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Proof of an Inequality

  1. Feb 16, 2014 #1
    Hello,

    I am given that β > α, which can be written as β - α > 0. What justification would I have to use in order to conclude that α - 2β < 0, given that the preceding propositions are true? Could someone possibly help me?
     
  2. jcsd
  3. Feb 16, 2014 #2

    Ray Vickson

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    This is not necessarily true: if α = -3 and β = -2 we have α < β but 2β < α.

    Anyway, you are supposed to show your work.
     
  4. Feb 16, 2014 #3
    This isn't actually a homework problem. It was something I came across and found curious, when I was drawing the slopes fields of a differential equation.

    I also forgot to mention that alpha and beta are both positive.
     
  5. Feb 16, 2014 #4
    Note: I can not provide my attempt at solving this problem, for I never did attempt at solving it; the reason being, that I do not know enough about inequalities to solve this problem. As I said, this is not a homework problem, it is a side-tracking.
     
  6. Feb 16, 2014 #5

    PeroK

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    So, to re-phrase the question. You have two positive numbers. You start with the smaller one and subtract twice the larger one and you want to prove that the answer is less than zero?
     
  7. Feb 16, 2014 #6
    PeroK, if I read your sentences correctly, yes that is what I would like to prove, rather, I would like to know how to prove, as I have never done any proves involving inequalities.

    α > 0 and β > 0, and β > α, and I would like to know if we can conclude that α - 2β < 0.
     
  8. Feb 16, 2014 #7

    PeroK

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    Well, if you had $α in the bank. And you withdrew $β. Then you withdrew $β again. Do you think you might be overdrawn? If β > α.
     
  9. Feb 16, 2014 #8
    I think there might be a misunderstanding. I understand that statement α - 2β < 0 is true; I am looking for a little more mathematically rigorous proof of the fact.
     
  10. Feb 16, 2014 #9

    PeroK

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    What about:

    α - 2β = α - β - β < α - β (as β > 0) < 0 (as α < β)
     
  11. Feb 16, 2014 #10
    I think if you take a step back and examine PeroK's last post (edit: the one talking about bank accounts), you might find that it is VERY suggestive of a proof. Very suggestive.
     
  12. Feb 16, 2014 #11

    PeroK

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    Actually, I like this proof better:

    α < β => α < 2β => α - 2β < 0
     
  13. Feb 16, 2014 #12

    Ray Vickson

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    You have assumed what he is trying to prove: he want to show that for a,b >0, a < b => a < 2b.
     
  14. Feb 17, 2014 #13

    PeroK

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    What about:

    α < β => α < β + β => α < 2β => α - 2β < 0

    I guess it's not that obvious that 2β = β + β. Hopefully that repairs the proof.
     
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