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Intro to Real Analysis

  1. Sep 15, 2011 #1
    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!
     
  2. jcsd
  3. Sep 15, 2011 #2
    What happens when b < epsilon?
     
  4. Sep 15, 2011 #3

    gb7nash

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    Homework Helper

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

    a < b−ϵ and plugging b in...
     
  5. Sep 15, 2011 #4

    HallsofIvy

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

    For a moment, I thought this was "[itex]a< b+ \epsilon[/itex] for all [itex]\epsilon> 0[/itex]". For that, it is NOT true that a< b but it is true that [itex]a\le b[/itex]
     
  6. Sep 15, 2011 #5
    Thanks for all the help! Wow that's easy now...*duh*
     
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