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Need clarification in proof (P9) from Chap 1 Spivak

  1. Aug 11, 2013 #1
    I am taking Calculus at the end of the month for the first time but I have started working through Spivak (wanted to get a headstart and didn't have the class text book yet) and I have been grasping his beautiful description of some the properties of real numbers yet there is a very minor detail in P9 in Chapter 1 in which a step is missing (or as I realized as I thought about it, I was missing something). I just need clarification to make sure I am understanding what happened.

    (P9) If a, b, and c are any numbers, then


    [itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c

    If [itex]a-b=b-a[/itex]
    then [itex](a-b)+b=(b-a)+b=b+(b-a)[/itex]
    hence [itex]a=b+b-a[/itex]
    hence [itex]a+a=(b+b-a)+a=b+b[/itex]
    Consequently [itex]a[/itex][itex]\cdot[/itex][itex](1+1)= b[/itex][itex]\cdot[/itex][itex](1+1)[/itex]
    and therefore [itex]a=b[/itex]

    Now does a+a=b+b become a(1+1)=b(1+1) because of p9 [itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c?

    I know this should be incredibly obvious and once I understand it I will kick myself but I couldn't move on without knowing exactly.
     
    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2

    micromass

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    Yes. And you also need ##a\cdot 1 = a##.
     
  4. Aug 11, 2013 #3
    Of course!

    [itex]a+a[/itex]

    [itex]a\cdot 1 = a[/itex]
    [itex]a\cdot 1[/itex][itex]+a\cdot 1[/itex][itex]=a(1+1)[/itex]

    Merci beaucoup mademoiselle! I knew it was something minor but opened my eyes more :)
     
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