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Proving order relation of real number

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data
    If a>0, then -a>0. Moreover, if a<b, then -a>-b


    2. Relevant equations
    axioms of real numbers can be applied


    3. The attempt at a solution
    I am not sure what am I suppose to prove. So what I did was trying to prove "if a<b, then -a>-b" given that "If a>0, then -a>0."

    I tried the direct proof method:
    Suppose a<b, since we are given -a<a. So, -a<a<b. But I don't know how to bring -b in as the question didn't specify anything about b. So do I do 2 cases as in when b>0 and b<0?

    I've also tried proof by contradiction:
    Assuming a<b and -a<-b
    But I don't know how to start ><

    Need help!! Thanks sooo much
     
  2. jcsd
  3. Aug 5, 2011 #2

    HallsofIvy

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    I believe those are two different problems:
    1) Prove that: If a> 0 then -a< 0.

    2) Prove that: if a> b then -a< -b.

    Now, how are "-a" and "-b" defined?
     
  4. Aug 5, 2011 #3
    So firstly I need to prove this statement "If a>0, -a<0" ?

    what do you mean by how is it defined?? Like -a is defined to be <0?
     
  5. Aug 5, 2011 #4
    Or like -a = -a + 0
    and -b = -b + 0
    -b = -b + (a+(-a))
    -b = (-b+a)+(-a) > -a??
     
  6. Aug 5, 2011 #5
    Oh! I found out a way but not sure about it

    -b = -b+0
    -b = -b+(a+(-a))
    -b = (-b+a)+(-a)
    Since a<b, (-b+a)<0
    So, -b = (-b+a)+(-a) < -a
    Therefore, -b < -a when a<b...

    Is this prove alright??
     
  7. Aug 5, 2011 #6

    HallsofIvy

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    Better. -a is defined as the "additive inverse" of a or simply a+ (-a)= 0.

    Pretty good. Slightly simpler would be
    a< b
    a+(-a)< b+ (-a)
    0< b+ -a

    Can you finish that?

    Proving "if 0< a then -a< 0" should be simpler.
     
  8. Aug 5, 2011 #7
    Yup yup ^^

    a< b
    a+(-a)< b+ (-a)
    0< b+ -a
    0+(-b)<b+ -a + -b
    -b < (b+(-b))+-a
    -b<-a

    yay! Thanks so much
     
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