Zero and Negative Indices
Any non-zero number to the power zero equals 1, and a negative index means a reciprocal: \(a^{-n}=\frac{1}{a^n}\). So \(\left(\frac{3}{5}\right)^{-2}=\frac{25}{9}\). In NSW Year 10 Advanced maths negative indices are used with pronumerals as well as numbers, and answers are written with positive indices (NESA outcome MA5‑IND‑P‑01).
You will evaluate numerical expressions with zero and negative indices without a calculator, simplify algebraic expressions such as \(\left(5p^{-2}q^{3}\right)^{-2}\), and write sums like \(2x^{-3}+x^{-1}\) as a single fraction. You will expand brackets, solve equations with reciprocals, and model a colony of insects that halves each week. Harder questions use the formula for resistors in parallel and simplify complex fractions. It is part of Year 10 Advanced Indices.
Theory
Any non-zero number to the power zero is \(1\), and a negative index means a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\). This Year 10 Advanced maths page covers evaluating numbers and fractions, simplifying algebraic terms with positive indices, single fractions, and expressions with negative indices on the top and bottom.
The index laws force two definitions. Since \(\dfrac{a^3}{a^3} = 1\) and also \(a^{3-3} = a^0\), we get \(a^0 = 1\) (for \(a \ne 0\)). Since \(\dfrac{a^2}{a^5} = \dfrac{1}{a^3}\) and also \(a^{2-5} = a^{-3}\), we get \(a^{-3} = \dfrac{1}{a^3}\).
So a negative index is not a negative number: it tells you to take the reciprocal. \(2^{-3} = \dfrac{1}{8}\), which is positive.
An index applies only to what it touches. \(4y^0 = 4 \times 1 = 4\), but \((4y)^0 = 1\). Likewise \(3x^{-2} = \dfrac{3}{x^2}\), while \((3x)^{-2} = \dfrac{1}{9x^2}\).
"Positive indices" in an answer means every power is written with a positive index: move factors with negative indices across the fraction line.
| Rule | In symbols | Example |
|---|---|---|
| Zero index | \(a^0 = 1\ (a \ne 0)\) | \((4y)^0 = 1\), but \(4y^0 = 4\) |
| Negative index | \(a^{-n} = \dfrac{1}{a^n}\) | \(6^{-2} = \dfrac{1}{36}\) |
| Moving across the line | \(\dfrac{1}{a^{-n}} = a^n\) | \(\dfrac{5}{k^{-3}} = 5k^3\) |
| Fraction base | \(\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}\) | \(\left(\dfrac{1}{3}\right)^{-2} = 9\) |
How to simplify with negative indices
- Expand brackets: raise each factor, multiplying the indices (a negative times a negative is positive).
- Combine each pronumeral with the multiplication and division laws.
- Rewrite any factor with a negative index on the other side of the fraction line with a positive index.
- Evaluate numbers last: \(2^{-3} = \dfrac{1}{8}\).
Other questions
- Decimals: change to a fraction first: \(0.25^{-1} = \left(\dfrac{1}{4}\right)^{-1} = 4\).
- Sums like \(x^{-1} + x^{-2}\): write as fractions and use a common denominator. Expanding \(x^{-1}(\ldots)\) uses the multiplication law on each term.
- Fractions of sums (complex fractions): multiply the top and bottom by the lowest common denominator; factorise if a difference of squares appears.
- Common factor: \(3^{-n} + 3^{-n-1} = 3^{-n}(1 + 3^{-1})\).
- Index equations: \(\dfrac{1}{27} = 3^{-3}\), then equate indices; the answer may be negative.
- Equations with reciprocals: \(x^{-1} = \dfrac{1}{x}\); solve for \(\dfrac{1}{x}\), then flip.
- Formulas: in \(P = P_0 \times 2^{-t}\), a negative \(t\) means a time in the past.
- Given \(b^{-x}\): \(b^{x} = \dfrac{1}{b^{-x}}\), then use the power-of-a-power law for other bases.
| \(\left(\dfrac{3}{4}\right)^{-2}\) | \(=\) | \(\left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}\) |
| \(5^0\) | \(=\) | \(1\) |
| \(3^{-1}\) | \(=\) | \(\dfrac{1}{3}\) |
| \(\text{total}\) | \(=\) | \(\dfrac{16}{9} + \dfrac{9}{9} - \dfrac{3}{9}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{22}{9}}\) |
A decimal base works the same way once it is a fraction: \(0.2^{-2} = \left(\dfrac{1}{5}\right)^{-2} = 25\).
Raise every factor to the power \(-2\) (multiply each index by \(-2\)), then move negative powers below the line.
| \(\left(-3a^{2}b^{-3}\right)^{-2}\) | \(=\) | \((-3)^{-2}\,a^{-4}\,b^{6}\) |
| \(\) | \(=\) | \(\dfrac{1}{9} \times \dfrac{1}{a^4} \times b^6\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{b^6}{9a^4}}\) |
\((-3)^{-2} = \dfrac{1}{(-3)^2} = \dfrac{1}{9}\): the even power makes it positive.
| \(5x^{-3} - 2x^{-1}\) | \(=\) | \(\dfrac{5}{x^3} - \dfrac{2}{x}\) |
| \(\) | \(=\) | \(\dfrac{5}{x^3} - \dfrac{2x^2}{x^3}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{5 - 2x^2}{x^3}}\) |
The lowest common denominator is \(x^3\), so multiply the second fraction by \(\dfrac{x^2}{x^2}\). Only the \(x\) carries the negative index: \(5x^{-3}\) is \(\dfrac{5}{x^3}\), not \(\dfrac{1}{5x^3}\).
Rewrite with fractions, then multiply the top and the bottom by the common denominator \(ab\) to clear the small fractions:
| \(\dfrac{\dfrac{1}{a} - \dfrac{1}{b}}{\dfrac{1}{a} + \dfrac{1}{b}}\) | \(=\) | \(\dfrac{\left(\dfrac{1}{a} - \dfrac{1}{b}\right) \times ab}{\left(\dfrac{1}{a} + \dfrac{1}{b}\right) \times ab}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{b - a}{b + a}}\) |
Check with \(a = 2\), \(b = 5\): \(\dfrac{0.5 - 0.2}{0.5 + 0.2} = \dfrac{3}{7}\) and \(\dfrac{5 - 2}{5 + 2} = \dfrac{3}{7}\) ✓
\(a^{-1} - b^{-1}\) is not \((a - b)^{-1}\).
Common pitfalls
Frequently asked questions
What does a negative index mean?
It means the reciprocal: \(a^{-n} = \dfrac{1}{a^n}\), so \(2^{-3} = \dfrac{1}{8}\).
Why is anything to the power of zero equal to 1?
Dividing a power by itself gives 1, and the index law gives a to the power zero. So a to the zero must be 1, for any number a that is not zero.
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 make its index positive: \(3a^{-2}b = \dfrac{3b}{a^2}\).
How do you raise a fraction to a negative power?
Flip the fraction and make the power positive: \(\left(\dfrac{3}{4}\right)^{-2} = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}\).
Is x to the minus 1 plus y to the minus 1 the same as (x plus y) to the minus 1?
No. \(x^{-1} + y^{-1} = \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x + y}{xy}\), which is not \(\dfrac{1}{x + y}\).