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Thread: Proving |a+b| = |a|+|b|

  1. #1
    Member rowe's Avatar
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    Proving |a+b| = |a|+|b|

    Hi guys, I'm doing a case by case proof of

    $\displaystyle |a+b| \leq |a|+|b|$

    from Spivak, and he says:

    When $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, we must prove that:

    $\displaystyle |a+b| \leq a-b$

    I'm a bit stuck on his line of reasoning, could someone explain why we have to prove the above?
    Last edited by rowe; Jun 5th 2010 at 05:28 AM.
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  2. #2
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    Quote Originally Posted by rowe View Post
    I'm doing a case by case proof of
    $\displaystyle |a+b| = |a|+|b|$ from Spivak, and he says
    You cannot prove $\displaystyle |a+b|=|a|+|b|$ because it is not true.
    So exactly what is the correct statement of this problem?
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  3. #3
    Member rowe's Avatar
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    Typo, sorry. Fixed original post
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  4. #4
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    Quote Originally Posted by rowe View Post
    Hi guys, I'm doing a case by case proof of

    $\displaystyle |a+b| \leq |a|+|b|$

    from Spivak, and he says:

    When $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, we must prove that:

    $\displaystyle |a+b| \leq a-b$

    I'm a bit stuck on his line of reasoning, could someone explain why we have to prove the above?
    Hi

    If $\displaystyle a \geq 0$ then $\displaystyle |a| = a$
    and if $\displaystyle b \leq 0$ then $\displaystyle |b| = -b$

    Therefore when $\displaystyle a \geq 0$ and $\displaystyle b \leq 0$, $\displaystyle |a|+|b| = a-b$
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