# CLEP College Algebra/Exponents

Imagine there was a natural number $a$ that was added to itself repeatedly over $b$ times. The integer $a$ is called the multiplicand (the thing being multiplied) while $b$ is the multiplier (the thing you are multiplying to the multiplicand). The result of this operation is the product. We refer to this specialized operation, $P(a,b)$, as multiplication, which is one of the definitions of multiplication:
    $P(a,b)=\underbrace {a+a+a+\cdots +a+a+a} _{b}{\overset {\underset {\mathrm {def} }}{=}}a\cdot b=a\times b=a(b)=ab$
This idea does not seem to work for integers. Let $b\in \mathbb {Z} ^{-}$ and $a\in \mathbb {Z} ^{+}$ be a part of, henceforth referred to, multiplication function $P(a,b)$. If $P(a,b)=\underbrace {a+a+a+\cdots +a+a+a} _{b}{\overset {\underset {\mathrm {def} }}{=}}ab$, then how can a number be added to itself negative times?
This dilemma requires us to work through a special case of $b\in \mathbb {Z} ^{+}$ and $a\in \mathbb {Z} ^{-}$. Let us say the following is true: $a=-p$ and $b=q$ whereby $p\in \mathbb {Z} ^{+}$ and $q\in \mathbb {Z} ^{+}$. Let us go ahead and input these two variables into our multiplication function:
    $P(a,b)=\underbrace {a+a+a+\cdots +a+a+a} _{b}=\underbrace {(-p)+(-p)+(-p)+\cdots +(-p)+(-p)+(-p)} _{q}{\overset {\underset {\mathrm {def} }}{=}}-pq$
The above definition tells us to simply subtract $a$ from itself $b$ times for $a\in \mathbb {Z} ^{-}$ while $b\in \mathbb {Z} ^{+}$. There is no problem since the number $a$ is subtracted $b$ times, despite it being negative. This is one of the ways we can say a negative times a positive number is negative (over the set of integers). However, this definition breaks down once we switch the roles of the number. Our next example shall attempt to prove a basic "theorem" most of us already know.
* Proof 1: Prove that multiplying $a\in \mathbb {Z} ^{+}$ to $b\in \mathbb {Z} ^{-}$, or $x(-y)$ for $a=x$ and $b=y$ given $x\in \mathbb {Z} ^{+}$ and $y\in \mathbb {Z} ^{+}$, is equivalent to $-xy$.
* First, we need to define what it means for a number to be negative. This is important to establish formally before we use it in our proof. Recall that one property of being a negative number is that it is the additive inverse of the magnitude of the number. That is, given a number $x\in \mathbb {Z} ^{+}$, the additive inverse is $-x$. Also, a number must equal itself; this is by definition true. This is the property of equality, and it is important in establishing our theorem.
* Let us say we are trying to solve the problem $x(-y)+xy$. By the property of equality,
* (2.0.0.1)
* $x(-y)+xy=x(-y)+xy$
* In (2.0.0.1), notice how $x$ is a factor of both terms on the right hand side. This means that we may rewrite the equation as the following:
* (2.0.0.2)
* $x(-y)+xy=x(-y+y)$
* Recall that we defined the negative number to be the inverse operation. By the additive inverse property, $-y+y=0$. Therefore, we know $x(-y+y)=x(0)=0$. We know any number times zero equals zero since we proved this was true in the previous chapter. Ergo, we have a new property:
* (2.0.0.3)
* $x(-y)+xy=0$
* By an extension of the property of equality, whatever is done to one side of an equation requires it must be done to the other. From this property of integers, we may define simply this:
* (2.0.0.4)
* $x(-y)=-xy$
* From this, we have proved a positive times a negative integer equals a negative integer. $\blacksquare$
A similar exercise can show that $-x=(-1)x$ and $(-x)(-y)=xy$. These two properties of multiplication allow us to keep the definition of multiplication the same for the set of integers. We will leave these two important "theorems" as non-trivial exercises for the reader.
Rational numbers start to break down the "repeated addition" idea of multiplication, since how can you add a number partially? Let us demonstrate the problem: Let $a\in \mathbb {Q}$ and $b\in \mathbb {Q}$. Allow four numbers to be defined as an integer that are relatively prime to the other — $m$, $n$, $p$, $q$. Let $a={\frac {m}{n}}$ and $b={\frac {p}{q}}$. If applied directly to our old definition, $P(a,b)$ would be asking us to a strange operation: add a fraction to itself "fractionally" many times. This is strange indeed. While we will not be discussing how we determined this formal definition below, keep in mind it took mathematicians more than simply writing this as true to be accepted into our lexicon, today:
    $P(a,b)=ab={\frac {m}{n}}\cdot {\frac {p}{q}}{\overset {\underset {\mathrm {def} }}{=}}{\frac {mn}{pq}}$.
The rational numbers definition does not contradict the old definition since each term therein is an integer, and thus, denominators for the terms are all $1$. This means $ab=mn=\underbrace {a+a+a+\cdots +a+a+a} _{b}=\underbrace {m+m+m+\cdots +m+m+m} _{n}$. We will skip defining real numbers since it involves introducing information that you will not learn unless you enter into some more pure math besides College Algebra or even Calculus.
This recursive addition defines a new operation, requiring us to define more and more specific cases. This idea of recursive addition can be extended to subtraction (which makes division) and then multiplication itself. Through recursive multiplication one reaches the idea of the exponent. As you have in the past referred back to properties of multiplication to help you operate expressions, you too will apply these ideas to learn about properties of exponents.

