AAR10

From Exampleproblems

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

$(1)(1)=1=(-1)(-1)\,$ so $\pm 1\,$ are two choices for $x\,$ . Suppose $\exists x_3 \isin R, x_3 \ne \pm 1, x_3^2=1\,$ and let $x=\pm 1\,$ . Then $1-1=x^2-x_3^2=(x-x_3)(x+x_3)=0\,$ . Since $x\ne \pm x_3, (x-x_3)\ne 0\,$ and $(x+x_3)\ne 0\,$ . But $R\,$ is an integral domain and has no zero divisors. This is a contradiction of the existence of $x_3\,$ .