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

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Theorem

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

Let $f: \Z \to \R$ be the real-valued function defined as:

$\map f k = a^k$

where $a^k$ denotes $a$ to the power of $k$.


Then $f$ is strictly decreasing.


Proof

Let $0 < a < 1$.

By Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer, the theorem is already proven for positive integers.

It remains to be proven over the negative integers.


Let $i, j$ be integers such that $i < j < 0$.


From Order of Real Numbers is Dual of Order of their Negatives:

$0 < -j < -i$

So:

\(\displaystyle a^{-j}\) \(>\) \(\displaystyle a^{-i}\) Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 {a^j}\) \(>\) \(\displaystyle \frac 1 {a^i}\) Real Number to Negative Power: Positive Integer
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^i\) \(>\) \(\displaystyle a^j\) Ordering of Reciprocals

Hence the result.

$\blacksquare$