# 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:

\(\ds r u\) | \(<\) | \(\ds t s\) | 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 |

$\blacksquare$