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Currently working through Apostol need a check on my logic

  1. Jul 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following theorem.
    If a≠0, then (a-1)-1=a

    Theorems proven before/axioms I am allowed to use:
    Existence of reciprocals axiom: there exist real numbers x and y where x≠0 such that xy=1
    Possibility of division Thm: basically a-1=1/a




    2. Relevant equations
    I just need a check on my proof since I am self studying the book. Any logical mistakes on my proof or suggestions etc? This is my first time writing proofs.


    3. The attempt at a solution
    Assume (a-1)-1=a is true.
    Then by the existence of reciprocals axiom there exists an x such that (a-1)-1x=1.
    Hence,
    (a-1)-1x=ax​
    ax=1​
    Now by the possibility of division theorem we have,
    x=1/a=a-1
    Finally since we assumed (a-1)-1=a was true it follows that:
    (a-1)-1x=ax=aa-1=1​
    Q.E.D​

    It looks fine to me, I'm just a noob at proofs. Thanks.
     
  2. jcsd
  3. Jul 30, 2012 #2

    SammyS

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    According to the first line of your proof,
    Assume (a-1)-1=a is true. ​
    you're assuming that the thing you're proving is true.

    That's a definite No-No .

    I suggest starting with:
    a-1 has a reciprocal, (a-1)-1.

    Therefore, (a-1)-1 a-1 = 1
    ...
    Now, multiply that by a, with a being on the right.
     
    Last edited: Jul 30, 2012
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