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I have done a proof by myself but not sure about it

proof: $|b-a|<\epsilon $ =$a-\epsilon $

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    $\begingroup$ Try \epsilon or \varepsilon to get $\epsilon$ or $\varepsilon$ respectively. $\endgroup$
    – Randall
    Mar 18 '19 at 12:44
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    $\begingroup$ Use dollar signs to wrap up math, like $\epsilon\gt 0$ for $\epsilon\gt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands. $\endgroup$
    – learner
    Mar 18 '19 at 12:52
  • $\begingroup$ Here's a hint to get you started: suppose $|a-b|=k\gt 0$, derive a contradiction by choosing suitable $\epsilon$; hence conclude that $|a-b|=0\iff a-b=0\iff a=b$ $\endgroup$
    – learner
    Mar 18 '19 at 13:01
  • $\begingroup$ Thanks for telling how to write epsilon .it worked $\endgroup$
    – newbie
    Mar 18 '19 at 13:09
  • $\begingroup$ learner will you please elaborate it more that how to choose suitable $ \epsilon $ $\endgroup$
    – newbie
    Mar 18 '19 at 13:10
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We have $\forall \epsilon > 0,\quad |a-b| < \epsilon$

If we suppose that $a \neq b$ then $|a-b| \neq 0$, we choose $\epsilon = \dfrac{|a-b|}{2} > 0$

Then $|a-b| < \dfrac{|a-b|}{2} \implies 1 < \dfrac{1}{2}$, contradiction!

Conclusion : $\forall \epsilon > 0,\quad |a-b| < \epsilon \implies a =b$

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  • $\begingroup$ Thanku for such quick answer @LAGRIDA $\endgroup$
    – newbie
    Mar 18 '19 at 13:58

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