Negative and Zero Exponents

Learning goals

Show why the quotient rule forces $a^0$ to equal one
Read $a^{-n}$ as the reciprocal of the matching positive power
Evaluate a zero or negative power to a definite number
Separate the sign of the exponent from the sign of the base
Extend the exponent laws to every integer exponent for a nonzero base

The gap left by the quotient rule

The quotient rule says that to divide powers of the same nonzero base you subtract the exponents: $\dfrac{a^m}{a^n} = a^{m-n}$ , for $a \neq 0$ . Last lesson we only allowed $m > n$ , so the answer always had a positive exponent and matched the “repeated multiplication” definition. But nothing stops you from writing a quotient where the exponents are equal or where the bottom is larger.

Take $\dfrac{2^3}{2^3}$ . The exponents are equal, so the rule produces $2^{3-3} = 2^0$ . Take $\dfrac{2^3}{2^5}$ . The bottom is larger, so the rule produces $2^{3-5} = 2^{-2}$ . Neither $2^0$ nor $2^{-2}$ means anything yet under the old definition.

We could declare these expressions off-limits, or we could define them so the quotient rule keeps holding. The second choice is far better, because each quotient above is also an ordinary fraction we can simplify directly. That simplification tells us the only value the new power is allowed to have.

Zero as an exponent

Start with the equal-exponent case, which forces the meaning of a zero exponent. The expression $\dfrac{2^3}{2^3}$ has the same thing on top and bottom, and any nonzero number over itself is $1$ . So the value is settled before any exponent rule is used:

\frac{2^3}{2^3} = \frac{8}{8} = 1.

Now read the same quotient through the quotient rule, which subtracts the exponents:

\frac{2^3}{2^3} = 2^{3-3} = 2^0.

The left sides are identical, so the right sides must be equal: $2^0 = 1$ . The same argument runs with any nonzero base, which gives the definition.

Why $a^0 = 1$ for every base $a \neq 0$ #

Take the quotient of $3^4$ with itself, and evaluate it both ways. First, directly: $3^4$ is $81$ , and $81$ divided by $81$ is $1$ . Second, through the quotient rule: subtracting the exponents makes that same quotient $3^{4-4} = 3^0$ .

\frac{3^4}{3^4} = \frac{81}{81} = 1 \qquad \text{and} \qquad \frac{3^4}{3^4} = 3^{4-4} = 3^0.

One quotient cannot have two different values, so $3^0 = 1$ follows from the arithmetic alone. Neither the $3$ nor the $4$ did any work of its own. The argument needed only a nonzero base and the same power above and below.

Take any nonzero number $a$ and any positive whole-number exponent $n$ . Look at the quotient of $a^n$ with itself, and evaluate it two ways.

First, directly. The numerator and denominator are the same number, and a nonzero number divided by itself is $1$ :

\frac{a^n}{a^n} = 1.

Second, through the quotient rule, which subtracts the exponents:

\frac{a^n}{a^n} = a^{n-n} = a^0.

Both expressions are the value of the one quotient $\dfrac{a^n}{a^n}$ , so they are equal to each other. That forces

a^0 = 1.

The base $a$ had to be nonzero for the very first step, since $\dfrac{0}{0}$ is not $1$ (it is not any number). So $a^0 = 1$ holds for every base except $0$ , and $0^0$ is left undefined.

There is a second way to see it that needs no algebra, just the staircase of powers. Each time the exponent drops by one, the value is divided by the base, because you remove one factor. Watch the powers of $2$ come down:

2^4 = 16, \quad 2^3 = 8, \quad 2^2 = 4, \quad 2^1 = 2.

Every step to the right divides by $2$ : from $16$ to $8$ , from $8$ to $4$ , from $4$ to $2$ . To stay on the pattern, the next step must divide $2$ by $2$ as well, and $2 \div 2 = 1$ . The slot below $2^1$ is $2^0$ , so $2^0 = 1$ , in perfect agreement with the algebra.

Check your understanding

What is $7^0$ ?

0

1

7

70

Negative exponents

Now push past zero. The case $\dfrac{2^3}{2^5}$ has a larger exponent on the bottom, and the quotient rule produces a negative exponent, $2^{3-5} = 2^{-2}$ . As before, the same quotient is also an ordinary fraction we can simplify by canceling, and that pins down the value.