## Definitions of Exponents (Form One)

As we referred to in the introduction to this page, exponent is an operation of repeated multiplication of the base $a$ over a period of $p$ times. The expression is often read aloud as $a$ to the power of $p$. Allowing the function $e_{a}(p)$, in which $a\in \mathbb {N}$ and $p\in \mathbb {N}$, to represent the exponent operation as follows,
    $e_{a}(p)=\overbrace {a\cdot a\cdots a\cdot a} ^{p}{\overset {\underset {\mathrm {def} }}{=}}a^{p}$
will let us see what properties might exist for natural numbers. Over more clarifications, the power function, so to speak, will lead to rules.

### Properties of Exponents (Under Form One)

The function $e_{a}(0)=a^{0}$ for non-zero $a$ results in the following
    $e_{a}(0)=a^{0}=1$
The above result is true by definition although there is another way to think about this function. This type of thinking will be given later. For now, simply think this is true.
Given the two exponents $e_{a}(n)=a^{n}$ and $e_{a}(m)=a^{m}$, multiplying the two gives the following:
    $a^{n}\cdot a^{m}=a^{n+m}$
* Proof 2: $a^{n}\cdot a^{m}=a^{n+m}$ for $a,n,m\in \mathbb {N}$.
* This property of exponents can be proven by using the definition of exponent created herein. By definition,
* $e_{a}(n)=a^{n}=\overbrace {a\cdot a\cdots a\cdot a} ^{n}$
* $e_{a}(m)=a^{m}=\overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{m}$
* Therefore, in multiplying the two different functions,
* $a^{n}\cdot a^{m}=\left(\overbrace {a\cdot a\cdots a\cdot a} ^{n}\right)\cdot \left(\overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{m}\right)$
* Because there are $n$ terms being multiplied to $m$ terms, every term therein has a base $a$, and multiplication is associative, it is safe to infer there are $n+m$ terms of $a$ being multiplied together. As such,
* $a^{n}\cdot a^{m}=\underbrace {\overbrace {a\cdot a\cdots a\cdot a} ^{n}\cdot \overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{m}} _{n+m}=a^{n+m}$
Given the two exponents $e_{a}(n)=a^{n}$ and $e_{a}(m)=a^{m}$, dividing $a^{n}$ by $a^{m}$ for $n>m$ will give the following:
    ${\frac {e_{a}(n)}{e_{a}(m)}}=a^{n-m}$
* Proof 3: ${\frac {a^{n}}{a^{m}}}=a^{n-m}$ for $a,n,m\in \mathbb {N}$ and $n>m$.
* Keep in mind that because we are working with natural numbers, $m\ngtr n$. To keep everything in the natural numbers, this restriction must apply. Either way, let us continue by proving this property. By definition,
* $e_{a}(n)=a^{n}=\overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{n}$
* $e_{a}(m)=a^{m}=\overbrace {a\cdot a\cdots a\cdot a} ^{m}$
* In dividing the two different functions,
* ${\frac {e_{a}(n)}{e_{a}(m)}}={\frac {\overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{n}}{\underbrace {a\cdot a\cdots a\cdot a} _{m}}}$
