us say \(m-1<nx<m\).

Now we write \(m \le nx+1<ny\).

Hence we have \(y > \frac{m}{n}\) and \(x < \frac{m}{n}\).So there exists a rational number between x and y.

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TopNewestWait, you didn't define \(y\) though! :P – Finn Hulse · 2 years, 4 months ago

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– Ethan Robinett · 2 years, 4 months ago

I thought the same thing. I'm sure he meant \(y \in \mathbb{Z}\) such that \(y >x\)Log in to reply

– Finn Hulse · 2 years, 4 months ago

That's what it seems like. :PLog in to reply