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

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

$0 < \dfrac 1 a < 1$

So:

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

Hence the result.

$\blacksquare$