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Math Help - [SOLVED] Abosolute Values, precalc

  1. #1
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    [SOLVED] Abosolute Values, precalc

    Hello; I currently working on absolute values and was doing good until I now.

    Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?

    Thanks a bunch

    Carolyn
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  2. #2
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    Quote Originally Posted by Carolyng66 View Post
    Hello; I currently working on absolute values and was doing good until I now.

    Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?

    Thanks a bunch

    Carolyn
     |a-b| = |b| - |a|
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  3. #3
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    Quote Originally Posted by Mush View Post
     |a-b| = |b| - |a|
    THAT IS FALSE!
    Prehaps, Mush means \left| {a - b} \right| = \left| {b - a} \right|?
    That is true for all a & b
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  4. #4
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    Quote Originally Posted by Plato View Post
    THAT IS FALSE!
    Prehaps, Mush means \left| {a - b} \right| = \left| {b - a} \right|?
    That is true for all a & b
    Now that I've gone over my notes gain, I understand this now. I believe this is right.

    Thank You both for your help.
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  5. #5
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    Quote Originally Posted by Carolyng66 View Post
    Hello; I currently working on absolute values and was doing good until I now. Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?
    By definition \left| a \right| = \left\{ {\begin{array}{*{20}c}<br />
   {a,} & {0 \leqslant a}  \\   { - a,} & {a < 0}  \\ \end{array} } \right..
    Now \sqrt 2  - 5 < 0\;\text{ so }\;\left| {\sqrt 2  - 5} \right| =  - \left( {\sqrt 2  - 5} \right) = 5 - \sqrt 2 .<br />
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  6. #6
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    Math symbols

    Quote Originally Posted by Plato View Post
    By definition \left| a \right| = \left\{ {\begin{array}{*{20}c}<br />
{a,} & {0 \leqslant a} \\ { - a,} & {a < 0} \\ \end{array} } \right..
    Now \sqrt 2 - 5 < 0\;\text{ so }\;\left| {\sqrt 2 - 5} \right| = - \left( {\sqrt 2 - 5} \right) = 5 - \sqrt 2 .<br />
    I'm finding all of this quite helpful. One more question about this sight (this is my first time using it). Can someone please explain to me how to make math symbols so I don't have to spell out words such as; "square root and absolute value". It would make future postings easier to read. I tried unsuccessfully locating a "cheat sheet" if you will.



    Regards
    Carolyn
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  7. #7
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    Quote Originally Posted by Plato View Post
    By definition \left| a \right| = \left\{ {\begin{array}{*{20}c}<br />
{a,} & {0 \leqslant a} \\ { - a,} & {a < 0} \\ \end{array} } \right..
    Now \sqrt 2 - 5 < 0\;\text{ so }\;\left| {\sqrt 2 - 5} \right| = - \left( {\sqrt 2 - 5} \right) = 5 - \sqrt 2 .<br />
    So would be the same as saying the absolute value of 1-3 is 3-1?
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  8. #8
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    Quote Originally Posted by Plato View Post
    THAT IS FALSE!
    Prehaps, Mush means \left| {a - b} \right| = \left| {b - a} \right|?
    That is true for all a & b
    Indeed, apologies.
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