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

Question

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.

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 $a>b$. — Jan, Feb 22 '17 at 11:19
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. — Martin R, Feb 22 '17 at 12:02
This is not a duplicate, the question asked here is different (strict inequality assumption) — qwr, Mar 29 '19 at 08:49

score 10 · Accepted Answer · edited Apr 01 '24 at 22:15

10

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 $a \le b$.

edited Apr 01 '24 at 22:15

Pedro

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answered Feb 22 '17 at 11:08

S.C.B.

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God thank you so much, i was becoming so confused. – Mr Bob Feb 22 '17 at 11:12
@MrBob You're welcome. I recommend you proceed with a proof by contradiction with problems like these. – S.C.B. Feb 22 '17 at 11:12
@MrBob Sorry, you're question is a duplicate. Sorry~ – S.C.B. Feb 22 '17 at 13:10
Yes ofcourse. Thank you – Mr Bob Feb 22 '17 at 16:24
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It's simpler to take $\epsilon=a-b$. – Pedro Apr 01 '24 at 22:16
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Indeed, and reasoning by contraposition is cleaner. – Anne Bauval Oct 10 '24 at 18:13

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

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