Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals
Jump to navigation
Jump to search
Theorem
- $\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$
Proof
From Reciprocal of Strictly Positive Real Number is Strictly Positive:
- $(1): \quad x > 0 \implies \dfrac 1 x > 0$
- $(2): \quad y > 0 \implies \dfrac 1 y > 0$
Then:
\(\ds x\) | \(>\) | \(\ds y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \times \frac 1 x\) | \(>\) | \(\ds y \times \frac 1 x\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication and from $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \times \frac 1 x \times \frac 1 y\) | \(>\) | \(\ds y \times \frac 1 x \times \frac 1 y\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication and from $(2)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \times \frac 1 x} \times \frac 1 y\) | \(>\) | \(\ds \paren {y \times \frac 1 y} \times \frac 1 x\) | Real Number Axiom $\R \text M1$: Associativity of Multiplication and Real Number Axiom $\R \text M2$: Commutativity of Multiplication | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 \times \frac 1 y\) | \(>\) | \(\ds 1 \times \frac 1 x\) | Real Number Axiom $\R \text A4$: Inverses for Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 y\) | \(>\) | \(\ds \frac 1 x\) | Real Number Axiom $\R \text M3$: Identity Element for Multiplication | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 x\) | \(<\) | \(\ds \frac 1 y\) | Definition of Dual Ordering |
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(j)}$