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Prove that $a<b+\epsilon$ for all $\epsilon>0$ implies $a\le b$

Can anybody help me with this question? I am new to this topic.

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Let us proceed with a proof by contradiction.

Assume that $a>b$. This implies $\frac{a-b}{2}>0$

Note that if we set $\epsilon=\frac{a-b}{2}$, then $$a<b+\frac{a-b}{2} \iff a<\frac{a+b}{2} \iff b>a$$ A contradiction to the assumption that $a>b$. Thus $a \le b$.

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  • $\begingroup$ God thank you so much, i was becoming so confused. $\endgroup$
    – Mr Bob
    Commented Feb 22, 2017 at 11:12
  • $\begingroup$ @MrBob You're welcome. I recommend you proceed with a proof by contradiction with problems like these. $\endgroup$
    – S.C.B.
    Commented Feb 22, 2017 at 11:12
  • $\begingroup$ @MrBob Sorry, you're question is a duplicate. Sorry~ $\endgroup$
    – S.C.B.
    Commented Feb 22, 2017 at 13:10
  • $\begingroup$ Yes ofcourse. Thank you $\endgroup$
    – Mr Bob
    Commented Feb 22, 2017 at 16:24
  • $\begingroup$ It's simpler to take $\epsilon=a-b$. $\endgroup$
    – Pedro
    Commented Apr 1 at 22:16

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