Non-Zero Real Numbers Closed under Multiplication/Proof 2

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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

Let $x \times y = 0$.

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

Then:

\(\displaystyle x \times y\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 x \times \paren {x \times y}\) \(=\) \(\displaystyle \frac 1 x \times 0\) as $x \ne 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\frac 1 x \times x} \times y\) \(=\) \(\displaystyle \frac 1 x \times 0\) Real Number Axioms: $\R M 1$: Associativity
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1 \times y\) \(=\) \(\displaystyle \frac 1 x \times 0\) Real Number Axioms: $\R M 4$: Inverse
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \frac 1 x \times 0\) Real Number Axioms: $\R M 3$: Identity
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle 0\) 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:

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

The result follows by the Rule of Transposition.

$\blacksquare$


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