Inverse of Positive Real is Positive

From ProofWiki
Jump to: navigation, search

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 $a^{-1} \le 0$.

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

The result follows from Proof by Contradiction.

$\blacksquare$


Sources