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:

\(\displaystyle x\) \(>\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \times \frac 1 x\) \(>\) \(\displaystyle y \times \frac 1 x\) Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \times \frac 1 x \times \frac 1 y\) \(>\) \(\displaystyle y \times \frac 1 x \times \frac 1 y\) Real Number Axioms: $\R O2$: compatibility with multiplication and from $(2)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x \times \frac 1 x} \times \frac 1 y\) \(>\) \(\displaystyle \paren {y \times \frac 1 y} \times \frac 1 x\) Real Number Axioms: $\R M1$ (Associativity) and $\R M2$ (Commutativity)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1 \times \frac 1 y\) \(>\) \(\displaystyle 1 \times \frac 1 x\) Real Number Axioms: $\R M4$ (Inverses)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 y\) \(>\) \(\displaystyle \frac 1 x\) Real Number Axioms: $\R M3$ (Identity)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 x\) \(<\) \(\displaystyle \frac 1 y\) Definition of Dual Ordering

$\blacksquare$


Sources