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

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:
 * $f \left({k}\right) = a^k$

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:

Hence the result.