Zero and Negative Indices
Zero and Negative Indices in NSW Year 10 Core maths explains what \(a^0\) and \(a^{-n}\) mean (NESA outcomes MA5‑IND‑C‑01 and MA5‑IND‑P‑01). Any non-zero base to the power zero equals 1, and a negative index means the reciprocal of the positive power, so \(7^{-2} = \frac{1}{49}\).
You will extend a table of powers to see where the rules come from, evaluate numbers such as \(4^{-3}\) and \(\left(\frac{2}{5}\right)^{-2}\), write fractions as powers, and rewrite algebraic terms like \(3k^{-5}\) with positive indices. You will also combine negative indices with the other index laws and use negative powers of 10 in a measurement problem. It is part of Year 10 Core Indices and Surds.
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 maths page shows where these rules come from, how to evaluate negative powers as fractions and decimals, and how to use them with the other index laws.
The zero index: for any number \(a \ne 0\), \(a^0 = 1\). It follows from the division law: \(\dfrac{a^3}{a^3} = a^{3 - 3} = a^0\), and anything divided by itself is \(1\).
A negative index means the reciprocal of the positive power: \(a^{-n} = \dfrac{1}{a^n}\). For example, \(\dfrac{a^2}{a^5} = a^{-3}\), and cancelling gives \(\dfrac{1}{a^3}\).
A negative index does not make the number negative. \(7^{-1} = \dfrac{1}{7}\), which is positive and less than \(1\).
The index belongs only to what it is attached to. In \(4m^{-2}\) only \(m\) has the negative index, so \(4m^{-2} = \dfrac{4}{m^2}\).
All the index laws still work with zero and negative indices.
For \(a \ne 0\):
A fraction to a negative power:
How to evaluate or simplify
- Combine powers of the same base first with the index laws, adding or subtracting the indices (with their signs).
- Replace any zero power with \(1\).
- Rewrite each negative power as a reciprocal with a positive index. For a fraction, flip it.
- Evaluate as a fraction, or as a decimal if asked.
Finding a negative index
- Write the denominator as a power of the base: \(\dfrac{1}{64} = \dfrac{1}{4^3}\).
- Move it to the numerator with a negative index: \(4^{-3}\).
(i) \(9^0 + (4x)^0\)
(ii) \(6y^0\) and \((6y)^0\)
Anything non-zero to the power \(0\) is \(1\). Check what the index belongs to.
| \(9^0 + (4x)^0\) | \(=\) | \(1 + 1 = \boldsymbol{2}\) |
| \(6y^0\) | \(=\) | \(6 \times 1 = \boldsymbol{6}\) |
| \((6y)^0\) | \(=\) | \(\boldsymbol{1}\) |
In \(6y^0\) only the \(y\) has index \(0\). In \((6y)^0\) the whole bracket does.
(i) \(4^{-2}\)
(ii) \(5^{-3}\), as a fraction and as a decimal
Write each with a positive index: \(a^{-n} = \dfrac{1}{a^n}\).
| \(4^{-2}\) | \(=\) | \(\dfrac{1}{4^2} = \boldsymbol{\dfrac{1}{16}}\) |
| \(5^{-3}\) | \(=\) | \(\dfrac{1}{5^3} = \boldsymbol{\dfrac{1}{125}}\) |
| \(\) | \(=\) | \(\boldsymbol{0.008}\) |
A negative index gives a small positive number, never a negative one.
(i) \(\left(\dfrac{2}{3}\right)^{-2}\)
(ii) \(\left(\dfrac{x^2}{5}\right)^{-2}\)
Take the reciprocal, then raise the top and bottom to the positive power.
| \(\left(\dfrac{2}{3}\right)^{-2}\) | \(=\) | \(\left(\dfrac{3}{2}\right)^2 = \boldsymbol{\dfrac{9}{4}}\) |
| \(\left(\dfrac{x^2}{5}\right)^{-2}\) | \(=\) | \(\left(\dfrac{5}{x^2}\right)^2\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{25}{x^4}}\) |
In (ii), square both parts: \(5^2 = 25\) and \((x^2)^2 = x^4\).
(i) Evaluate \(10^4 \times 10^{-6}\).
(ii) Find \(n\) if \(3^n = \dfrac{1}{81}\).
(iii) Write \(6x^{-3}\) with a positive index.
The index laws work the same way with negative indices.
| \(10^4 \times 10^{-6}\) | \(=\) | \(10^{4 + (-6)} = 10^{-2}\) |
| \(\) | \(=\) | \(\dfrac{1}{100} = \boldsymbol{0.01}\) |
(ii) \(81 = 3^4\), so \(\dfrac{1}{81} = 3^{-4}\) and \(n = \boldsymbol{-4}\).
(iii) Only \(x\) has the negative index, so only \(x\) moves to the denominator: \(6x^{-3} = \boldsymbol{\dfrac{6}{x^3}}\).
Common pitfalls
Frequently asked questions
Why is any number to the power of 0 equal to 1?
Because \(\dfrac{a^n}{a^n} = a^0\) by the division law, and any non-zero number divided by itself is \(1\). The pattern \(8, 4, 2, 1\) for \(2^3, 2^2, 2^1, 2^0\) shows the same thing.
What does a negative index mean?
It means one over the positive power, the reciprocal: \(4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}\).
Does a negative index make a number negative?
No. A negative index makes a reciprocal, which is still positive for a positive base: \(7^{-2} = \dfrac{1}{49}\).
How do you write a negative power as a decimal?
Write it as a fraction with a positive index, then divide: \(5^{-3} = \dfrac{1}{125} = 0.008\).
What is a fraction to a negative power?
Flip the fraction and make the power positive: \(\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\).