Rational Power of Product of Real Numbers

Theorem

Let $r, s \in \R_{> 0}$ be (strictly) positive real numbers.

Let $x \in \Q$ be a rational number.

Let $r^x$ be defined as $r$ to the power of $x$.

Then:

$\paren {r s}^x = r^x s^x$

Proof

Let $x = \dfrac p q$ where $p, q \in \Z$ and $q > 0$.

We have:

\(\ds r^x s^x\)

\(=\)

\(\ds \paren {r^p}^{1 / q} \paren {s^p}^{1 / q}\)

\(\ds \)

\(=\)

\(\ds \paren {r^p s^p}^{1 / q}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {r s}^p}^{1 / q}\)

\(\ds \)

\(=\)

\(\ds \paren {r s}^{p / q}\)

\(\ds \)

\(=\)

\(\ds \paren {r s}^x\)

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.12 \ (3) \, \text{(iii)}$