Proving a < b for All Positive Epsilon in Real Analysis

[doubleaxel195]

Homework Statement

If $$a < b-\epsilon$$ for all $$\epsilon >0$$, then $$a<0$$

Homework Equations

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

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!

 

[daveb]

What happens when b < epsilon?

 

[gb7nash]

But I'm stuck when $$b>0$$

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

a < b−ϵ and plugging b in...

 

[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 [itex]a< b- \epsilon[/b], it follows immediately that $a< b$.

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 $a\le b$

 

[doubleaxel195]

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