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How To Prove it Inequality Proof

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that a and b are nonzero real numbers. Prove that if a<1/a<b<1/b then a<-1.


    3. The attempt at a solution
    So after a while I realized that I could prove that a<-1 by contradiction but first I have to prove that a<0. I figured out how to prove it but I'm not sure if my wording is convincing enough and I feel that it might be redundant at points. Anyway, here's the proof:

    Proof: Suppose a<1/a<b<1/b. It then follows that 1/a<1/b. Multiplying both sides of the inequality yields b<a, but this contradicts a<b. Therefore, one and only one variable a or b must be negative, and it follows that since b>a then a<0. Now suppose a[itex]\geq[/itex]-1. Plugging the value a=-1 into a<1/a yields -1<1/-1 or -1<-1, which contradicts a<1/a. Therefore a<-1.

    Now I feel like the wording of that proof is a mess but I'm not sure how I would reword it. Also am I being redundant in saying "one and only one variable a or b must be negative" or does that need to be there? Or am I crazy and the proof is acceptable as is?
     
  2. jcsd
  3. Jul 26, 2011 #2

    SammyS

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    What is it you multiply by? What if ab is negative?

     
  4. Jul 26, 2011 #3
    Oops yeah I multiply both sides by ab. If ab is negative then the statement doesn't contradict because we then get a<b, but didn't I state that by pointing out the contradiction of if ab is positive?

    If I modified that part to say:

    Suppose ab is a positive real number. Multiplying both sides of the inequality 1/a<1/b by ab yields b<a, but this contradicts a<b therefore ab must be negative and thus because a<b a<0.

    Is that better?
     
  5. Jul 26, 2011 #4

    SammyS

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    Yes, that's better.
     
  6. Jul 26, 2011 #5
    Awesome, thanks!
     
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