Zero and Negative Exponents

The four laws of exponents

The four laws of exponents used in the preceding chapters are:

Law I may be stated more completely thus: x^a \cdot x^b \cdot x^c \cdots = x^{a+b+c+\cdots}. Law III includes the more general forms (x^ay^b)^c = x^{ac}y^{bc} and \left(\left((x^a)^b\right)^c\right)\cdots = x^{abc\cdots}.

When Laws I, II, and III were used in multiplication and division, a and b were positive integers, and in Law II, a was greater than b. In the work on radicals the meaning of an exponent was extended to fractional exponents, as defined by Law IV. Though the laws were first used only for positive integers and fractions, they hold for any rational values of a and b. This fact will be assumed without proof. We shall now explain the meaning that these laws force on a zero exponent and on negative exponents.

The zero exponent

From Law II, x^4 \div x^4 = x^{4-4} = x^0. But x^4 \div x^4 = \dfrac{x^4}{x^4} = 1. Therefore x^0 = 1. More generally, x^a \div x^a = x^0 and also = 1, so

x^0 = 1.

That is, any number (except zero) whose exponent is zero is equal to 1. Hence if x is not 0 or 1, 4^0 = \left(\tfrac23\right)^0 = (-6)^0 = (5x)^0 = (x^2 - 2x + 1)^0, for each equals 1.

Negative exponents

From Law II, a^3 \div a^5 = a^{3-5} = a^{-2}. Obviously a^3 \div a^5 = \dfrac{a^3}{a^5} = \dfrac{1}{a^2}. Therefore a^{-2} is another way of writing \dfrac{1}{a^2}. Then

4^{-3} = \frac{1}{4^3} = \frac{1}{64}, \qquad a^{-\frac23} = \frac{1}{a^{\frac23}} = \frac{1}{\sqrt[3]{a^2}}, \qquad 16^{-\frac34} = \frac{1}{16^{\frac34}} = \frac{1}{\sqrt[4]{16^3}} = \frac18.

In general terms,

x^{-a} = \frac{1}{x^a}, \qquad \frac{1}{x^{-a}} = x^a, \qquad bx^{-a} = \frac{b}{x^a}, \qquad \frac{b}{x^{-a}} = bx^a.

Therefore: any factor of the numerator of a fraction may be omitted from the numerator and written as a factor of the denominator, and vice versa, if the sign of the exponent of the factor be changed.

It follows that any expression involving negative exponents may be written as an expression involving only positive exponents. Negative exponents are not a mathematical necessity, but merely a convenience. The extension of the laws of exponents which brings with it the zero and the negative exponent is another illustration of the Law of Permanence of Form.

Example. Simplify \left(\tfrac23\right)^{-2} \cdot 8^{-\frac23}.

\left(\tfrac23\right)^{-2} = \left(\tfrac32\right)^{2} = \tfrac94, \qquad 8^{-\frac23} = \frac{1}{\sqrt[3]{8^2}} = \frac14, \qquad \text{so the product is } \tfrac{9}{16}.

A negative base under an odd root is fine: (-8)^{\frac13} = \sqrt[3]{-8} = -2, so (-8)^{-\frac23} = \dfrac{1}{(-2)^2} = \dfrac14.

Typing answers. The answers in this lesson are numbers: type 9/16, 0.008 or -2. The checker wants the result worked out, so 3^-2 is marked "equal, but work out the numbers".