Product of Indices of Real Number/Positive Integers

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Product_of_Indices_of_Real_Number/Positive_Integers&oldid=497645"