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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$\begingroup$ What are your thoughts on the problem? People will be happy to help if you show you put some effort into answering your own question. Hint: Assume, towards a contradiction, that $a<b+\epsilon$ $\forall\epsilon>0$ and $a>b$. $\endgroup$– JanFeb 22, 2017 at 11:19
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$\begingroup$ Also here: math.stackexchange.com/questions/1906981/…, math.stackexchange.com/questions/1027284/…, math.stackexchange.com/questions/1559389/…. $\endgroup$– Martin RFeb 22, 2017 at 12:02
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1$\begingroup$ This is not a duplicate, the question asked here is different (strict inequality assumption) $\endgroup$– qwrMar 29, 2019 at 8:49
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1 Answer
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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 $e=\frac{a-b}{2}$, 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 BobFeb 22, 2017 at 11:12
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$\begingroup$ @MrBob You're welcome. I recommend you proceed with a proof by contradiction with problems like these. $\endgroup$– S.C.B.Feb 22, 2017 at 11:12
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$\begingroup$ @MrBob Sorry, you're question is a duplicate. Sorry~ $\endgroup$– S.C.B.Feb 22, 2017 at 13:10
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