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Multiplying by 1 adds a restriction?

  1. Nov 1, 2012 #1
    [tex](a-b)=(a-b)*1\\ =(a-b)\frac { a+b }{ a+b } \\ =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b } [/tex]

    But now we have the restriction that a+b≠0?
     
  2. jcsd
  3. Nov 1, 2012 #2

    arildno

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    It is the identification of 1=(a+b)/(a+b) that introduces that restriction, not multiplying with 1 as such.
     
  4. Nov 1, 2012 #3
    So how come we do that all the time in algebra (multiplying by the conjugate) to get rid of annoying radicals? Doesn't that make it not equal to the previous line?
     
  5. Nov 1, 2012 #4

    arildno

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    Basically because the case a+b=0 can usually be treated as an uninteresting special case.
     
  6. Nov 1, 2012 #5
  7. Nov 1, 2012 #6

    Mark44

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    You can always multiply by 1 to get a new expression with the same value. The problem here is that what you multiplied by [ (a + b)/(a + b) ] isn't identically 1, and it is those values of a and b that cause a problem.
     
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