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Section 1.4 Exponentiation

In Definition 1.3.10, we introduced the concept of multiplication as repeated addition, and we build upon that idea here. We define exponentiation as repeated multiplication.

Subsection 1.4.1 Definition of exponentiation

Definition 1.4.1.

Let \(b\) be an integer and \(n\) be a positive integer. We define the \(n\)-th power of \(b\) to be the product of \(n\) copies of \(b\text{.}\)

We denote the \(n\)-th power of \(b\) by \(b^n\text{.}\) In short we also read \(b^n\) as “\(b\) to the \(n\)-th”.

Written as a formula our definition is

\begin{equation*} b^n:=\underbrace{b\cdot b\cdot\ldots \cdot b}_{n \text{ copies of }b} \text{.} \end{equation*}

We call \(b\) the base of \(b^n\) and \(n\) the exponent of \(b^n\text{.}\)

By Definition 1.4.1 an integer to the first power is the integer itself. That is for any integer \(b\) we have \(b^1=b\text{.}\)

Powers with the common exponents two and three are also read differently. Let \(b\) be an integer. We often read \(b^2\) as \(b\) squared (also see Definition 1.3.8) and \(b^3\) as “\(b\) cubed”.

In the video in Figure 1.4.2 the definition of exponentiation is followed by an overview of the topics in the remainder of this section.

Figure 1.4.2. Exponentiation by Matt Farmer and Stephen Steward

For examples of powers we show how they are read.

  1. \(3^2=9\) is read “\(3\) squared is equal to \(9\)” or “3 to the 2-nd is equal to 9”

  2. \(2^3=8\) is read “2 to the third is equal to 8”

  3. \(2^4=16\) is read “2 to the 4th is equal to 16”

For examples of powers we identify the base and exponent.

  1. In \(3^2\) the base is 3 and the exponent is 2.

  2. In \(2^3\) the base is 2 and the exponent is 3.

  3. In \(2^4\) the base is 2 and the exponent is 4.

We compute powers using the definition.

  1. \(\displaystyle 2^2=2\cdot 2=4\)

  2. \(\displaystyle 2^3=2\cdot 2\cdot 2=8\)

  3. \(\displaystyle 2^4=2\cdot 2\cdot 2\cdot 2=16\)

  4. \(\displaystyle 3^2=3\cdot 3=9\)

  5. \(\displaystyle 3^3=3\cdot 3\cdot 3=27\)

  6. \(\displaystyle (-2)^3 = (-2)\cdot(-2)\cdot(-2) = -8\)

  7. \(\displaystyle (-2)^4=(-2)\cdot(-2)\cdot(-2)\cdot(-2) =16\)

Subsection 1.4.2 Properties of exponentiation

We present properties of exponents and prove them using the idea that exponentiation is repeated multiplication.

Because \(b^m\) is the product of \(m\) copies of \(b\) and \(b^n\) is the product of \(n\) copies of \(b\) we have that \((b^m)\cdot (b^n)\) is the product of \(m+n\) copies of \(b\text{.}\) Writing the latter as a power we get

\begin{equation*} (b^m)\cdot (b^n)=b^{m+n}. \end{equation*}

Written as formula the argument above becomes:

\begin{equation*} (b^m)\cdot( b^n) = \underbrace{b\cdot b\cdot\ldots \cdot b}_{m \text{ copies of }} \cdot \underbrace{b\cdot b\cdot\ldots \cdot b}_{n \text{ copies of }b} = \underbrace{b\cdot b\cdot\ldots\cdot b}_{(m + n) \text{ copies of } b} = b^{(m+n)} \end{equation*}

Because \({(b^m)}^n\) is the product of \(n\) copies of \(b^m\) and \(b^m\) is the product of \(m\) copies of \(b\) we have that \({(b^m)}^n\) is the product of \(m\cdot n\) copies of \(b\text{.}\) Writing the latter as a power we get

\begin{equation*} (b^m)^n=b^{m\cdot n}. \end{equation*}

Written as formula the argument above becomes:

\begin{equation*} {(b^m)}^n = (\underbrace{b\cdot b\cdot \ldots \cdot b}_{m\text{ copies of }b})^n = \underbrace{\underbrace{b\cdot b\cdot \ldots \cdot b}_{{m \text{ copies of } b}}\cdot \ldots \cdot \underbrace{b\cdot b\cdot \ldots \cdot b}_{m \text{ copies of }b}}_{n \text{ copies of } \underbrace{b\cdot b\cdot \ldots \cdot b}_{m \text{ copies of } b}} = \underbrace{b\cdot b\cdot \ldots \cdot b}_{(m \cdot n) \text{ copies of } b} = b^{(m\cdot n)} \end{equation*}

We illustrate the proofs of the properties of exponentiation with examples.

