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Prove 2ab<= a^2+b^2 using order axioms

  1. Oct 5, 2012 #1
    1. The problem statement, all variables and given/known data

    prove 2ab<= a^2+b^2 using order axioms
    Of course use of other basic axioms for real numbers are also okay.

    2. Relevant equations
    axioms for set of real numbers.


    3. The attempt at a solution
    The easy way to do this would be just subtract 2ab from both sides, factor, and see that (a-b)^2 is greater or equal to zero.

    But we have to use the basic axioms. So I tried constructing a^2+b^2-2ab by multiplying (a-b)(a-b) using the axioms. I'm not sure I was rigorous enough. Also I'm not sure whether I can just say a square of a real number is greater than zero. Though it is easy to prove.
     
  2. jcsd
  3. Oct 5, 2012 #2
    You can use only the order axioms and the field axioms. I don't think you can use squared numbers are greater than zero.
     
  4. Oct 5, 2012 #3
    well then I can just prove that a^2 is always greater or equal to zero.

    a>0
    Then a(a)>(a)0
    a^3>0

    a<0
    a+(-a)<0+(-a)
    0<-a
    -a(0)<(-a)(-a)
    0<(-1)(-1)(a)(a)
    0<a

    a=0
    a(a)=0(a)
    a^2=0

    So a^2 >=0
     
  5. Oct 8, 2012 #4
    Hmmm I guess so. did you prove that (-x)(-y)=-(xy)?
     
  6. Oct 8, 2012 #5
    yeah. I also solved this by using three lemmas. =) Don't know how to close the thread though.
     
  7. Oct 8, 2012 #6

    Mentallic

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    Homework Helper

    I sure hope he didn't.
     
  8. Oct 8, 2012 #7
    woops I mean (x)(-y) = -(xy)
     
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