Product of Indices of Real Number/Positive Integers
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Theorem
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
- $\paren {r^n}^m = r^{n m}$
Proof
Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
- $\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
$\map P 0$ is true, as this just says:
- $\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$
Basis for the Induction
$\map P 1$ is true, by definition of power to an integer:
- $\paren {r^n}^1 = r^n = r^{n \times 1}$
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\forall n \in \Z_{\ge 0}: \paren {r^n}^k = r^{n k}$
Then we need to show:
- $\forall n \in \Z_{\ge 0}: \paren {r^n}^{\paren {k + 1} } = r^{n \paren {k + 1} }$
Induction Step
This is our induction step:
\(\ds \paren {r^n}^{\paren {k + 1} }\) | \(=\) | \(\ds \paren {\paren {r^n}^k} \times r^n\) | Definition of Integer Power | |||||||||||
\(\ds \) | \(=\) | \(\ds r^{n k} \times r^n\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds r^{n k + n}\) | Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds r^{n \paren {k + 1} }\) | Integer Multiplication Distributes over Addition |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n, m \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 0.1$. Arithmetic: Example $1: \ \text{II}$