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Prove that $a0$ implies $a\le b$

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

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    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 $a0$ and $a>b$. – 2017-02-22
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    Also here: http://math.stackexchange.com/questions/1906981/if-a-leq-b-epsilon-for-each-epsilon0-then-a-le-b, http://math.stackexchange.com/questions/1027284/x-y-are-real-xy-varepsilon-with-varepsilon0-how-to-prove-x-le-y, http://math.stackexchange.com/questions/1559389/prove-that-if-a-leq-b-varepsilon-forall-varepsilon0-then-a-leq-b. – 2017-02-22

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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 $$aa$$ A contradiction to the assumption that $a>b$. Thus $a \le b$.

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    God thank you so much, i was becoming so confused. – 2017-02-22
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    @MrBob You're welcome. I recommend you proceed with a proof by contradiction with problems like these. – 2017-02-22
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    @MrBob Sorry, you're question is a duplicate. Sorry~ – 2017-02-22
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    Yes ofcourse. Thank you – 2017-02-22