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

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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

$\blacksquare$