In Definition 1.31, we introduced the concept of multiplication as repeated addition, and we build upon that idea here. We define exponentiation as repeated multiplication.
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 1.29) and \(b^3\) as “\(b\) cubed”.
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
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
We apply Theorem 1.59 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
We apply Theorem 1.59 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
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 1.40) 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
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 1.59. In particular Theorem 1.59 which states that for all natural numbers \(a\) and \(c\) we have
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
The argument in Example 1.67 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 1.59 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 1.59 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.
Figure1.70.Powers of integers. The rows contain the base \(b\) for \(0\le b\le 10\) and the columns contain the exponent \(n\) for \(0\le n\le 9\text{.}\)
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 \(a^2=b\text{.}\)
Example1.72.Square roots of small perfect squares.
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 \(\sqrt{16} = 4\text{.}\)
