Epsilon form of inequalities

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This is not an important page. It was created so anyone unsure of what it was supposed to be from the title can see what the statement is. The proof of it is very easy


For [ilmath]a,b\in\mathbb{R} [/ilmath] then:

  • [ilmath]\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff(a\le b)[/ilmath]


  • To prove [ilmath]\implies[/ilmath] it's easiest to use the contrapositive
  • To prove [ilmath]\impliedby[/ilmath] is straightforward.
Grade: F
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Extremely easy

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