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Case analysis

  1. Sep 7, 2008 #1
    If (|k1|)(|k2|)=1, show k1, k2= +/- 1

    I don't know how to show that abs value of k1=abs value k2=1.
  2. jcsd
  3. Sep 7, 2008 #2
    You don't need to do that. You have to consider four cases: k1 = 1, k2 = 1; k1 = -1, k2 = 1; etc. For each case, you have to show that the equality holds.
  4. Sep 7, 2008 #3
    I'm assuming k1 and k2 are integers by the way. Otherwise, the statement is false.
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