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Proof of ax=a then x=1

  1. Jan 17, 2014 #1
    I haven't written a lot of proofs so I need the opinion of the experts on my proof of a simple proposition. Here's the various properties I used: (P10) (Trichotomy law) For every number a, one and only one of the following holds: (i) a = 0, (ii) a is in the collection P, (iii) —a is in the collection P.

    (P7) For every number a not equal to 0, there is a number a^-1 such that a • a^-1 = a^-1 • a = 1.

    (P6) If a is any number, then a • 1 = 1 • a = a.

    If ax=a for some number "a" different from 0, then x=1.(Spivak's calculus.)

    I consider two cases: a>0 or a<0. By definition (Given in Spivak's calculus) :

    a>b if a-b is in the collection P (P being the collection of all positive numbers.)

    a>0 because a-o is in collection P by trichotomy law(P10)

    So a*x=a

    by P7 a*a^-1*x= a*a^-1

    1*x=1

    by P6 x=1

    Second case :

    By definition :

    a< to b if b>a

    a<0 because 0>a Now, we do the same thing as the previous case.

    Proposition proven !

    Here's Spivak's answer

    1=a^-1*a=a^-1*(a*x)=(a^-1*a)*x=1*x=x

    Here's my interpretation line by line (just to be sure im understanding it)

    a^-1*a=1
    a^-1*(a*x)=1 (Is my interpretation correct in saying that the number "a" can be factorized in a way that makes a=a*x, "x" being a variable which the value is to be determined)
    (a^-1*a)*x=1
    1*x=1
    x=1

    So, basically, Spivak is constructing his proof. Is it correct ?

    Any opinions ? Thank you!
     
    Last edited: Jan 17, 2014
  2. jcsd
  3. Jan 17, 2014 #2

    mfb

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    Staff: Mentor

    None of your rules would lead to the left-hand side where a^-1 appears in the middle of the previous multiplication.

    Second case of what?

    That does not make sense. This equation is not proven at this step.

    This is right, but it hides the important step: take the initial equation a*x=x and multiply both sides by a^-1 on the left, afterwards simplify.
     
  4. Jan 17, 2014 #3
    -Wait, what do you mean that my rules won't result in a*a^-1*x= a*a^-1 ? Is it only because I've put a^-1 in the middle that it isn't correct ?

    -The way I thought about the proof was to justify that it would work wether we were talking about positive numbers or negative numbers, hence the the second case.

    -Then did he just put the the variable "x" just like that ?

    -You meant a*x=a. Also, "multiply both sides by a^-1 on the left" you meant to multiply both sides and to simplify on only the left side ? Could you rephrase?
     
  5. Jan 17, 2014 #4
    No need to answer, I understand now. THank you!
     
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