# Exponentiation

## Introduction

To understand how exponents arise, let's first review how we can build multiplication from addition. Let's say we wanted to capture the notion of "the amount equal to 3, ten times." We could write this out as $3 + 3 +3 + 3 +3 + 3 +3 + 3 +3 + 3$, but this gets burdensome quickly: if we wanted to capture the idea of "the amount equal to two hundred 3s." Thus, we define the multiplication function, usually denoted $\times$ or $\cdot$, such that $3\times 200=3+3+\ldots+3$ where there are 200 threes in the sum. This process (actually an inductive definition) defines the operation of "multiplication by positive integers." We can then extend the notion of multiplication to non-integers.

Similarly, the exponentiation is defined as the repetition of multiplication. For example, writing out $3\cdot 3\cdot 3\cdot 3\cdot 3$ can get boring fast, so we define the exponential function to express this in a much more compact form so that the preceeding example can be written as $3^5$ (read 3 to the 5th or 3 to the 5 power). What this means is that we are multiplying 3 by itself 5 times.

Before we proceed, we define 3 terms:

• exponent or power - In $4^6$, the exponent is 6; this tells us how many times we multiply the 4.
• base - In $10^9$, the base is 10; this tells us what we will be multiplying 9 times.

Our definition of exponentiation makes sense if the exponent is a positive integer. How about negative integers such as $2^{-4}$? How do we multiply 2 by itself -4 times!? Let's think about what a negative sign means a little more. When we append a negative sign to a number (say 4, for example), we are basically saying go four units in the opposite direction. We want to do the opposite of multiplication four times. In other words, we want to divide by 2 four times. Therefore, $2^{-4}=\frac 1{2^4}.$

It is also possible to extend the exponential function to all non-integers.

## Basic Properties

Listed below are some important properties of exponents:

1. $b^x\cdot b^y = b^{x+y}$
2. $b^{-x}=\frac 1{b^x}$
3. $\frac{b^x}{b^y}=b^{x-y}$
4. $(b^x)^y = b^{xy}$
5. $(ab)^x = a^x b^x$
6. $b^0 = 1$ (if $b \neq 0$. $0^0$ is undefined.)

Here are explanations of the properties listed above:

1. On both sides, we are multiplying b together x+y times. Thus, they are equivalent.
2. This is described in the previous section.
3. This results from using the previous two properties.
4. We are multiplying $b^x$ by itself y times, which is the same as multiplying b by itself xy times.
5. After multiplying ab by itself x times, we can collect a and b terms, thus establishing the property.
6. Hoping that property #1 will be true when $y=0$, we see that $b^x\cdot b^0$ should (hopefully) be equal to $b^x$. Thus, we define $b^0$ to be equal to $1$ in order to make this be true.

## Fractional exponents

If $b$ is a number and each of $x$ and $y$ is a positive integer, then, as explained above (property 1), $b^x b^y = b^{x+y}$. For example, $b^2 b^3 = (b\cdot b)(b\cdot b \cdot b) = b^5$.

How could we make sense of an expression like "$b^0$"? Well, hoping that property 1 will remain true when $y=0$, we see that $b^x b^0$ should (hopefully) be equal to $b^{x+0}=b^x$. For that reason, we define $b^0 = 1$, in order to make that be true. (And we only make this definition in the case where $b \neq 0$. We choose to leave $0^0$ undefined.)

We can make sense of an expression like "$b^{-5}$" in a similar way. Hoping that property 1 will remain true even if $x$ or $y$ is negative, we see that $b^5 b^{-5}$ should (hopefully) be equal to $b^{5 + (-5)} = b^0 = 1$. Thus, we define $b^{-5}$ to be $\frac{1}{b^5}$, in order to make this be true. Similarly, if $x$ is a positive integer, we define $b^{-x}$ to be $\frac{1}{b^x}$. (This depends on having $b \neq 0$. Otherwise we'd be dividing by $0$.)

How could we make sense of an expression like $b^{\frac{1}{2}}$? If you don't already know the answer, this is a good exercise; I recommend puzzling over it for awhile.

Answer: Hoping that property 1 will remain true when $x$ or $y$ is a fraction, we see that $b^{\frac{1}{2}} b^{\frac{1}{2}}$ should (hopefully) be equal to $b^{\frac{1}{2} + \frac{1}{2}} = b^1 = b$. Thus, we define $b^{\frac{1}{2}}$ to be $\sqrt{b}$, in order to make this be true.

For the time being, how to deal with other fractions in the exponent can be an exercise for the reader. Hint: What would $b^{\frac{1}{3}}$ be? What about $b^{\frac{1}{4}}$? Do you notice anything? Try to figure out $b^{\frac{2}{3}}$ -- how does it relate to $b^{\frac{1}{1}}$