Prove that for rational $x$,$y$ and $\epsilon$ if $|x-y| \leq \epsilon$ , $ \forall \epsilon > 0$ then $x=y$ I know that this is a repeated question, but I wanted to show my attempt. Suppose $x \neq y$ and (wlog) $x > y$, then $x$ can be written as $x = y + \delta$, for some ($\delta > 0$ and $\delta \in \mathbb Q$) Therefore $|x-y| = \delta$. But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $ , and we assumed that $\delta > 0$ ; therefore contradiction arises. Hence $x=y$. Is this a correct proof? I want to know what is missing or how to add more rigour in the last statement.

> _...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $..._

This part is not right.

Make sure to use the _right_ argument with the _right_ inequality!

> Since $\delta >0$, there **exists** $\epsilon \in \Bbb Q$ such that $0 < \epsilon < \delta$ and therefore $\epsilon < |x-y| = \delta$, which is absurd.