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Math Help - Absolute value proof

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    Member Ranger SVO's Avatar
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    Absolute value proof

    Using distance what would be the simplest way to prove that the

    abs(-a) = abs(a) for all a
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ranger SVO View Post
    Using distance what would be the simplest way to prove that the

    abs(-a) = abs(a) for all a
    i'd probably represent a and -a on a number line. it is "obvious" that the distance from 0 to -a is the same as the distance from 0 to a. if you want to be rigorous, you can apply the distance formula with coordinates and the whole shabang
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    Quote Originally Posted by Ranger SVO View Post
    Using distance what would be the simplest way to prove that the

    abs(-a) = abs(a) for all a
    By consider two cases a>0 and a<0.

    Case 1 if a>0 then abs(-a) = -(-a)=a because -a <0. And abs(a) = a because a>0. Q.E.D.

    Case 2 if a<0 now continue ...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    By consider two cases a>0 and a<0.

    Case 1 if a>0 then abs(-a) = -(-a)=a because -a <0. And abs(a) = a because a>0. Q.E.D.

    Case 2 if a<0 now continue ...
    TPH is right. i guess the simplest way would be to use the definition. recall that:

    |a| = \left\{\begin{array}{cc}a,&\mbox{ if }<br />
a \geq 0\\-a, & \mbox{ if } a<0\end{array}\right.
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    Member Ranger SVO's Avatar
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    Thank you to all. All of you have been helpful.

    It hard to believe that I will graduate after this class and Abstract Algebra next semester. Its been a long 12 years (I'm not real bright).
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ranger SVO View Post
    Thank you to all. All of you have been helpful.

    It hard to believe that I will graduate after this class and Abstract Algebra next semester. Its been a long 12 years (I'm not real bright).
    what class is this that you are taking?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ranger SVO View Post
    Using distance what would be the simplest way to prove that the

    abs(-a) = abs(a) for all a
    I'm probably going to get into trouble saying this (because my comment is over-generalized) but here goes:

    The way I introduce the absolute value function is to define it as the distance from the point x = a to the origin. Note that when you get to the modulus of a complex number this same exact definition will hold. (This idea is my own, but is spurred on by the fact that the absolute value and the complex modulus operators are given using the same symbol. At least in the texts I've seen. So presumably someone else out there might have had similar ideas.)

    -Dan
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    Quote Originally Posted by topsquark View Post
    The way I introduce the absolute value function is to define it as the distance from the point x = a to the origin. Note that when you get to the modulus of a complex number this same exact definition will hold. (This idea is my own, but is spurred on by the fact that the absolute value and the complex modulus operators are given using the same symbol. At least in the texts I've seen. So presumably someone else out there might have had similar ideas.)
    In my Real Multivariable Class we define | \bold{x} | = \sqrt{x_1^2+...+x_2^2}. For any \bold{x}\in \mathbb{R}^n. Hence \bold{x-y}=\bold{y-x} is trivial. And so |a|=|-a| because this norm on 1 dimensional space is \sqrt{x_1^2}=|x_1|. Maybe this is what you are saying (because the complex plane is just C=\mathbb{R}^2).
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    In my Real Multivariable Class we define | \bold{x} | = \sqrt{x_1^2+...+x_2^2}. For any \bold{x}\in \mathbb{R}^n. Hence \bold{x-y}=\bold{y-x} is trivial. And so |a|=|-a| because this norm on 1 dimensional space is \sqrt{x_1^2}=|x_1|. Maybe this is what you are saying (because the complex plane is just C=\mathbb{R}^2).
    Yeah, that's it. I just didn't have a "formal" justification for doing it. (Or maybe the right word is "analytic." Whatever.)

    -Dan
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