# 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 $\left({\R_{\ne 0}, \times}\right)$ is a group.

Thus, by definition, $\times$ is closed in $\left({\R_{\ne 0}, \times}\right)$.

$\blacksquare$

## 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 Axioms: $\R M 1$: Associativity $\ds \leadsto \ \$ $\ds 1 \times y$ $=$ $\ds \frac 1 x \times 0$ Real Number Axioms: $\R M 4$: Inverse $\ds \leadsto \ \$ $\ds y$ $=$ $\ds \frac 1 x \times 0$ Real Number Axioms: $\R M 3$: Identity $\ds \leadsto \ \$ $\ds y$ $=$ $\ds 0$ Real Zero is Zero Element

Thus:

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

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.

$\blacksquare$