* Because there are $n$ terms being divided by $m$ terms of $a$, every term therein has a base $a$, and multiplication of the inverse operation is associative, it is safe to infer there are $n-m$ terms of $a$ after the division. This can further be shown through the operation as follows: all red terms are canceled and what is left over is the terms that are not canceled, $n-m$.
* ${\frac {e_{a}(n)}{e_{a}(m)}}={\frac {{\color {red}\overbrace {a\cdot a\cdots a\cdot a} ^{m}}\cdot \overbrace {a\cdot a\cdots a} ^{n-m}}{\color {red}\underbrace {a\cdot a\cdots a\cdot a} _{m}}}$
* ${\frac {e_{a}(n)}{e_{a}(m)}}=\overbrace {a\cdot a\cdots a} ^{n-m}$
* As such,
* ${\frac {a^{n}}{a^{m}}}={\frac {\overbrace {a\cdot a\cdot a\cdots a\cdot a\cdot a} ^{n}}{\underbrace {a\cdot a\cdots a\cdot a} _{m}}}=\overbrace {a\cdot a\cdots a} ^{n-m}=a^{n-m}$
Given the exponent operation $e_{a}(n)=a^{n}$, for any $n,m\in \mathbb {N}$ such that $n,m>0$
    $\left(e_{a}(n)\right)^{m}=\left(a^{n}\right)^{m}=a^{n\cdot m}$.
For any operation that results in $n\cdot m=0$, $a\neq 0$.
* Proof 4: $\left(e_{a}(n)\right)^{m}=\left(a^{n}\right)^{m}=a^{n\cdot m}$ for $a,n,m\in \mathbb {N}$ and $n,m>0$.
* This second-to-last rule has a pretty involved intuition behind it. Be sure to follow along:
* $e_{a}(n)=a^{n}=\overbrace {a\cdot a\cdots a\cdot a} ^{n}$
* $e_{b}(m)=b^{m}=\overbrace {b\cdot b\cdot b\cdots b\cdot b\cdot b} ^{m}$
* If $b=a^{n}$, then $e_{a^{n}}(m)=\left(a^{n}\right)^{m}$. By the second bullet point above:
* $e_{a^{n}}(m)=\left(a^{n}\right)^{m}=\overbrace {a^{n}\cdot a^{n}\cdot a^{n}\cdots a^{n}\cdot a^{n}\cdot a^{n}} ^{m}=\overbrace {\left(\overbrace {a\cdot a\cdots a\cdot a} ^{n}\right)\cdots \left(\overbrace {a\cdot a\cdots a\cdot a} ^{n}\right)} ^{m}$
* Pay close attention to what this operation is telling us: there $n$ many constants $a$ multiplied to itself, for which this operation repeats $m$ times. Because $n$ factors of $a$ are multiplied again to itself as its operation $m$ many times, there $n\cdot m$ of those factors in total. As such:
* $\left(a^{n}\right)^{m}=\overbrace {a^{n}\cdot a^{n}\cdot a^{n}\cdots a^{n}\cdot a^{n}\cdot a^{n}} ^{m}=\overbrace {\left(\overbrace {a\cdot a\cdots a\cdot a} ^{n}\right)\cdots \left(\overbrace {a\cdot a\cdots a\cdot a} ^{n}\right)} ^{m}=a^{n\cdot m}$
Given the exponent operation $e_{ab}(n)=(ab)^{n}$, for any $a,b,n\in \mathbb {N}$
    $e_{ab}(n)=a^{n}b^{n}$.
* Proof 5: $e_{ab}(n)=a^{n}b^{n}$ for $a,b,n\in \mathbb {N}$.
* This will be the simplest proof in this chapter. All that will be needed is the definition of an exponent; specifically, the one that states the operation is simply repeated multiplication. We are given the following:
* $e_{ab}(n)=(ab)^{n}=\overbrace {ab\cdot ab\cdots ab\cdot ab} ^{n}$
* Because multiplication is associative,
* $\overbrace {ab\cdot ab\cdots ab\cdot ab} ^{n}=\overbrace {a\cdot a\cdots a\cdot a} ^{n}\cdot \overbrace {b\cdot b\cdots b\cdot b} ^{n}$
* From this, we learn that
* $(ab)^{n}=\overbrace {ab\cdot ab\cdots ab\cdot ab} ^{n}=\overbrace {a\cdot a\cdots a\cdot a} ^{n}\cdot \overbrace {b\cdot b\cdots b\cdot b} ^{n}=a^{n}b^{n}$