# 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.

$\blacksquare$