Negative Indices
A negative index means a reciprocal: \(a^{-n} = \frac{1}{a^n}\), so \(2^{-3} = \frac{1}{8}\). This NSW Year 9 Advanced maths subtopic uses negative indices with algebraic bases (NESA outcome MA5‑IND‑P‑01) and gives answers with positive indices.
You will simplify expressions such as \((3m^4)^{-2}\), \(\left(\frac{2m}{5n}\right)^{-2}\) and \(\frac{15x^{-2}y^3}{5x^4y^{-1}}\), and evaluate \(2^{-1} + 3^{-2}\) as a fraction. You will see how a negative index turns a fraction upside down, write \(x^{-1} + y^{-1}\) as one fraction, solve \(3^{2n+2} = \frac{1}{81}\), find unknown indices, and prove a result that uses negative indices. It is part of Year 9 Advanced Indices and Scientific Notation.
Theory
A negative index means a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\). This Year 9 Advanced maths page covers numbers and pronumerals, fractions to a negative power, answers with positive indices, equations and sums of reciprocals.
For any base \(a \neq 0\), \(a^{-n} = \dfrac{1}{a^n}\). A negative index tells you to take the reciprocal; it does not make the number negative. \(3^{-2} = \dfrac{1}{9}\), which is positive.
It fits the division law: \(\dfrac{a^2}{a^5}\) cancels to \(\dfrac{1}{a^3}\), and subtracting the indices gives \(a^{2-5} = a^{-3}\).
All the index laws still hold with negative indices. The usual final step is to rewrite the answer with positive indices: a factor with a negative index moves to the other side of the fraction line, and its index becomes positive.
A fraction to a negative power flips: \(\left(\dfrac{a}{b}\right)^{-n}\) \(= \left(\dfrac{b}{a}\right)^{n}\).
Using them with pronumerals, as in Examples 2 and 4, leads on to Year 11 Advanced.
| Expression | With positive indices |
|---|---|
| \(10^{-3}\) | \(\dfrac{1}{1000}\) |
| \(5k^{-4}\) | \(\dfrac{5}{k^4}\) |
| \((5k)^{-2}\) | \(\dfrac{1}{25k^2}\) |
| \(\dfrac{7}{w^{-2}}\) | \(7w^2\) |
| \(\left(\dfrac{c}{4}\right)^{-3}\) | \(\dfrac{64}{c^3}\) |
| \(2x^{-1} + y^{-2}\) | \(\dfrac{2y^2 + x}{xy^2}\) |
How to simplify with negative indices
- Brackets first: raise every factor in the bracket to the power, multiplying the indices. A negative outer index on a fraction flips the fraction.
- Combine: add indices when multiplying and subtract when dividing, treating negative indices like any other integers.
- Tidy up: move each factor with a negative index across the fraction line and make its index positive. Coefficients with no negative index stay where they are.
- Signs: a negative coefficient raised to an odd power stays negative; to an even power it becomes positive.
Other question types
- Evaluate numbers: rewrite each power as a fraction, then use a common denominator to add or subtract (Example 1).
- Equations: write a fraction such as \(\dfrac{1}{81}\) as a negative power of the base, then equate indices (Example 3).
- Find the indices: if an expression simplifies to \(x^ay^b\), the answers \(a\) and \(b\) may be negative. Leave them as they come.
- Single fraction: rewrite each term as a fraction and use a common denominator (Example 4).
- Showing an identity: multiplying the top and bottom of a fraction by the same expression, such as \(xy\), clears negative indices without changing its value.
(i) \(4^{-2} - 2^{-3}\)
(ii) \(\left(\dfrac{2}{5}\right)^{-2}\).
Rewrite each negative index as a reciprocal, then do the arithmetic.
