Sum of Indices of Real Number/Integers

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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

Proof by induction on $m$:

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

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


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

$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$


Basis for the Induction

$P \left({1}\right)$ is true, as follows:


When $n \ge 0$:

$r^{n + 1} = r^n \times r = r^n \times r^1$

by definition of integer power.


When $n < 0$:

\(\displaystyle \frac {r^{n + 1} } r\) \(=\) \(\displaystyle r^n\) $\quad$ Definition of Integer Power $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle r^{n + 1}\) \(=\) \(\displaystyle r^n \times r\) $\quad$ multiplying both sides by $r$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^n \times r^1\) $\quad$ Definition of Integer Power $\quad$


This is our basis for the induction.


Induction Hypothesis

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

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


Induction Step

This is our induction step:


\(\displaystyle r^n \times r^{k + 1}\) \(=\) \(\displaystyle r^n \times \left({r^k \times r}\right)\) $\quad$ Definition of Integer Power $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({r^n \times r^k}\right) \times r\) $\quad$ Real Multiplication is Associative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^{n + k} \times r\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r^{n + k + 1}\) $\quad$ Definition of Integer Power $\quad$

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


Therefore:

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

$\Box$


It remains to be shown that:

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

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


For all $p \in \Z_{> 0}$, let $Q \left({p}\right)$ be the proposition:

$\forall n \in \Z: r^{n + \left({- p}\right)} = r^n \times r^{-p}$

that is:

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


Basis for the Induction (Negative Index)

$Q \left({1}\right)$ is true, as follows:


When $n > 0$:

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


When $n \le 0$:

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


This is our basis for the induction.


Induction Hypothesis (Negative Index)

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

$\forall n \in \Z: r^{n - \left({k + 1}\right)} = r^n \times r^{- \left({k + 1}\right)}$


Induction Step

This is our induction step:


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

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


Therefore:

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

$\blacksquare$


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