Inverse of Positive Real 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

Suppose $a > 0$ but $a^{-1} \le 0$.

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

The result follows from Proof by Contradiction.

$\blacksquare$


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