# 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:

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

$\blacksquare$