Reciprocal of Real Number is Non-Zero
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Theorem
- $\forall x \in \R: x \ne 0 \implies \dfrac 1 x \ne 0$
Proof
Aiming for a contradiction, suppose that:
- $\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$
From Real Zero is Zero Element
- $\dfrac 1 x \times x = 0$
But from Real Number Axiom $\R \text M4$: Inverses for Multiplication:
- $\dfrac 1 x \times x = 1$
The result follows by Proof by Contradiction.
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(p)}$