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title: Some thoughts on exponentiation

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# Some thoughts on exponentiation

## Exponentiation when it is integral power

We were taught that according to the arithmetic rule, we want to, and should define recursive operations in a clear way. Recall that $$a\times b = a+a\times (b-1)$$ is a first principle of multiplication, we have also for the recursive formula of integral exponents that $$a^b = a\times a^{b-1}$$ where $a$ and $b$ are positive integers.

Later we found that by extending the addition rule to all integers, we defined the negative exponents by considering $a-b=a+(-b)$ such that $$x^{a+(-b)}=x^{a-b}=\frac{x^a}{x^b}$$ yielding the result when $a=b$, we have $$x^0=x^{a-a}=\frac{x^a}{x^a}=1$$ and consequently $$x^{-a}=x^{0-a}=\frac{x^0}{x^a}=\frac{1}{x^a}.$$ Therefore, the negative integral exponent and in general, all negative exponentiation means the reciprocal of its positive exponentiation.

One more important result is that we can join exponentiation as a product of exponents, i.e. $$(x^a)^b=x^{a+a+a+\dots+a}=x^{ab}$$ to make everything simple.

## Exponentiation with rational power

In terms of positive real numbers, all computations are explicitly defined. We know finding square-root of $x$ ($\sqrt{x}$) is easy when $x$ is positive. We ought to extend the result in terms of exponentiation, as we see $$\sqrt{x^2}=x$$ when $x$ is positive. Moreover, we found earlier in the integral exponentiation that $$(x^a)^b=x^{ab}$$ supports the identification that if $x^{1/2}=\sqrt{x}$ then $$\sqrt{x^2}=x=(x^{1/2})^2$$ naturally exists. In general, we ought to write $$x^{1/n}=\sqrt[n]{x}$$ as a simplified expression.

Therefore, we were taught that $$x^{p/q}=\sqrt[q]{x^p}$$ when $x$ is positive, $p$ and $q$ are integers.

However, the identification on n-th root is valid only when $x$ is a positive real number.

## Domain and range of exponentiation

To be rigorously discuss what is exponentiation, we shall use the concept of functions, identifying the writing of exponentiation as a two-variable function from domain $B$ (for base) and $P$ (for powers) to the resulting codomain $R$ (for redult, not to be confused with real number set $\mathbb{R}$). Let $f$ be such function and define it by $$f:B\times P\to R,\ (b,p)\mapsto b^p$$ for the following discussions.

Let us assume $P=1/\mathbb{Z}^{\times}$ for instant

### Exponentiation when $B=\mathbb{R}_{\geq 0}$ and $R=\mathbb{R}_{\geq 0}$

This is the initial intuition of exponentiation. Given $x>0$ and any integer $p$, we have $$x^{1/p}=\sqrt[p]{x}.$$

Formally, we shall not treat rational power as the same function as integral power. In fact, when we are considering the equation $$x^2=2,$$ one solution given by the traditional writing of surds is that $x=\sqrt{2}$, however, $x=-\sqrt{2}$ is also a case for solving the quadratic equation. We shall see with the properties of exponentiation, we have $$x=2^{1/2}=\pm \sqrt{2}$$ in full extent. Hence, our previous 'definition' is not enough to describe the charateristics of fractional indices. It is indeed a restriction under the given codomain $\mathbb{R}_{\geq 0}$.

### Exponentiation when $B=\mathbb{R}_{\geq 0}$ and $R=\mathbb{R}$

Notice the discussion above introduced the actual meaning of rational exponents. $$x^{1/2}=\pm\sqrt{x}$$

It seems quite natural to extend the writing so that $$x^{1/n}=\pm \sqrt[n]{x}$$ but the case of odd denominator is not the same as even power. Note that for the equation $$x^3=8,$$ we can explicitly find only one real solution to the equation, that is $x=2$. This comes up with the following case consideration: $$x^{1/n}=\cases{\pm\sqrt[n]{x},\ n\textrm{ is even}\\\sqrt[n]{x},n\textrm{ is odd}}$$ where $x$ is a non-negative real number.

One step forward, we don't need to care about the mechanism so much as we are theoretically developing what is rational exponents. Our work above has been focusing on solving equation and we are trying to express the **set of solution** to any exponentiated equations. Let us write down a new definition so that we can upgrade the tool later.