Write top and bottom as factors and cancel the three $2$ ‘s they share:

\frac{2^3}{2^5} = \frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2} = \frac{\cancel{2} \times \cancel{2} \times \cancel{2}}{\cancel{2} \times \cancel{2} \times \cancel{2} \times 2 \times 2} = \frac{1}{2 \times 2} = \frac{1}{2^2}.

Three factors cancel from both, leaving nothing on top but $1$ and two factors of $2$ on the bottom. So the plain arithmetic gives $\dfrac{1}{2^2}$ , while the quotient rule gives $2^{-2}$ , and the two must agree:

2^{-2} = \frac{1}{2^2} = \frac{1}{4}.

A negative exponent is an instruction to take the reciprocal of the matching positive power. The reciprocal of a number is $1$ divided by that number. The minus sign in the exponent flips the power between the numerator and the denominator; it does nothing to the sign of the value. The base has to be something other than zero, because $\tfrac{1}{0^n}$ would ask you to divide by zero.

Why $a^{-n} = \dfrac{1}{a^n}$ for every base $a \neq 0$ #

Divide $4^2$ by $4^5$ , where the bottom carries three more factors of $4$ than the top does. Through the quotient rule, subtracting the exponents gives $4^{2-5} = 4^{-3}$ . Directly, expand and cancel: the two factors on top cancel two of the five below, leaving three factors of $4$ under a $1$ .

\frac{4^2}{4^5} = \frac{4 \times 4}{4 \times 4 \times 4 \times 4 \times 4} = \frac{1}{4 \times 4 \times 4} = \frac{1}{4^3} = \frac{1}{64}.

One quotient cannot have two different values, so $4^{-3} = \tfrac{1}{4^3}$ . The $4$ , the $2$ and the $3$ were not special. Every factor on top cancels because there are fewer of them, and the base only had to be nonzero.

Now take a nonzero base $a$ and a positive whole-number exponent $n$ . Build a quotient whose exponents differ by exactly $n$ , with the larger exponent on the bottom: divide $a^m$ by $a^{m+n}$ for any positive $m$ .

Through the quotient rule, subtract the exponents:

\frac{a^m}{a^{m+n}} = a^{m-(m+n)} = a^{-n}.

Directly, expand and cancel. The numerator has $m$ factors of $a$ and the denominator has $m + n$ factors of $a$ . So all $m$ factors on top cancel with $m$ of the factors on the bottom, leaving $1$ above and the remaining $n$ factors below:

\frac{a^m}{a^{m+n}} = \frac{\overbrace{a \times \cdots \times a}^{m}}{\underbrace{a \times \cdots \times a}_{m+n}} = \frac{1}{\underbrace{a \times \cdots \times a}_{n}} = \frac{1}{a^n}.

Both readings describe the same quotient, so they are equal:

a^{-n} = \frac{1}{a^n}.

The base must be nonzero so the canceling is honest (each $\tfrac{a}{a} = 1$ ) and so the reciprocal $\tfrac{1}{a^n}$ is not a division by zero. Reading the equation in reverse is just as useful: $\dfrac{1}{a^n} = a^{-n}$ , so a reciprocal power can always be rewritten with a negative exponent.

The staircase confirms this too. Keep dividing by the base as the exponent drops below zero, starting from $2^1 = 2$ and $2^0 = 1$ :

2^1 = 2, \quad 2^0 = 1, \quad 2^{-1} = \tfrac{1}{2}, \quad 2^{-2} = \tfrac{1}{4}, \quad 2^{-3} = \tfrac{1}{8}.

Each step right still divides by $2$ : $1 \div 2 = \tfrac{1}{2}$ , then $\tfrac{1}{2} \div 2 = \tfrac{1}{4}$ , then $\tfrac{1}{4} \div 2 = \tfrac{1}{8}$ . The values never turn negative; they turn into fractions, shrinking toward zero, which is exactly what reciprocals of growing powers do.

One unbroken staircase: each step right divides by the base. To the left of the marked exponent zero the values are whole-number powers shrinking toward one; to its right, where the exponent has gone negative, the values keep halving into smaller positive reciprocals. The values stay positive throughout; only the exponent goes negative.

Check your understanding

Write $3^{-2}$ as a fraction.

