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.

  • 1
    $\begingroup$ The step "$\delta<\epsilon,\forall \epsilon >0$ [...] and therefore a contradiction arises" could be improved, by simply choosing explicitly $\epsilon=\frac{\delta}{2}<\delta$. For the rest your proof is correct (although I don't see why you need the rationality assumption here) $\endgroup$ – b00n heT Jun 30 '16 at 5:40
  • $\begingroup$ I haven't constructed the reals yet. $\endgroup$ – the_dude Jun 30 '16 at 5:41
  • $\begingroup$ Sure. Makes sense :) $\endgroup$ – b00n heT Jun 30 '16 at 5:41
  • $\begingroup$ But is it wrong / immature the way I wrote it? That since $\delta < \epsilon$ for all $\epsilon > 0$, and also that $\delta$ is positive; Hence the contradiction. $\endgroup$ – the_dude Jun 30 '16 at 5:44
  • $\begingroup$ Yes and no: You are claiming that this is a contradiction without giving a proof of it, thus it is mathematically incorrect. $\endgroup$ – b00n heT Jun 30 '16 at 5:46

...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.

  • $\begingroup$ Yeah right. I get the absurdity. $\endgroup$ – the_dude Jun 30 '16 at 5:53

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