Self-Inverse Element of Integral Domain is Unity or its Negative

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Theorem

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.

Let $x \in D$ such that $x^2 = 1_D$.


Then either $x = 1_D$ or $x = -1_D$.


Proof

\(\ds x^2\) \(=\) \(\ds 1_D\)
\(\ds \leadsto \ \ \) \(\ds \paren {x + 1_D} \paren {x + \paren {-1_D} }\) \(=\) \(\ds 0_D\)
\(\ds \leadsto \ \ \) \(\ds x + 1_D\) \(=\) \(\ds 0_D\)
\(\, \ds \lor \, \) \(\ds x + \paren {-1_D}\) \(=\) \(\ds 0_D\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds -1_D\)
\(\, \ds \lor \, \) \(\ds x\) \(=\) \(\ds 1_D\)

$\blacksquare$


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