Zero and negative exponents

So far every expo­nent has been a count­ing num­ber: $1$, $2$, $3$ and so on. But what could $5^0$ mean, or $5^{-2}$? You can­not mul­ti­ply $5$ by itself zero times, or minus two times. Yet these pow­ers have a clear mean­ing, and a sim­ple pat­tern shows what it must be. This les­son gives you laws 6, 7 and 8.

A pat­tern#

Start with $5^3$ and keep low­er­ing the expo­nent by $1$. Each step down divides by $5$.

$$\displaystyle \begin{aligned} 5 \times 5 \times 5 &= 5^3 = 125 \\ 5 \times 5 &= 5^2 = 25 \\ 5 &= 5^1 = 5 \\ 1 &= 5^0 \\ \frac{1}{5} &= 5^{-1} \\ \frac{1}{25} = \frac{1}{5^2} &= 5^{-2} \end{aligned}$$

Going down from $5^1 = 5$, divid­ing by $5$ gives $1$. So $5^0$ must be $1$. Divid­ing again gives $\displaystyle \frac{1}{5}$, so $\displaystyle 5^{-1} = \frac{1}{5}$. Once more gives $\displaystyle \frac{1}{25}$, so $\displaystyle 5^{-2} = \frac{1}{5^2}$.

Law 6: the zero expo­nent#

Law 6. $a^0 = 1$, where $a \neq 0$. Any num­ber except $0$ to the power $0$ is $1$.

Law 2 says the same thing. $\displaystyle \frac{a^3}{a^3}$ is $1$, because any­thing divided by itself is $1$. But by Law 2 it is also $a^{3-3} = a^0$. So $a^0 = 1$.

Note that $0^0$ is unde­fined: it has no value.

Exam­ple 1#

$$8^0 = 1 \qquad 10^0 = 1 \qquad (100000)^0 = 1 \qquad 2^0 \times 3^0 \times 4^0 = 1 \times 1 \times 1 = 1$$

The base can be a whole expres­sion. As long as it is not $0$, the answer is $1$.

$$(500 + 700)^0 = 1 \qquad (3^{-1} + 5^{-1} + 7^{-1})^0 = 1$$

Law 7: neg­a­tive expo­nents#

Law 7. $\displaystyle a^{-n} = \frac{1}{a^n}$, where $a \neq 0$. In par­tic­u­lar, $\displaystyle a^{-1} = \frac{1}{a}$.

A neg­a­tive expo­nent means "one over". It turns the power into a frac­tion. The law also works back­wards: $\displaystyle \frac{1}{a^n} = a^{-n}$.

Exam­ple 2#

$$\displaystyle 5^{-10} = \frac{1}{5^{10}} \qquad 8^{-5} = \frac{1}{8^5} \qquad \frac{1}{9^2} = 9^{-2}$$

Exam­ple 3: find­ing the value#

$$\displaystyle 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \qquad 5^{-3} = \frac{1}{5^3} = \frac{1}{125} \qquad 2^{-7} = \frac{1}{2^7} = \frac{1}{128}$$

Exam­ple 4#

Find $5^{-1} \times 2^{-1}$.

$$\displaystyle 5^{-1} \times 2^{-1} = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10}$$

A neg­a­tive expo­nent never makes the answer neg­a­tive. $2^{-2}$ is $\displaystyle \frac{1}{4}$, a pos­i­tive num­ber.

Law 8: a frac­tion with a neg­a­tive expo­nent#

Look at $\displaystyle \left(\frac{2}{3}\right)^{-2}$. Use Law 7, then Law 5.

$$\displaystyle \begin{aligned} \left(\frac{2}{3}\right)^{-2} &= \frac{1}{\left(\frac{2}{3}\right)^2} \\ &= \frac{1}{\;\frac{2^2}{3^2}\;} \\ &= \frac{3^2}{2^2} = \left(\frac{3}{2}\right)^2 \end{aligned}$$

The frac­tion turned upside down, and the minus sign in the expo­nent dis­ap­peared.

Law 8. $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$. Turn the frac­tion upside down and make the expo­nent pos­i­tive.

Exam­ple 5#

$$\displaystyle \left(\frac{4}{5}\right)^{-8} = \left(\frac{5}{4}\right)^8 \qquad \left(\frac{27}{28}\right)^{-29} = \left(\frac{28}{27}\right)^{29}$$

$$\displaystyle \left(\frac{3}{8}\right)^{-2} = \left(\frac{8}{3}\right)^2 = \frac{64}{9} \qquad \left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$

Exam­ple 6#

$$\displaystyle \left(\frac{6}{4}\right)^{-2} = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 \qquad \left(\frac{1}{2}\right)^{-1} = \left(\frac{2}{1}\right)^1 = 2$$

A whole num­ber can be writ­ten over $1$: $\displaystyle (15)^{-3} = \left(\frac{15}{1}\right)^{-3} = \left(\frac{1}{15}\right)^3$.

The other laws still work#

Laws 1 to 5 work with neg­a­tive expo­nents too. Just be care­ful with the signs.

Exam­ple 7#

$$\displaystyle \begin{aligned} 3^{-4} \times 3^7 &= 3^{-4+7} = 3^3 = 27 \\ \frac{6^{15}}{6^{-8}} &= 6^{15-(-8)} = 6^{15+8} = 6^{23} \\ (5^{-3})^{-2} &= 5^{(-3) \times (-2)} = 5^6 \\ (6^2)^{-2} &= 6^{2 \times (-2)} = 6^{-4} = \frac{1}{6^4} \\ 9^{-3} \times 7^{-3} &= (9 \times 7)^{-3} = 63^{-3} \end{aligned}$$

In the sec­ond line, tak­ing away $-8$ is the same as adding $8$.

Remem­ber. $a^0 = 1$. A neg­a­tive expo­nent means "one over": $\displaystyle a^{-n} = \frac{1}{a^n}$. A frac­tion with a neg­a­tive expo­nent turns upside down.

Com­mon mis­takes#

• Writ­ing $8^0 = 0$ or $8^0 = 8$. It is $1$.
• Think­ing a neg­a­tive expo­nent gives a neg­a­tive num­ber. $\displaystyle 2^{-3} = \frac{1}{8}$, not $-8$.
• Turn­ing a frac­tion upside down but keep­ing the minus sign. $\displaystyle \left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2$, not $\displaystyle \left(\frac{3}{2}\right)^{-2}$.
• Get­ting the sign wrong in $6^{15-(-8)}$. Minus a minus is a plus.

Prac­tice#

1. Find the value of $7^0$.
2. Find the value of $(a + 900)^0$, where $a + 900$ is not $0$.
3. Find the value of $2^{-5}$.
4. Write $\displaystyle \frac{1}{7^4}$ with a neg­a­tive expo­nent.
5. Find the value of $\displaystyle \left(\frac{1}{3}\right)^{-2}$.
6. Find the value of $\displaystyle \left(\frac{2}{5}\right)^{-3}$.
7. Write $\displaystyle \left(\frac{9}{10}\right)^{-5}$ with a pos­i­tive expo­nent.
8. Find the value of $3^{-4} \times 3^7$.
9. Sim­plify $2^{-3} \times 2^{-2}$ and find its value.
10. Sim­plify $\displaystyle \frac{15^{16}}{15^{-74}}$.

Answers#