-9

-6

\dfrac{1}{9}

\dfrac{1}{6}

Evaluating zero and negative powers

To turn a negative power into a plain number, do two separate jobs in order. First take the reciprocal the minus sign calls for, then evaluate the positive power that remains. Keeping the steps apart is what prevents the usual sign slip.

Worked example 1 Evaluate $2^{-3}$ , $5^0$ , and $10^{-2}$

For $2^{-3}$ , the negative exponent flips it to a reciprocal, then evaluate the cube:

2^{-3} = \frac{1}{2^3} = \frac{1}{8}.

For $5^0$ , a zero exponent on a nonzero base is $1$ outright:

5^0 = 1.

For $10^{-2}$ , take the reciprocal of $10^2$ :

10^{-2} = \frac{1}{10^2} = \frac{1}{100} = 0.01.

When the base is a fraction, taking its reciprocal simply turns the fraction upside down.

Worked example 2 Evaluate $\left(\tfrac{1}{2}\right)^{-1}$ and $\left(\tfrac{2}{3}\right)^{-2}$

A power of $-1$ asks for one reciprocal and nothing more, so it just flips the base over.

For $\left(\tfrac{1}{2}\right)^{-1}$ , take the reciprocal of $\tfrac{1}{2}$ :

\left(\tfrac{1}{2}\right)^{-1} = \frac{1}{\left(\tfrac{1}{2}\right)^{1}} = \frac{1}{\tfrac{1}{2}} = 2.

Dividing $1$ by $\tfrac{1}{2}$ gives $2$ , so a positive base smaller than $1$ raised to a negative power comes out larger than $1$ . For $\left(\tfrac{2}{3}\right)^{-2}$ , flip the fraction first, then square it:

\left(\tfrac{2}{3}\right)^{-2} = \left(\tfrac{3}{2}\right)^{2} = \frac{3^2}{2^2} = \frac{9}{4}.

Flipping the base turns the $-2$ into a $+2$ , which is the quickest route for a fraction raised to a negative power.

Check your understanding

Evaluate $10^{-3}$ as a fraction.

-1000

\dfrac{1}{30}

\dfrac{1}{300}

\dfrac{1}{1000}

The sign of the base is a separate question

A negative exponent and a negative base are two different things that beginners often blur together. The exponent’s minus sign decides whether you take a reciprocal; the base’s minus sign decides whether the factors carry a negative.

Consider $(-2)^{-2}$ . The parentheses make $-2$ the base, and the $-2$ exponent calls for the reciprocal of $(-2)^2$ . First square the base, where two negatives cancel to a positive, then take the reciprocal:

(-2)^{-2} = \frac{1}{(-2)^2} = \frac{1}{(-2)\times(-2)} = \frac{1}{4}.

The answer is positive $\tfrac{1}{4}$ , even though both the base and the exponent wore a minus sign. An odd power keeps the base’s sign, so $(-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}$ . That value is negative, but it is negative because the base is negative and the power is odd, never because the exponent is negative.

Two minus signs, two jobs

The minus sign in the exponent flips the power into its reciprocal: $a^{-n} = \tfrac{1}{a^n}$ . The minus sign on the base is what can make a value negative, and only when the exponent is odd. So $2^{-2} = \tfrac{1}{4}$ (positive), $(-2)^{2} = 4$ (positive), $(-2)^{-2} = \tfrac{1}{4}$ (positive), and $(-2)^{-3} = -\tfrac{1}{8}$ (negative, from the odd power of a negative base). Parentheses still matter just as they did for positive powers. Without them the leading minus stays outside the power, so $-2^{-2} = -(2^{-2}) = -\tfrac{1}{4}$ , the opposite sign from $(-2)^{-2} = \tfrac{1}{4}$ .

Check your understanding

Evaluate $(-3)^{-2}$ .

\dfrac{1}{9}

-\dfrac{1}{9}

-9

9

Every law still holds

The whole reason these definitions were chosen is that they keep the laws of exponents intact. Those laws now hold for a nonzero base and every integer exponent, rather than only the positive ones. Both $a^0$ and $a^{-n}$ were defined to be exactly the values the quotient rule forces.

Worked example 3 Show $2^3 \times 2^{-5} = \tfrac{1}{4}$ two ways

First use the product rule, adding the exponents straight through, including the negative one:

2^3 \times 2^{-5} = 2^{3 + (-5)} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}.

