Is it true? I see a lot of inequalities like that in proofs, but i don't understand why it should be true.

  • $\begingroup$ Your question is equivalent at $$\forall \varepsilon>0, a\leq \varepsilon\implies a\leq 0.$$ If $a>0$, take $\varepsilon=\frac{a}{2}$, then $a>\varepsilon$ which is a contradiction. $\endgroup$
    – Surb
    Oct 31, 2015 at 22:42

1 Answer 1


If $a\not\le b$, then $a>b$. Let $\epsilon=\frac{a-b}2$: then $\epsilon>0$, and $a=b+2\epsilon>b+\epsilon$, contradicting the hypothesis.


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