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

## Theorem

$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$

## Proof

$(1): \quad x > 0 \implies \dfrac 1 x > 0$
$(2): \quad y > 0 \implies \dfrac 1 y > 0$

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

$\blacksquare$