For every $\epsilon\gt 0$, $|a-b|<\epsilon$ ,then b=a .

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

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

• Try \epsilon or \varepsilon to get $\epsilon$ or $\varepsilon$ respectively. Mar 18 '19 at 12:44
• 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. Mar 18 '19 at 12:52
• Here's a hint to get you started: suppose $|a-b|=k\gt 0$, derive a contradiction by choosing suitable $\epsilon$; hence conclude that $|a-b|=0\iff a-b=0\iff a=b$ Mar 18 '19 at 13:01
• Thanks for telling how to write epsilon .it worked Mar 18 '19 at 13:09
• learner will you please elaborate it more that how to choose suitable $\epsilon$ Mar 18 '19 at 13:10

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