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!