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Thread: Prove elementary rules in a field

  1. #1
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    Prove elementary rules in a field

    Hey.

    Given the field F, Prove:
    * For any a, b in F, if a*a=b*b, then a=b or a=-b
    * For any a, b in F, if a*a*a = b*b*b, then a=b

    Given the field F where 1+1=0, Prove:
    * For any a in F, -a=a
    * For any a, b in F, a+b=a-b

    The proofs have to be supported by the axioms of fields rigorously.

    I don't know actually how even start proving it.
    Thanks.
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  2. #2
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    Re: Prove elementary rules in a field

    I'll start you off on the first one:

    As a preliminary Lemma, I will prove that if xy = 0, then x=0 or y=0.

    Indeed, if x is not 0, then y = 1y = [(1/x)*x]y = (1/x)(xy) = (1/x)(0) = 0 (You will need to prove that multiplying by 0 in a field always produces 0).

    [Notice how I used the existence of a multiplicative identity (namely 1), the multiplicative inverse property, and associativity of multiplication.]

    Now, if a*a=b*b, then a*a - b*b = b*b - b*b = 0.

    Next, prove that a*a - b*b = (a - b)(a + b) (to prove this you will need to use several field properties and it will require several steps).

    By the preliminary Lemma above, a - b = 0 or a + b = 0.

    Now, show that this last statement is equivalent to a = b or a = -b.

    That should help to get you started.
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  3. #3
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    Re: Prove elementary rules in a field

    For any a, b in F, if a*a*a = b*b*b, then a=b


    counterexample:

    in the field of complex numbers, let

    $\displaystyle a=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$ and $\displaystyle b=1$

    $\displaystyle a^3=b^3$ but $\displaystyle a \neq b$
    Last edited by Idea; Nov 18th 2018 at 07:14 AM.
    Thanks from topsquark and CStudent
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  4. #4
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    Re: Prove elementary rules in a field

    Quote Originally Posted by Idea View Post
    For any a, b in F, if a*a*a = b*b*b, then a=b


    counterexample:

    in the field of complex numbers, let

    $\displaystyle a=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$ and $\displaystyle b=1$

    $\displaystyle a^3=b^3$ but $\displaystyle a \neq b$
    Thank you!
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