**Definition**. Let $n$ be a non-zero integer and $a>0$. The expression $$a^{1/n}$$ defines the solution set to the equation $x^n=a$.

In terms of real number solution, we can make the following examples.
*Example*. $2^{1/2}=\pm\sqrt{2}$.
*Example*. $3^{1/3}=\sqrt[3]{3}$.
*Example*. $5^{1/5}=\sqrt[5]{5}$.
*Example*. $10^{1/10}=\pm\sqrt[10]{10}$.

From this point of view, we know that $\sqrt[n]{x}\in x^{1/n}$ is a true statement for all $x>0$ and all $n\in\mathbb{Z}\setminus\{0\}$. We may call $\sqrt[n]{x}$ the **representative** of $x^{1/n}$.

### Exponentiation when $B=\mathbb{R}$ and $R=\mathbb{R}$

A critical point to the development is that we have no even-root for negative base. The history of mathematics provides us the number $i$ to tackle this problem such that $$i^2=-1.$$ Our discussion is now jumping to have $R=\mathbb{C}$ so that negative base can be included.

### Exponentiation when $B=\mathbb{R}$ and $R=\mathbb{C}$

Once we could use complex numbers, we have to acknowledge the famous formula for complex number - the Euler's formula for complex numbers.

We just write down the definitions and proceed to the representation of rational exponents.

**Definition**. $\displaystyle e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$
**Definition**. $\displaystyle \sin(z)=\sum_{n=0}^\infty \frac{z^{2n}}{(2n)!}$
**Definition**. $\displaystyle \cos(z)=\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}$

**De Moivre's Theorem**. For all $z:=x+yi\in\mathbb{C}$ where $x,y\in\mathbb{R}$, the modulus of $z$ is defined by $|z|=\sqrt{x^2+y^2}$, then there exists $\theta\in\mathbb{R}$ such that $$z=|z|(\cos{\theta}+i\sin{\theta}).$$ Such $\theta$ is called the argument of $z$, denoted by $\theta=\arg{z}$, and could be found by $\displaystyle \arctan(\frac{y}{x})$.

**Euler's formula**. For all $z\in\mathbb{C}$, $$z=|z|e^{i\arg{z}}.$$

We can therefore write all real numbers as $x=|x|e^{2\pi ki}$ for $k\in\mathbb{Z}$. By this, rational power over all real number base could be computed by $$x^{1/n}=\{\sqrt[n]{|x|}e^{2\pi ki/n}:k\in\mathbb{Z}\}.$$

With the fact that $e^{i(\theta+2\pi n)}=e^{i\theta}$, we can consider only the set $$x^{1/n}=\{\sqrt[n]{|x|}e^{2\pi ki/n}:k = 0, 1, 2, \dots, n-1\}.$$

## The $n$-th root of unity

We are nearly the end of the discussion. We have achieved the basic thought of 'what exponentiation means' - solving equations of any power. We also acknowledge that for all $z\in\mathbb{C}$, $$z=|z|e^{i\arg{z}}.$$

By the periodicity of sine and cosine, we may further write $$z=|z|e^{i\arg{z}+2\pi ni}.$$ Similar to the case of real number, $$z^{1/n}=\{\sqrt[n]{|z|}e^{(\arg{z}+2\pi ki)/n}:k = 0, 1, 2, \dots, n-1\}$$ defines the rational power over complex numbers. Notice that this is in general true for all complex numbers and always computable.

Finally, observe the number of elements in the set $z^{1/n}$ for all $z\in\mathbb{C}$ and $n\in\mathbb{N}$, we may conclude the following:

**Theorem**. For all equation in the form of $z^n=a$, there are $n$ solutions to the equation.

## Further exploration

Question 1: How many roots are there to a degree $n$ polynomial eqaution?
Reading: [The fundamental theorem of Algebra](/afOdW1PSSyWxT105_Dx7oQ)

Question 2: We have discussed about rational exponents, how about 
- irrational exponents? Do they exists and what is their properties? 
- How about Complex exponents? Any restrictions when we want to use complex index? 
- What are the geometric meaning of real and complex exponentiation?

Feel free to comment your thought.
