Sum of Indices of Real Number/Rational Numbers
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Theorem
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $x, y \in \Q$ be rational numbers.
Let $r^x$ be defined as $r$ to the power of $n$.
Then:
- $r^{x + y} = r^x \times r^y$
Proof
Let $x = \dfrac p q, y = \dfrac u v$.
Then:
\(\ds r^\paren {x + y}\) | \(=\) | \(\ds r^\paren {\paren {p / q} + \paren {u / v} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds r^\paren {\paren {p v + u q} / q v}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {r^\paren {1 / q v} }^\paren {p v + u q}\) | Definition of Rational Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {r^\paren {1 / q v} }^\paren {p v} \times \paren {r^\paren {1 / q v} }^\paren {u q}\) | Sum of Indices of Real Number: Integers | |||||||||||
\(\ds \) | \(=\) | \(\ds r^\paren {p v / q v} \times r^\paren {u q / q v}\) | Definition of Rational Power | |||||||||||
\(\ds \) | \(=\) | \(\ds r^\paren {p / q} \times r^\paren {u / v}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds r^x \times r^y\) |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.12 \ (3) \, \text{(i)}$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $9$