AAR10

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Prove: If R\, is an integral domain and \exists x\in R\, s.t. x^{2}=1\, then x=\pm 1\,.

(1)(1)=1=(-1)(-1)\, so \pm 1\, are two choices for x\,. Suppose \exists x_{3}\in R,x_{3}\neq \pm 1,x_{3}^{2}=1\, and let x=\pm 1\,. Then 1-1=x^{2}-x_{3}^{2}=(x-x_{3})(x+x_{3})=0\,. Since x\neq \pm x_{3},(x-x_{3})\neq 0\, and (x+x_{3})\neq 0\,. But R\, is an integral domain and has no zero divisors. This is a contradiction of the existence of x_{3}\,.


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