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:
\(\ds x \times y\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 x \times \paren {x \times y}\) | \(=\) | \(\ds \frac 1 x \times 0\) | 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:
- $x \times y = 0, x \ne 0 \implies y = 0$
- $x \times y = 0, y \ne 0 \implies x = 0$
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
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(l)}$