  1. \((7^2)\cdot (7^3) = (7\cdot 7) \cdot (7\cdot7\cdot7)\)\(= 7\cdot 7 \cdot 7\cdot7\cdot7 = 7^5\)

  2. \((7^2)^3 = (7^2)\cdot(7^2)\cdot(7^2)\)\(= (7\cdot7)\cdot(7\cdot7)\cdot(7\cdot7)\)\(= 7\cdot7\cdot7\cdot7\cdot7\cdot7=7^6\)

Use the properties of exponentiation to simplify \(1256^3\cdot 1256^{11}\text{.}\)

Solution.

We apply Theorem 1.4.6 which states that for all integers \(b\) and for all non-negative integers \(m\) and \(n\) we have \(b^m\cdot b^n=b^{m+n}\text{.}\) With \(b=1256\) and \(m=3\) and \(n=11\) we get

\begin{equation*} 1256^3\cdot 1256^{11}= 1256^{3+11} = 1256^{14} \end{equation*}

Let \(d\) be an integer. Use the properties of exponentiation to simplify \(d^9\cdot d^7\cdot d^3\text{.}\)

Solution.

We apply Theorem 1.4.6 which states that for non-negative integers \(m\) and \(n\) we have \(d^m\cdot d^m=d^{m+n}\text{.}\) With \(m=9\) and \(n=7\) we get

\begin{equation*} d^9\cdot d^7\cdot d^3 = d^{9+7} \cdot d^3= d^{16}\cdot d^3\text{.} \end{equation*}

Applying the theorem again (this time with \(m=16\) and \(n=3\)) we obtain

\begin{equation*} d^{16} \cdot d^3 = d^{16+3} =d^{19}\text{.} \end{equation*}

We have found

\begin{equation*} d^9\cdot d^7\cdot d^3 = d^{9+7} \cdot d^3= d^{16}\cdot d^3 = d^{16+3} =d^{19}\text{.} \end{equation*}

Thus \(d^9\cdot d^7\cdot d^3\) simplifies to \(d^{19}\text{.}\)

Let \(d\) be an integer. Use the properties of exponentiation to simplify \({(d^3)}^5\text{.}\)

Solution.

Apply Theorem 1.4.7 we get

\begin{equation*} {(d^3)}^5 = d^{3\cdot 5} = d^{15} \end{equation*}

Another property of exponentiation follows from the commutative property of multiplication.

By the definition of powers \((a\cdot b)^n\) is the product of \(m\) copies of \(a\cdot b\text{.}\) Because of the commutative property of multiplication (see Example 1.3.19) we can reorder the product of \(m\) copies of \(a\cdot b\) copies such that we have the product of \(m\) copies of \(a\) times the product of \(m\) copies of \(b\text{.}\) Writing the latter as a power with base \(a\) times a power with base \(b\) we get

\begin{equation*} (a\cdot b)^n=(a^n)\cdot (b^n). \end{equation*}

Written as a formula the argument above becomes:

\begin{equation*} (a\cdot b)^n =\underbrace{(a\cdot b)\cdot (a\cdot b)\cdot \ldots \cdot (a\cdot b)}_{n \text{ copies of }a\cdot b } = \underbrace{a\cdot a\cdot \ldots \cdot a}_{n \text{ copies of }a} \cdot \underbrace{b\cdot b\cdot \ldots \cdot b}_{n \text{ copies of }b } = a^n \cdot b^n\text{,} \end{equation*}

where the middle equal sign holds by the commutative property of multiplication.

We illustrate the proof of the property with an example.

We have

\begin{align*} (5\cdot 7)^3 \amp = (5\cdot 7)\cdot (5 \cdot 7)\cdot (5 \cdot 7) \\ \amp = 5\cdot 7\cdot 5 \cdot 7\cdot 5 \cdot 7 \\ \amp = 5\cdot 5\cdot 5 \cdot 7\cdot 7 \cdot 7 = 5^3 \cdot 7^3 \end{align*}

To extend our definition of exponentiation to all non-negative integer exponents, we must determine how to define the 0th power of an integer. We first consider an example.