Now check it the long way, with no negative-exponent rule at all. Rewrite $2^{-5}$ as its reciprocal first, then multiply the fractions:

2^3 \times 2^{-5} = 8 \times \frac{1}{2^5} = \frac{8}{32} = \frac{1}{4}.

Both routes land on $\tfrac{1}{4}$ . The product rule still says “add the exponents,” and adding $3 + (-5) = -2$ gives the same answer as grinding through the arithmetic.

The quotient rule now works with no restriction on which exponent is larger, because a negative result is a legitimate power. So $\dfrac{3^2}{3^6} = 3^{2-6} = 3^{-4} = \dfrac{1}{3^4} = \dfrac{1}{81}$ , with the rule applied in one clean step instead of the side condition $m > n$ we needed before.

Worked example 4 Simplify $\dfrac{a^2}{a^5}$ using only a positive exponent (with $a \neq 0$ )

The base is shared, so subtract the exponents, even though the bottom one is larger:

\frac{a^2}{a^5} = a^{2-5} = a^{-3}.

That is already a single power, but a negative exponent is usually rewritten with a positive one by moving the power to the denominator:

a^{-3} = \frac{1}{a^3}.

So $\dfrac{a^2}{a^5} = \dfrac{1}{a^3}$ . Reading it directly agrees: two factors of $a$ over five factors cancel two pairs, leaving three factors of $a$ on the bottom.

Check your understanding

Write $5^2 \times 5^{-4}$ using only a positive exponent.

5^2

\dfrac{1}{5^2}

\dfrac{1}{5^6}

\dfrac{1}{5^8}

Common mistakes

Reading $a^0$ as $0$ . A zero exponent gives $1$ , not $0$ : $9^0 = 1$ and $1000^0 = 1$ . “No factors of the base” leaves the multiplication at its starting value, which is $1$ . Only $0^0$ is undefined.
Thinking a negative exponent makes the value negative. $2^{-3} = \tfrac{1}{8}$ , a positive number, not $-8$ and not $-\tfrac{1}{8}$ . The minus sign in the exponent flips the power into a reciprocal; it never touches the sign of the result.
Putting the wrong piece in the denominator. $a^{-n} = \tfrac{1}{a^n}$ , so only the power moves below the line. For example $2^{-3} = \tfrac{1}{2^3} = \tfrac{1}{8}$ , not $\tfrac{1}{2 \times 3}$ or $\tfrac{3}{2}$ . Evaluate the positive power, then put it under $1$ .
Confusing a negative base with a negative exponent. They are separate decisions: $(-2)^{-2} = \tfrac{1}{4}$ is positive, because squaring a negative base is positive and the reciprocal keeps it positive. A negative result needs a negative base and an odd exponent, as in $(-2)^{-3} = -\tfrac{1}{8}$ .
Subtracting exponents the wrong way past zero. In $\tfrac{3^2}{3^6}$ the answer is $3^{2-6} = 3^{-4}$ , not $3^{6-2} = 3^4$ . Always subtract the denominator’s exponent from the numerator’s, and let the result be negative when it is.

Key takeaways

$a^0 = 1$ for every base $a \neq 0$ , because $\tfrac{a^n}{a^n}$ is $1$ directly while the quotient rule makes the same quotient $a^0$ . The value $0^0$ is left undefined.
$a^{-n} = \dfrac{1}{a^n}$ for every base $a \neq 0$ : a negative exponent means the reciprocal of the matching positive power.
A negative exponent never makes a value negative. It moves the power between numerator and denominator; the sign of the result is decided only by the base and whether the exponent is odd.
For a fraction base with $a \neq 0$ and $b \neq 0$ , a negative exponent flips the fraction: $\left(\tfrac{a}{b}\right)^{-n} = \left(\tfrac{b}{a}\right)^{n}$ , so $\left(\tfrac{2}{3}\right)^{-2} = \tfrac{9}{4}$ .
With $a^0$ and $a^{-n}$ defined this way, every law of exponents holds for a nonzero base and all integer exponents. So $2^3 \times 2^{-5} = 2^{-2} = \tfrac{1}{4}$ , and the quotient rule no longer needs $m > n$ .

Negative and Zero Exponents

Learning goals

The gap left by the quotient rule

Zero as an exponent

Negative exponents

Evaluating zero and negative powers

The sign of the base is a separate question

Every law still holds

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice