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

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

Let $f: \R \to \R$ be the real function defined as:
 * $f \left({x}\right) = a^x$

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

Then $f$ is strictly decreasing.

Proof
Let $x, y \in \R$ be such that $x < y$.

Since $0 \lt a \lt 1$ then $\dfrac 1 a > 1$.

By Real Power Function on Base Greater than One is Strictly Increasing then:
 * $\paren{\dfrac 1 a}^x \lt \paren{\dfrac 1 a}^y$

But:

The result follows.