We try to find out what \((-6)^0\) should be. Our definition of \((-6)^0\) should be consistent with the properties of exponentiation in Theorem 1.4.6. In particular Theorem 1.4.6 which states that for all natural numbers \(a\) and \(c\) we have

\begin{equation*} (-6)^a\cdot (-6)^c = (-6)^{(a+c)} \end{equation*}

should also hold for \(a=0\text{.}\) We want

\begin{equation*} (-6)^0\cdot (-6)^c = (-6)^{(0+c)} \end{equation*}

to be true. As for all natural number c we have \(0+c = 0\) we get

\begin{equation*} (-6)^{(0+c)}=(-6)^c\text{.} \end{equation*}

So the equality we want to be true can be written as

\begin{equation*} (-6)^0\cdot (-6)^c = (-6)^c\text{.} \end{equation*}

That is we want \((-6)^0\) multiplied by \((-6)^c\) to be equal to \((-6)^c\text{.}\) The only number by which we can multiply a (non-zero) number and get the number as a result is \(1\text{.}\) So for our equation to be true we must set

\begin{equation*} (-6)^0:=1\text{.} \end{equation*}

The argument in Example 1.4.14 holds not only for \((-6)\text{,}\) but for all integers (except for 0). Let \(b\) be an integer. To extend our definition of exponentiation to all non-negative integer exponents, we must determine how to define \(b^0\text{.}\) Let \(n\) be a positive integer. If we want the property in Theorem 1.4.6 to include the possibility of an exponent of zero, we must have \(b^0 \cdot b^n = b^{0+n} = b^n\text{.}\) If \(b\ne 0\text{,}\) the only choice for \(b^0\) that works is \(b^0 = 1\text{.}\)

When the base is \(0\text{,}\) there are multiple possibilities for \(b^0\) that would keep the properties in Theorem 1.4.6 correct. One possibility is defining \(0^0 := 1\text{.}\) As it does not break anything, that it does not build a contradiction into the system of mathematics, and it matches what we have found for non-zero bases, we go with this choice.

Definition 1.4.15.

For all integers \(b\) we set \(b^0:=1\text{.}\)

We remark that some authors leave \(0^0\) undefined, while with our definition we have \(0^0=1\text{.}\)

In Checkpoint 1.4.16 compute some powers yourself.

We end out discussion of exponentiation with a table of powers (Table 1.4.17).

Table 1.4.17. Powers of integers. The rows contain the base \(b\) for \(0\le b\le 10\) and the columns contain the exponent \(n\) for \(0\le n\le 9\text{.}\)
\(b^n\) 0 1 2 3 4 5 6 7 8 9
0 1 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1
2 1 2 4 8 16 32 64 128 256 512
3 1 3 9 27 81 243 729 2 187 6 561 19 683
4 1 4 16 64 256 1 024 4 096 16 384 65 536 262 144
5 1 5 25 125 625 3 125 15 625 78 125 390 625 1 953 125
6 1 6 36 216 1 296 7 776 46 656 279 936 1 679 616 10 077 696
7 1 7 49 343 2 401 16 807 117 649 823 543 5 764 801 40 353 607
8 1 8 64 512 4 096 32 768 262 144 2 097 152 16777216 13 421 7728
9 1 9 81 729 6 561 59 049 531 441 4 782 969 43046721 387 420 489
10 1 10 100 1 000 10 000 100 000 1 000 000 10 000 000 100 000 000 1 000 000 000

Subsection 1.4.3 Square Roots

Definition 1.4.18.

Let \(b\) be a non-negative integer. By the square root of \(b\text{,}\) written as \(\sqrt{b}\text{,}\) we mean the non-negative number \(a\) such that \(a^2=b\text{.}\)

Some, but not all, square roots are integers. If the square root of \(b\) is an integer, we call \(b\) a perfect square.

Some examples of perfect squares are \(1 = 1^2\text{,}\) \(4 = 2^2\text{,}\) \(9 = 3^2\text{,}\) and \(16 = 4^2\text{.}\) Their square roots are integers: \(\sqrt{1} = 1\text{,}\) \(\sqrt{4} = 2\text{,}\) \(\sqrt{9} = 3\text{,}\) and \(\sqrt{16} = 4\text{.}\)

If a number is given in a convenient form, it is easy to find its square root.

We give some more square roots of perfect squares.

  1. \(\sqrt{25} = \sqrt{5^2} = 5\text{.}\)

  2. \(\sqrt{144} = \sqrt{12^2} = 12\text{.}\)

  3. \(\displaystyle \sqrt{169} = \sqrt{13^2}=13\)

  4. \(\displaystyle \sqrt{24372634816267643286^2}=24372634816267643286\)

When an integer is given as a square it is always easy to find its square root.

What is \(\displaystyle{\sqrt{77^2}}\text{?}\)

Solution.

The square root of \(77^2\) is \(77\text{.}\)

Even for integers that are too large for most calculators to handle, it is always easy to find its square root when the number is given as a square.

What is \(\displaystyle{\sqrt{667848628784687^2}}\text{?}\)

Solution.

The square root of \(667848628784687^2\) is \(667848628784687\text{.}\)

In Checkpoint 1.4.23 find the square root of a perfect square.

\(\sqrt{7709469804289804^ 2}\) is:

Answer.

\(7709469804289804\)