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
\(\ds x^2\) | \(=\) | \(\ds 1_D\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + 1_D} \paren {x + \paren {-1_D} }\) | \(=\) | \(\ds 0_D\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + 1_D\) | \(=\) | \(\ds 0_D\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds x + \paren {-1_D}\) | \(=\) | \(\ds 0_D\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -1_D\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds x\) | \(=\) | \(\ds 1_D\) |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $12$