# Sum of Indices of Real Number/Integers

## 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:

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

$\blacksquare$