Power Function on Base Greater than One is Strictly Increasing/Rational Number

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Theorem

Let $a \in \R$ be a real number such that $a > 1$.

Let $f: \Q \to \R$ be the real-valued function defined as:

$\map f q = a^q$

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


Then $f$ is strictly increasing.


Proof

Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z$ are integers and $s, u \in \Z_{>0}$ are strictly positive integers.

Let $\dfrac r s < \dfrac t u$.


From Ordering of Reciprocals:

$0 < \dfrac 1 a < 1$


So:

\(\displaystyle \paren {\frac 1 a}^{t / u}\) \(<\) \(\displaystyle \paren {\frac 1 a}^{r / s}\) Power Function on Base between Zero and One is Strictly Decreasing: Rational Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt [u] {\paren {\frac 1 a}^t}\) \(<\) \(\displaystyle \sqrt [s] {\paren {\frac 1 a}^r}\) Definition of Rational Power
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt [u] {\paren {\frac 1 {a^t} } }\) \(<\) \(\displaystyle \sqrt [s] {\paren {\frac 1 {a^r} } }\) Real Number to Negative Power: Integer
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 {\sqrt [u] {a^t} }\) \(<\) \(\displaystyle \frac 1 {\sqrt [s] {a^r} }\) Root of Reciprocal is Reciprocal of Root
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 {a^{t / u} }\) \(<\) \(\displaystyle \frac 1 {a^{r / s} }\) Definition of Rational Power
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{r / s}\) \(<\) \(\displaystyle a^{t / u}\) Ordering of Reciprocals


Hence the result.

$\blacksquare$