Gaussian Integers are not Closed under Division

Theorem

The set of Gaussian integers $\Z \sqbrk i$ is not closed under division.

Proof

Proof by Counterexample:

Let:

\(\ds x\)

\(=\)

\(\ds 1 + 2 i\)

\(\ds y\)

\(=\)

\(\ds 3 + 4 i\)

Then:

\(\ds \leadsto \ \ \)

\(\ds x \div y\)

\(=\)

\(\ds \dfrac {1 + 2 i} {3 + 4 i}\)

Definition of Complex Division

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {1 + 2 i} \paren {3 - 4 i} } {\paren {3 + 4 i} \paren {3 - 4 i} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {11 + 2 i} {\sqrt {3^2 + 4^2} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {11} 5 + \dfrac {2 i} 5\)

which is not a Gaussian integer

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Gaussian integer
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Gaussian integer