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:

$\left({r^n}\right)^m = r^{n m}$


Proof

Proof by induction on $m$:

For all $m \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the proposition:

$\forall n \in \Z_{\ge 0}: \left({r^n}\right)^m = r^{n m}$


$P \left({0}\right)$ is true, as this just says:

$\left({r^n}\right)^0 = 1 = r^0 = r^{n \times 0}$


Basis for the Induction

$P \left({1}\right)$ is true, by definition of power to an integer:

$\left({r^n}\right)^1 = r^n = r^{n \times 1}$

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is our induction hypothesis:

$\forall n \in \Z_{\ge 0}: \left({r^n}\right)^k = r^{n k}$


Then we need to show:

$\forall n \in \Z_{\ge 0}: \left({r^n}\right)^{\left({k + 1}\right)} = r^{n \left({k + 1}\right)}$


Induction Step

This is our induction step:


\(\displaystyle \left({r^n}\right)^{\left({k + 1}\right)}\) \(=\) \(\displaystyle \left({\left({r^n}\right)^k}\right) \times r^n\) $\quad$ Definition of Integer Power $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^{n k} \times r^n\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^{n k + n}\) $\quad$ Sum of Indices $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^{n \left({k + 1}\right)}\) $\quad$ Integer Multiplication Distributes over Addition $\quad$

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n, m \in \Z_{\ge 0}: \left({r^n}\right)^m = r^{n m}$

$\blacksquare$


Sources