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I'm new to proofs. Would someone please give me an opinion on my proof?

  1. Sep 15, 2009 #1
    My prof gave us twelve basic properties of numbers, and I think I'm supposed to use those in my proof, but I'm not sure how to incorporate them.

    The properties are:
    P1 Associative law for addition
    P2 Additive identity
    P3 Additive inverse
    P4 Commutative law for addition
    P5 Associative law for multiplication
    P6 Multiplicative identity
    P7 Multiplicative inverse
    P8 Commutative law for multiplication
    P9 Distributive law
    P10 Trichotomy law
    P11 Closure under addition
    P12 Closure under multiplication

    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

    N/A

    3. The attempt at a solution

    Part I:

    [tex]
    \begin{align*}
    a&<b\\
    a^2&<ab\\
    \sqrt{a^2}&<\sqrt{ab}\\
    a&<\sqrt{ab}
    \end{align*}
    [/tex]

    Part II:

    [tex]
    Suppose:
    \begin{align*}
    \sqrt{ab}&\geq\frac{a+b}{2}\\
    ab&\geq\left(\frac{a+b}{2}\right)^2\\
    ab&\geq\frac{a^2+2ab+b^2}{4}\\
    4ab&\geq a^2+2ab+b^2\\
    0&\geq a^2-2ab+b^2\\
    0&\geq (a-b)^2\\
    \end{align*}
    [/tex]

    [tex]
    but:
    \begin{align*}
    0&<(a-b)^2\\
    \therefore \sqrt{ab}&<\frac{a+b}{2}
    \end{align*}
    [/tex]

    Part III:

    [tex]
    \begin{align*}
    a&<b\\
    a+b &< b+b\\
    a+b &< 2b\\
    \frac{a+b}{2} &< b
    \end{align*}
    [/tex]

    So I'm able to prove them, but I don't know if I used the properties correctly (if at all). Any opinions or suggestions?
     
  2. jcsd
  3. Sep 15, 2009 #2

    Mark44

    Staff: Mentor

    By "using the properties" I think your prof means for you to indicate which property allows you to do each step. However, in some of your steps you are using operations that aren't listed amongst the properties you show. For example, in your 2nd inequality, when you multiply both members of an inequality by a positive number, the direction of the inequality stays the same.
    For the one above, instead of doing a proof by contradiction, as you have done, it would be simpler to start with (a - b)2 >= 0 (the square of any real number is always nonnegative). Then expand the left side and you should be able to get to the conclusion you need.
    In the one above, b + b = b(1 + 1) = b*2 = 2b. The properties used are the distributive property and the commutative property of multiplication.
     
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