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Homework Help: Easy Analysis Proof

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data
    The positive part of an a in R is defined by:

    [tex]a^{+} = (|a| + a) / 2[/tex]

    and the negative by:

    [tex]a^{-} = (|a| - a) / 2[/tex].

    Prove that [tex]a = a^{+} - a^{-}[/tex] and [tex]|a| = a^{+} + a^{-}[/tex]

    2. Relevant equations
    The field axioms(closure, associativity,...)
    The order axioms
    Definition of the absolute value (and a few theorems)



    3. The attempt at a solution

    Now, I'm new at proof writing, but this seems too simple (and I don't know if providing all these steps is too much or not?) : /

    [tex]a = a^{+} - a^{-}
    = (|a| + a)/2 - (|a| - a)/2
    = 2^{-1}(|a| + a) - 2^{-1}(|a| - a)
    = 2^{-1}(|a| + a - |a| + a)
    = 2^{-1}(2a)
    = a
    [/tex]


    (btw, how can I separate my proof line by line with LaTeX?)
     
  2. jcsd
  3. Jan 12, 2010 #2

    Mark44

    Staff: Mentor

    It looks like you have proved that a = a, which is obvious to the most casual observer, and for this reason, not very interesting.

    For the first part, start with a+ - a- and use your definitions. What do you get?
    For the second part, start with a+ + a-. What do you get this time?

    For your last questions, use separate LaTeX lines.
     
  4. Jan 12, 2010 #3
    alright, so we'd have something like this:

    if [tex]a \geq 0[/tex] then [tex]|a| = a[/tex]
    [tex](|a| + a) / 2 - (|a| - a) / 2 = [/tex]
    [tex]= (a + a) / 2 - (a - a) / 2 = [/tex]
    [tex] = 2(a)/2 - (0)/2 = [/tex]
    [tex]= a[/tex]

    if [tex]a < 0[/tex] then[tex] |a| = -a[/tex]
    and then the same procedure
     
  5. Jan 13, 2010 #4

    Mark44

    Staff: Mentor

    This is correct, but is more complicated than it needs to be.

    a+ - a-
    = (1/2)(|a| + a) - (1/2)(|a| - a)
    = (1/2)[|a| - |a| + a -(-a)]
    = (1/2)(2a) = a
    This is true for any real value of a, not just for a >= 0.

    You should be able to do something similar for the other part.
     
  6. Jan 13, 2010 #5
    Alright, thanks Mark44. I included it just because I'm not used to writing proofs yet (ie I don't exactly know what to include/omit yet).
     
  7. Jan 13, 2010 #6

    Mark44

    Staff: Mentor

    Include everything that needs to be there, and omit all the rest. Start at one side of the equation you are trying to prove. Use definitions to rewrite the expression on that side and work toward making what you start with look like the other side you're trying to get to.
     
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