Non-Zero Complex Numbers are Closed under Multiplication/Proof 3

From ProofWiki
Jump to navigation Jump to search

Theorem

The set of non-zero complex numbers is closed under multiplication.


Proof

Equivalently this is to say:

$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$


Let $z_1 z_2 = 0$.

\(\ds z_1\) \(=\) \(\ds \tuple {x_1, y_1}\) Definition 2 of Complex Number: for some $x_1, y_1 \in \R$
\(\ds z_2\) \(=\) \(\ds \tuple {x_2, y_2}\) Definition 2 of Complex Number: for some $x_2, y_2 \in \R$
\(\ds \leadsto \ \ \) \(\ds z_1 z_2\) \(=\) \(\ds \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2}\) Definition of Complex Multiplication
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x_1 x_2\) \(=\) \(\ds y_1 y_2\)
\(\text {(2)}: \quad\) \(\, \ds \land \, \) \(\ds x_1 y_2\) \(=\) \(\ds -y_1 x_2\) as $z_1 z_2 = 0$


Without loss of generality, let $\tuple {x_2, y_2} \ne \tuple {0, 0}$.

Aiming for a contradiction, suppose also that $\tuple {x_1, y_1} \ne \tuple {0, 0}$.

Then:

\(\ds x_1\) \(=\) \(\ds \frac {y_1 y_2} {x_2}\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \frac {y_1 y_2} {x_2} y_2\) \(=\) \(\ds -y_1 x_2\) substituting in $(2)$
\(\ds \leadsto \ \ \) \(\ds y_2^2\) \(=\) \(\ds - x_2^2\)
\(\ds \leadsto \ \ \) \(\ds x_2 = y_2\) \(=\) \(\ds 0\) as both $x_2$ and $y_2$ are real

From this contradiction it follows that:

$\tuple {x_1, y_1} = \tuple {0, 0}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Non-Zero_Complex_Numbers_are_Closed_under_Multiplication/Proof_3&oldid=537178"