Originally Posted by

**JaneBennet** Pick an integer *b* such that $\displaystyle \frac{1}{y-x}<b$. Then *b* is positive and $\displaystyle by-bx>1$. Since the difference between $\displaystyle by$ and $\displaystyle bx$ is greater than 1, there must be an integer *a* between them*, i.e. $\displaystyle bx<a<by$. Hence $\displaystyle x<\frac{a}{b}<y$, showing that there is a rational number between *x* and *y*.

*The set of integers $\displaystyle \{n\in\mathbb{Z}:n\leq bx\}$ is nonempty (by the Archimedean principle) and bounded above, so it has a maximum member *c*. It follows that $\displaystyle bx<c+1<by$.

For irrational, choose $\displaystyle b>\frac{j}{y-x}$, where *j* is any irrational number greater than 1 (e.g. $\displaystyle \sqrt{2}$). This should lead to $\displaystyle x<\frac{aj}{b}<y$.