# Power Function on Base between Zero and One is Strictly Decreasing/Rational Number

## Theorem

Let $a \in \R$ be a real number such that $0 < a < 1$.

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 decreasing.

## Proof

Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z, s, u \in \Z_{>0}$.

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

Then:

 $\ds r u$ $<$ $\ds t s$ Real Number Ordering is Compatible with Multiplication $\ds \leadsto \ \$ $\ds a^{r u}$ $>$ $\ds a^{t s}$ Power Function on Base between Zero and One is Strictly Decreasing: Integer $\ds \leadsto \ \$ $\ds \paren {a^r}^u$ $>$ $\ds \paren {a^t}^s$ Product of Indices of Real Number: Integers $\ds \leadsto \ \$ $\ds \sqrt [u] {\paren {a^r}^u}$ $>$ $\ds \sqrt [u] {\paren {a^t}^s }$ Root is Strictly Increasing $\ds \leadsto \ \$ $\ds a^r$ $>$ $\ds \sqrt [u] {\paren {a^t}^s }$ Definition of $u$th Root $\ds \leadsto \ \$ $\ds \sqrt [s] {\paren {a^r} }$ $>$ $\ds \sqrt [s] {\sqrt [u] {\paren {a^t}^s } }$ Root is Strictly Increasing $\ds \leadsto \ \$ $\ds \sqrt [s] {\paren {a^r} }$ $>$ $\ds \sqrt [u] {\sqrt [s] {\paren {a^t}^s } }$ Root is Commutative $\ds \leadsto \ \$ $\ds \sqrt [s] {\paren {a^r} }$ $>$ $\ds \sqrt [u] {\paren {a^t} }$ Definition of $s$th Root $\ds \leadsto \ \$ $\ds a^{r / s}$ $>$ $\ds a^{t / u}$ Definition of Rational Power

$\blacksquare$