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Proof using Rule of Signs

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data
    In an ordered ring, show that
    If a>0, b>0, then a>b <--> (a^2)>(b^2).

    Hint: [(b^2)-(a^2)] = (b-a)(b+a). Use Rule of Signs.


    2. Relevant equations
    Rule of Signs????


    3. The attempt at a solution
    I don't know how to use the hint and I'm having trouble with the proof, as this is my first proofs class.
     
  2. jcsd
  3. Sep 9, 2008 #2
    Since we know that both a and b must be positive value
    Therefore, if we take square root on both side of equation (a^2) > (b^2).
    it would make a > b always. Would this work?
     
  4. Sep 9, 2008 #3
    Are you asking me? That looks like it would work for me but I don't know if it is sufficient enough in "proof language."

    Also, the hint isn't being used in your answer...
     
  5. Sep 9, 2008 #4

    HallsofIvy

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    Because this is "if and only if" you need to show two things:

    1) If a> b then a2> b2. Since this is an ordered ring, the fact that a> b and a> 0 tells you that a*a> a*b (why?); now do the same with a> b and b> 0.

    2) if a2> b2 then a> b. Okay, a2> b2 tells you that (a+ b)(a-b)> 0. What do the "signs" of a+ b and a- b have to be? I put "signs" in quotes because you cannot just assume facts from real numbers. You are working in an arbitrary ordered ring and you have to use the axioms and definitions for an order ring. In particular, a is "positive" if and only if a> 0. Since you are given that a> 0 and b>0, can you show that a+b> 0? What does that tell you about a- b?
     
  6. Sep 9, 2008 #5

    Dick

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    I think the 'rule of signs' just means positive*negative=negative. (b-a)<0 and (b+a)>0. So (b-a)*(b+a)<0.
     
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