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Math Help - Confusion with absolute value proof

  1. #1
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    Confusion with absolute value proof

    Hi, I hope this is the correct forum, if not my apologies

    I've been confused with this for a while and have decided I am really stuck.

    I have to prove that |-a| = |a| using the absolute value definition which was given to us as:

    |a|=\left\{\begin{array}{cc}a,&\mbox{ if }<br />
x\geq 0\\-a, & \mbox{ if } x<0\end{array}\right

    Now I understand the following by definition:

    if a>0 then

    |a| = a

    if a<0 then

    |a| = -(-a) = a

    But with |-a| = |a| I get really confused! My lecturer explained it as follows:

    if a > 0, then,

    LHS = |-a| = -(-a) = a
    RHS = |a| = a = LHS

    and if a < 0, then

    LHS = |-a| = -a
    RHS = |a| = -a = LHS

    Firstly, |-a| = -(-a) = a Is this because originally we treat a as positive, but because it is made negative within the absolute value, then we treat it as if it is negative, thus -(-a)?

    Then with a < 0, is it so because, we originally treat it as a<0, thus -(-a) but then as a negative, thus -a? Also with the RHS, -(-a) then -a?

    Some clarification on this would be really appreciated, here is my attempt at proving it:

    if a>0, then

    <br />
|-a| = |a| \Rightarrow -(-a) = a \Rightarrow a = a, therefore |-a| = |a|

    if a<0, then

    |-a| = |a| \Rightarrow -(-(-a)) = -(-a) \Rightarrow -a = a

    As you can see, I'm missing some small fact to make this true or I am making a really silly mistake somewhere as this seems like it should be very easy!
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  2. #2
    MHF Contributor

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    Do three cases.
    If a=0 then -a=0 so 0=|a|=|-a|

    If a>0 then -a<0 so |a|=a~\&~|-a|=-(-a)=a. Again |a|=|-a|

    If a<0 then -a>0 so |a|=-a~\&~|-a|=-a. Again |a|=|-a|
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