Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals

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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