Intro to Real Analysis

1. Sep 15, 2011

doubleaxel195

1. The problem statement, all variables and given/known data
If $$a < b-\epsilon$$ for all $$\epsilon >0$$, then $$a<0$$

2. Relevant equations

All I really have are the field axioms of the real numbers and the order axioms (trichotomoy, transitive, additive property, multiplication property).

3. The attempt at a solution
Well I broke this proof into three cases: $$b<0, b=0, b>0$$. When $$b<=0$$, I'm fine.

But I'm stuck when $$b>0$$. I know that $$-\epsilon < 0 \implies b-\epsilon < b \implies a<b$$

To me it seems like this is saying that no matter what number you have, there is always a negative number that is smaller. Can anyone see a better way to do this without cases? Any help is appreciated!

2. Sep 15, 2011

daveb

What happens when b < epsilon?

3. Sep 15, 2011

gb7nash

If b > 0, set ϵ = b (since the inequality holds for all ϵ > 0). Looking at:

a < b−ϵ and plugging b in...

4. Sep 15, 2011

HallsofIvy

I can't imagine why you would think that b positive or negative makes any difference. For all $\epsilon> 0$ $b-\epsilon< b$. If it also true that $a< b- \epsilon[/b], it follows immediately that [itex]a< b$.

For a moment, I thought this was "$a< b+ \epsilon$ for all $\epsilon> 0$". For that, it is NOT true that a< b but it is true that $a\le b$

5. Sep 15, 2011

doubleaxel195

Thanks for all the help! Wow that's easy now...*duh*