Zero and negative exponents

If $a^3 = a \cdot a \cdot a$ counts three factors, what could $a^0$ count? Or $a^{-2}$? The answers are not arbitrary , they are forced on us by the laws of exponents we already trust. Once you accept those laws, $a^0 = 1$ and $a^{-n} = 1/a^n$ become inevitable.

Concept

Why $a^0 = 1$ (for $a \ne 0$).

Apply the quotient law $\frac{a^m}{a^n} = a^{m-n}$ to the case $m = n$. The left side is $\frac{a^n}{a^n} = 1$. The right side is $a^{n-n} = a^0$. So $a^0$ must equal $1$.

A second way to see it: look at a descending pattern.

$2^4 = 16, \quad 2^3 = 8, \quad 2^2 = 4, \quad 2^1 = 2, \quad 2^0 = ?$

Each step divides by $2$. Continuing the pattern: $2^0 = 1$. The pattern continues past zero:

$2^{-1} = \tfrac{1}{2}, \quad 2^{-2} = \tfrac{1}{4}, \quad 2^{-3} = \tfrac{1}{8}, \ldots$

The negative-exponent rule.

$a^{-n} = \frac{1}{a^n} \quad (a \ne 0).$

So $7^{-2} = \frac{1}{49}$, $\big(\tfrac{3}{4}\big)^{-2} = \big(\tfrac{4}{3}\big)^2 = \tfrac{16}{9}$, and $10^{-3} = 0.001$.

A useful consequence: a negative exponent in the denominator becomes a positive exponent in the numerator, and vice versa. For example,

$\frac{1}{a^{-3}} = a^3, \qquad \frac{x^{-2}}{y^{-5}} = \frac{y^5}{x^2}.$

All laws still hold. Negative exponents play perfectly with Laws 1-5 from the previous topic. For instance,

$a^{-3} \cdot a^{5} = a^{-3+5} = a^{2}, \qquad (a^{-2})^{4} = a^{-8} = \frac{1}{a^{8}}.$

Note on $0^0$. The expression $0^0$ has no fixed value in elementary mathematics. Different conventions are used in different settings. For us, the rule $a^0 = 1$ requires $a \ne 0$.

Worked examples

Example 1. Evaluate $5^0 + 3^{-2}$.

• $5^0 = 1$, $3^{-2} = \frac{1}{9}$.
• Sum: $1 + \frac{1}{9} = \frac{10}{9}$.

Example 2. Simplify $\dfrac{2^{-3} \cdot 2^{5}}{2^{-1}}$.

• Top: $2^{-3+5} = 2^{2}$.
• Divide: $2^{2 - (-1)} = 2^{3} = 8$.

Example 3. Write $\left(\dfrac{3}{5}\right)^{-2}$ as a positive-power fraction.

• $\left(\dfrac{3}{5}\right)^{-2} = \left(\dfrac{5}{3}\right)^{2} = \dfrac{25}{9}$.

Example 4. Find $x$ if $5^{x} = \dfrac{1}{125}$.

• $\frac{1}{125} = 5^{-3}$.
• So $x = -3$.

Try it yourself

1. Evaluate $4^0$, $(-7)^0$, and $\left(\tfrac{2}{3}\right)^0$.
2. Compute $2^{-4}$ and $(-3)^{-2}$.
3. Simplify $\dfrac{6^{-3} \cdot 6^{5}}{6^{4}}$.
4. Write $\left(\tfrac{7}{2}\right)^{-3}$ as a positive-power fraction.
5. Find $x$ if $3^{x} = \tfrac{1}{81}$.
6. Find $x$ if $\left(\tfrac{2}{5}\right)^{x} = \tfrac{125}{8}$.
7. Express $0.000\,000\,1$ as a power of $10$.
8. Show that $a^{-m} \cdot a^{m} = 1$ for any $a \ne 0$.

Activity

Build a power table. On graph paper, list the powers of $2$ from $2^{-5}$ to $2^{5}$. In one column write the exponent, in another the value. Notice that the table is "symmetric in a multiplicative sense": $2^{n}$ and $2^{-n}$ are reciprocals of each other. Use this table to estimate things like $\frac{1}{32}$ as a power and convert it back to a fraction.