Non-Zero Real Numbers Closed under Multiplication

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Theorem

The set of non-zero real numbers is closed under multiplication:

$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$


Proof 1

Recall that Real Numbers form Field under the operations of addition and multiplication.

By definition of a field, the algebraic structure $\struct {\R_{\ne 0}, \times}$ is a group.

Thus, by definition, $\times$ is closed in $\struct {\R_{\ne 0}, \times}$.

$\blacksquare$


Proof 2

Let $x \times y = 0$.

Without loss of generality, suppose that $x \ne 0$.

Then:

\(\ds x \times y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 x \times \paren {x \times y}\) \(=\) \(\ds \frac 1 x \times 0\) as $x \ne 0$
\(\ds \leadsto \ \ \) \(\ds \paren {\frac 1 x \times x} \times y\) \(=\) \(\ds \frac 1 x \times 0\) Real Number Axiom $\R \text M1$: Associativity of Multiplication
\(\ds \leadsto \ \ \) \(\ds 1 \times y\) \(=\) \(\ds \frac 1 x \times 0\) Real Number Axiom $\R \text M4$: Inverses for Multiplication
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \frac 1 x \times 0\) Real Number Axiom $\R \text M3$: Identity Element for Multiplication
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds 0\) Real Zero is Zero Element


Thus:

$x \times y = 0, x \ne 0 \implies y = 0$

Mutatis mutandis

$x \times y = 0, y \ne 0 \implies x = 0$

and so:

$x \times y = 0 \implies y = 0 \lor x = 0$


So:

\(\ds \) \(\) \(\ds x \times y = 0\)
\(\ds \) \(\leadsto\) \(\ds \paren {x = 0} \lor \paren {y = 0}\)
\(\ds \) \(\leadsto\) \(\ds \neg \paren {x \ne 0 \land y \ne 0}\) De Morgan's Laws: Disjunction

The result follows by the Rule of Transposition.

$\blacksquare$


Sources

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