Power of Positive Real Number is Positive/Rational Number

Theorem

Let $x \in \R_{>0}$ be a (strictly) positive real number.

Let $q \in \Q$ be a rational number.

Then:

$x^q > 0$

where $x^q$ denotes the $x$ to the power of $q$.

Proof

Let $q = \dfrac r s$, where $r \in \Z$, $s \in \Z \setminus \set 0$.

Then:

\(\ds x > 0\)

\(\leadsto\)

\(\ds x^r > 0\)

Power of Positive Real Number is Positive: Integer

\(\ds \)

\(\leadsto\)

\(\ds \sqrt [s] {\paren {x^r} } > 0\)

Existence of Positive Root of Positive Real Number

\(\ds \)

\(\leadsto\)

\(\ds x^{r / s}\)

Definition of Rational Power

Hence the result.

$\blacksquare$