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:

$\map f x = a^x$

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 $0 < a < 1$, we have that:

$\dfrac 1 a > 1$

Then we have that:

\(\ds \paren {\dfrac 1 a}^x\)

\(<\)

\(\ds \paren {\dfrac 1 a}^y\)

Real Power Function on Base Greater than One is Strictly Increasing

\(\ds \leadstoandfrom \ \ \)

\(\ds \dfrac 1 {a^x}\)

\(<\)

\(\ds \dfrac 1 {a^y}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds a^x\)

\(>\)

\(\ds a^y\)

Reciprocal Function is Strictly Decreasing

The result follows.

$\blacksquare$