Proving a < b for All Positive Epsilon in Real Analysis

[doubleaxel195]

Homework Statement

If a &lt; b-\epsilon for all \epsilon &gt;0, then a&lt;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&lt;0, b=0, b&gt;0. When b&lt;=0, I'm fine.

But I'm stuck when b&gt;0. I know that -\epsilon &lt; 0 \implies b-\epsilon &lt; b \implies a&lt;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&gt;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&gt; 0 b-\epsilon&lt; b. If it also true that a&lt; b- \epsilon[/b], it follows immediately that a&amp;lt; b.<br /> <br /> For a moment, I thought this was &quot;a&amp;lt; b+ \epsilon for all \epsilon&amp;gt; 0&quot;. For that, it is NOT true that a&lt; b but it is true that a\le b

 

[doubleaxel195]

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