Reciprocal of Positive Real Number is Positive

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Theorem

Let $a \in \R$ such that $a > 0$.

Then $a^{-1} = \dfrac 1 a > 0$.


It follows directly that $a < 0 \implies a^{-1} < 0$.


Proof

Aiming for a contradiction, suppose $a > 0$ but $a^{-1} \le 0$.

\(\displaystyle a^{-1}\) \(\le\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \times a^{-1}\) \(\le\) \(\displaystyle a \times 0\) Real Number Ordering is Compatible with Multiplication
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(\le\) \(\displaystyle 0\)

The result follows from Proof by Contradiction.

$\blacksquare$


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