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Math Help - Proof that x^2 = 1 implies x = 1 or x = -1

  1. #1
    Newbie driegert's Avatar
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    Proof that x^2 = 1 implies x = 1 or x = -1

    I know this should be really easy, but I can't seem to figure it out. I don't really want an out an out answer if possible, just a starting point to work from.

    The question reads as follows:

    "Let R be an integral domain, and suppose x \in R satisfies x^2 = 1. Prove that we must either have x = 1 or x = -1."

    I started off by saying that if we have:

    x \cdot x = 1

    Then x has an inverse, x^{-1} where x^{-1} = x.

    This means that x is also in the group of units of R, but then I just don't know where to go... part of me wants to ignore the commutative law and say something like:

    x^2 = 1
    x \cdot x = 1
    x \cdot 1 = 1 \cdot x^{-1}

    So by right multiplication then x = 1 and x^{-1} = 1. But that just seems off.

    Thanks for any help.
    --
    Dave
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hmmm...

    x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

    ?
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Hmmm...

    x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

    ?
    Judging by the type of question you'll probably have to prove  x^2 - y^2 = (x-y)(x+y) to justify the second step.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Hmmm...

    x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

    ?
    I'm afraid it's not as easy like that: Factorisation is not necessarily unique in a ID. For example Z/4Z is an ID but factorisation (x-2)^2= x^2 is not unique.
    You probably need to use the fact in this case that roots of x^2=1 must be units.
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  5. #5
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    Quote Originally Posted by Dinkydoe View Post
    I'm afraid it's not as easy like that: Factorisation is not necessarily unique in a ID. For example Z/4Z is an ID but factorisation (x-2)^2= x^2 is not unique.
    You probably need to use the fact in this case that roots of x^2=1 must be units.

    Well, it is true that x^2=1\Longrightarrow (x-1)(x+1)=0 in any commutative ring, whether this decomposition is unique or not. So we're forced

    to deduce that (x-1)(x+1)=0 and now we use that the ring is an ID to reach the conclusion.

    And \mathbb{Z}/4\mathbb{Z} is not an ID , since in it 2\cdot 2=0\,,\,but\,\,2\neq 0

    Tonio
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  6. #6
    Senior Member Dinkydoe's Avatar
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    Quote Originally Posted by tonio View Post
    Well, it is true that x^2=1\Longrightarrow (x-1)(x+1)=0 in any commutative ring, whether this decomposition is unique or not. So we're forced

    to deduce that (x-1)(x+1)=0 and now we use that the ring is an ID to reach the conclusion.

    And \mathbb{Z}/4\mathbb{Z} is not an ID , since in it 2\cdot 2=0\,,\,but\,\,2\neq 0

    Tonio
    Ai, yes...I mixed up things. We must have ab=0\Rightarrow a=0, or  b=0.

    Thus now (x-1) =0, or (x+1)=0 gives the desired conclusion.
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  7. #7
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    It is simple:
    x^2=1
    x^2-1=0
    x^2+x-x-1=0
    x(x+1)-(x+1)=0
    (x+1)(x-1)=0
    x=1 or x=-1
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  8. #8
    Newbie driegert's Avatar
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    Thanks everyone for your help! I made this a lot more difficult for myself than I needed to.

    I really appreciate all the responses.

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    Dave
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