Sum of Indices of Real Number/Integers

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $r \in \R_{> 0}$ be a positive real number.

Let $n, m \in \Z$ be integers.

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


Then:

$r^{n + m} = r^n \times r^m$


Proof

From Sum of Indices of Real Number: Positive Integers, we have that:

$m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$


It remains to be shown that:

$\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$


The proof will proceed by induction on $m$.

As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.


For all $p \in \Z_{>0}$, let $\map P p$ be the proposition:

$\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$

that is:

$\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$


Basis for the Induction

$\map P 1$ is true, as follows:


When $n > 0$:

\(\ds r^{n - 1} \times r\) \(=\) \(\ds r^n\) Definition of Integer Power
\(\ds \leadsto \ \ \) \(\ds r^{n - 1}\) \(=\) \(\ds r^n \dfrac 1 r\) multiplying both sides by $\dfrac 1 r$
\(\ds \) \(=\) \(\ds r^n \times r^{-1}\) Real Number to Negative Power: Integer


When $n \le 0$:

\(\ds r^{n - 1}\) \(=\) \(\ds \dfrac {r^n} r\) Definition of Integer Power
\(\ds \) \(=\) \(\ds r^n \times r^{-1}\) Real Number to Negative Power: Integer


This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$\forall n \in \Z: r^{n - k} = r^n \times r^{- k}$


Then we need to show:

$\forall n \in \Z: r^{n - \paren {k + 1} } = r^n \times r^{-\paren {k + 1} }$


Induction Step

This is our induction step:


\(\ds r^n \times r^{-\paren {k + 1} }\) \(=\) \(\ds r^n \times \dfrac {r^{-k} } r\) Definition of Integer Power
\(\ds \) \(=\) \(\ds \paren {r^n \times r^{-k} } \times \dfrac 1 r\) Real Multiplication is Associative
\(\ds \) \(=\) \(\ds r^{n - k} \times \dfrac 1 r\) Induction Hypothesis
\(\ds \) \(=\) \(\ds r^{n - k} \times r^{-1}\) Real Number to Negative Power: Integer
\(\ds \) \(=\) \(\ds r^{n - k - 1}\) Basis for the Induction
\(\ds \) \(=\) \(\ds r^{n - \paren {k + 1} }\)

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


Therefore:

$\forall n, m \in \Z: r^{n + m} = r^n \times r^m$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Indices_of_Real_Number/Integers&oldid=406900"