(i)
| \(4^{-2} - 2^{-3}\) | \(=\) | \(\dfrac{1}{16} - \dfrac{1}{8}\) |
| \(\) | \(=\) | \(\dfrac{1}{16} - \dfrac{2}{16}\) |
| \(\) | \(=\) | \(\mathbf{-\dfrac{1}{16}}\) |
(ii) Flip the fraction and make the index positive:
| \(\left(\dfrac{2}{5}\right)^{-2}\) | \(=\) | \(\left(\dfrac{5}{2}\right)^{2}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{25}{4}}\) |
A negative index never makes the value negative by itself. In (i) the answer is negative only because \(\dfrac{1}{16} \lt \dfrac{1}{8}\).
(i) \(\left(\dfrac{-2a^3}{b}\right)^{-3}\)
(ii) \(\dfrac{12p^{-3}q^{2}}{8p^{2}q^{-3}}\).
(i) Flip the fraction, then cube every factor. The power \(3\) is odd, so the minus sign stays.
| \(\left(\dfrac{-2a^3}{b}\right)^{-3}\) | \(=\) | \(\left(\dfrac{b}{-2a^3}\right)^{3}\) |
| \(\) | \(=\) | \(\dfrac{b^3}{(-2)^3a^9}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{b^3}{8a^9}}\) |
(ii) Subtract indices, top minus bottom. Subtracting a negative adds.
| \(\dfrac{12p^{-3}q^{2}}{8p^{2}q^{-3}}\) | \(=\) | \(\dfrac{3}{2}\,p^{-3-2}\,q^{2-(-3)}\) |
| \(\) | \(=\) | \(\dfrac{3}{2}\,p^{-5}q^{5}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3q^5}{2p^5}}\) |
Write the right side as a power of \(2\). Since \(128 = 2^7\):
| \(\dfrac{1}{128}\) | \(=\) | \(\dfrac{1}{2^7} = 2^{-7}\) |
| \(2^{5-2k}\) | \(=\) | \(2^{-7}\) |
| \(5 - 2k\) | \(=\) | \(-7\) |
| \(-2k\) | \(=\) | \(-12\) |
| \(k\) | \(=\) | \(\mathbf{6}\) |
Check: \(2^{5-12} = 2^{-7} = \dfrac{1}{128}\).
(i) \(m^{-2} + m^{-1}\)
(ii) \(\dfrac{1}{2x^{-1} + 3y^{-1}}\).
Index laws do not join terms that are added. Write each term as a fraction and use a common denominator.
(i)
| \(m^{-2} + m^{-1}\) | \(=\) | \(\dfrac{1}{m^2} + \dfrac{1}{m}\) |
| \(\) | \(=\) | \(\dfrac{1}{m^2} + \dfrac{m}{m^2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1 + m}{m^2}}\) |
(ii) Simplify the bottom first:
| \(2x^{-1} + 3y^{-1}\) | \(=\) | \(\dfrac{2}{x} + \dfrac{3}{y} = \dfrac{2y + 3x}{xy}\) |
| \(\dfrac{1}{2x^{-1} + 3y^{-1}}\) | \(=\) | \(1 \div \dfrac{3x + 2y}{xy}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{xy}{3x + 2y}}\) |
Dividing by a fraction is multiplying by its reciprocal. Note the answer is not \(\dfrac{x}{2} + \dfrac{y}{3}\): the reciprocal of a sum is not the sum of the reciprocals.
Common pitfalls
Frequently asked questions
What does a negative index mean?
It means a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\). For example \(5^{-2} = \dfrac{1}{25}\).
Does a negative index make the answer negative?
No. A negative index gives a reciprocal, which is positive for a positive base. The sign depends on the base and the coefficient, not on the index being negative.
How do you write an answer with positive indices?
Move every factor that has a negative index to the other side of the fraction line and change its index to positive. Factors with positive indices stay where they are.
What is a fraction to a negative power?
Flip the fraction and make the index positive: \(\left(\dfrac{a}{b}\right)^{-n}\) \(= \left(\dfrac{b}{a}\right)^{n}\). Then raise the top and bottom.
Why does a to the power negative 1 equal 1 over a?
Divide \(a^2\) by \(a^3\). Cancelling gives \(\dfrac{1}{a}\), and subtracting the indices gives \(a^{-1}\). Both describe the same number.