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

$\blacksquare$