(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

If [tex] a < b-\epsilon[/tex] for all [tex]\epsilon >0[/tex], then [tex]a<0[/tex]

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: [tex]b<0, b=0, b>0[/tex]. When [tex]b<=0[/tex], I'm fine.

But I'm stuck when [tex] b>0[/tex]. I know that [tex]-\epsilon < 0 \implies b-\epsilon < b \implies a<b[/tex]

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Intro to Real Analysis

**Physics Forums | Science Articles, Homework Help, Discussion**