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Let $b, L \in \Bbb R$. Prove that if $b \ge L - \varepsilon$ for all positive $\varepsilon$, then $b \ge L$

I started of by assuming the "if" part. But, are you supposed to use cases to prove this? Is there an epsilon you can pick? Not sure what the approach should be here.

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1 Answer 1

up vote 5 down vote accepted

Suppose $b\geq L-\epsilon$ for all $\epsilon>0$.

If on the contrary $b<L$, we may set $\epsilon=\frac{L-b}{2}>0$. Then $L-\epsilon=L-\frac{L-b}{2}=\frac{L+b}{2}>b$, a contradiction.

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So, do you mean use the contrapositive of the statement to prove it? –  icanc Dec 7 '12 at 0:43
    
Using the contradiction method, it's still a little unclear to me. So I substituted $\frac{L-b}{2}$ for $\epsilon$, which gives me $b \ge \frac{L-b}{2}$. But, I can't seem to wrap my around the reason this becomes a contradiction. –  icanc Dec 7 '12 at 1:26
    
@JasperLoy - If you solve the (L+b)/2>b, you end up with L>b, which is what you assumed originally and not a contradiction –  devcoder Dec 7 '12 at 1:37
    
@JasperLoy It makes sense now since we assumed that $b \ge L - \epsilon$ and $b \lt L$. Thank you! –  icanc Dec 7 '12 at 1:46
    
I've run out of answers to upvote :( –  amWhy Dec 30 '12 at 0:28

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