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Prove the following (using some basic axioms)

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if 0 < a < b, then

    [tex]a < \sqrt{ab} < \frac{a+b}{2} < b[/tex]


    2. Relevant equations
    Axioms (Properties), courtesy of Wikipedia:

    Addition:
    P1: For all a, b, and c in F, a + (b + c) = (a + b) + c
    P2: There exists an element of F, called the additive identity element and denoted by 0, such that for all a in F, a + 0 = a
    P3: For every a in F, there exists an element −a in F, such that a + (−a) = 0
    P4: For all a and b in F, a + b = b + a

    Multiplication:
    P5: For all a, b, and c in F, a · (b · c) = (a · b) · c
    P6: There exists an element of F, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a
    P7: For any a in F other than 0, there exists an element a^(−1) in F, such that a · a^(−1) = 1
    P8: For all a and b in F, a · b = b · a
    P9: For all a, b and c in F, the following equality holds: a · (b + c) = (a · b) + (a · c)

    3. The attempt at a solution


    I really just don't know what to do. A small push in the right direction could make all the difference. Thanks to anyone that replies and helps.
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2
    Go through each inequality one at a time. The ordered field axioms (and consequences) are important here. The idea is to take one inequality, say (a+b)/2 < b, and perform algebra until you can see why the statement has to be true (first multiply both sides by 2, etc.). Although this may seem like taking the conclusion to be true, if you can reverse your steps so that you start with the true statement and end with the inequality you want to prove (you can usually do this if it's basic algebra and there's no loss of information between steps), then you have a rigorous proof.
     
  4. Sep 13, 2009 #3
    I was in the midst of doing that, but wasn't sure if I was allowed to say

    "since [tex]a < b, a^{2} < a \cdot b[/tex]

    similarly, [tex]\sqrt{a^{2}} < \sqrt{a \cdot b}[/tex] "


    On top of that, I get stuck when trying to prove the "middle" part: [tex]\sqrt{ab} < \frac{a+b}{2}[/tex]
     
  5. Sep 13, 2009 #4
    Sure you can, provided that you have proved that x^2 < y^2 implies x < y if x and y are both positive. If you're using the textbook I think you're using, you should have proved this fact already, even though technically you probably haven't shown that square roots actually exist (which is fine at this point).

    For the middle part, see if you can multiply by 2, square, bring everything to one side, and factor, and make you all your steps are reversible.
     
  6. Sep 13, 2009 #5
    Suppose

    [tex] \frac{a+b}{2} < \sqrt{ab} [/tex]

    [tex] \Rightarrow (a+b)^2 < 4ab [/tex]

    [tex] \Rightarrow (a+b)^2-4ab < 0 [/tex]

    [tex] \Rightarrow a^2 -2ab +b^2 < 0 [/tex]

    [tex] \Rightarrow (a-b)^2<0 [/tex]

    Which is not possible for real a and b.
     
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