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Spivak's Calculus proof explain this please

  1. Sep 17, 2013 #1
    In Spivak's Calculus, Chapter 1, pg 12, theorem 1, it says that in order to prove that theorem: |a + b| <= |a| + |b|...there are 4 cases of what could happen...blah, blah.

    So one of the cases supposes that a >= 0 and b <= 0. Then it goes on to say something like "In this case, we must prove that: |a + b| <= a-b. This case may therefore be divided into two subcases. If a + b >= 0, then we must prove that a + b <= a -b, [and vice versa]."

    My question is: Where the heck did the |a + b| <= a - b come from and how does it have to do with the theorem, and how will proving this statement then prove the theorem? Also, why are we dividing into subcases and where did those come from?

  2. jcsd
  3. Sep 17, 2013 #2


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    If a > 0, then |a| = a. If b<0, then |b| = -b. Therefore if a>0 and b<0, to prove |a+b| <= |a| + |b| you have to prove |a+b| <= a-b
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