Negative Indices
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Theory
A negative index means a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\). This Year 9 maths page shows where the rule comes from, how to evaluate negative powers, how to write answers with positive indices, and why a fraction to a negative power turns upside down.
A negative index tells you to take the reciprocal. The reciprocal of a number is \(1\) divided by it, so the reciprocal of \(5\) is \(\dfrac{1}{5}\).
So \(a^{-n} = \dfrac{1}{a^n}\). For example \(3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}\). The answer is a fraction, not a negative number.
The index only moves what it touches. In \(5m^{-2}\) only the \(m\) goes under the fraction line: \(\dfrac{5}{m^2}\). In \((5m)^{-2}\) the whole bracket goes: \(\dfrac{1}{25m^2}\).
A fraction to a negative power turns upside down and the index becomes positive: \(\left(\dfrac{2}{7}\right)^{-1} = \dfrac{7}{2}\).
"Write with positive indices" means rewrite the answer so that no index is negative.
For any base \(a\) except \(0\):
Why. By the division law, \(\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}\). Cancelling gives \(\dfrac{1}{a^3}\). Both are the same thing.
A fraction to a negative power:
| Expression | What moves | Answer |
|---|---|---|
| \(y^{-4}\) | the \(y^4\) goes underneath | \(\dfrac{1}{y^4}\) |
| \(6y^{-4}\) | only the \(y^4\) moves | \(\dfrac{6}{y^4}\) |
| \(6xy^{-4}\) | only the \(y^4\) moves | \(\dfrac{6x}{y^4}\) |
| \((6y)^{-4}\) | the whole bracket moves | \(\dfrac{1}{(6y)^4} = \dfrac{1}{1296y^4}\) |
How to write an answer with positive indices
- Look for brackets. Decide what each negative index touches.
- Move that part to the other side of the fraction line and make its index positive. A fraction in a bracket just flips.
- Raise every factor in any bracket to its power.
- Tidy up: numbers that did not move stay where they are. Check no index is negative.
(i) \(5^{-2}\)
(ii) \(\left(\dfrac{3}{4}\right)^{-2}\)
(i) Write the reciprocal to make the index positive, then work out the power.
| \(5^{-2}\) | \(=\) | \(\dfrac{1}{5^2}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{1}{25}}\) |
(ii) A negative index on a fraction turns it upside down.
| \(\left(\dfrac{3}{4}\right)^{-2}\) | \(=\) | \(\left(\dfrac{4}{3}\right)^{2}\) |
| \(\) | \(=\) | \(\dfrac{4^2}{3^2}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{16}{9}}\) |
Both answers are positive. The minus in the index does not make the answer negative.
(i) \(3x^{-4}\)
(ii) \((2y)^{-3}\)
(iii) \((4k^3)^{-2}\)
(i) No bracket, so only the \(x\) moves. The \(3\) stays on top.
| \(3x^{-4}\) | \(=\) | \(3 \times \dfrac{1}{x^4}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3}{x^4}}\) |
(ii) The bracket moves underneath as a whole. Then expand it.
| \((2y)^{-3}\) | \(=\) | \(\dfrac{1}{(2y)^3}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{8y^3}}\) |
(iii) Move the bracket, then raise every factor. Multiply the indices for \(k^3\).
| \((4k^3)^{-2}\) | \(=\) | \(\dfrac{1}{(4k^3)^2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{16k^6}}\) |
(i) \(\left(\dfrac{a}{4}\right)^{-2}\)
(ii) \(\left(\dfrac{-2}{p}\right)^{-3}\)
(i) Flip the fraction and make the index positive. Then raise the top and the bottom.
| \(\left(\dfrac{a}{4}\right)^{-2}\) | \(=\) | \(\left(\dfrac{4}{a}\right)^{2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{16}{a^2}}\) |
(ii) Flip it the same way. The index \(3\) is odd, so the minus sign stays.
| \(\left(\dfrac{-2}{p}\right)^{-3}\) | \(=\) | \(\left(\dfrac{p}{-2}\right)^{3}\) |
| \(\) | \(=\) | \(\dfrac{p^3}{(-2)^3}\) |
| \(\) | \(=\) | \(\dfrac{p^3}{-8}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{p^3}{8}}\) |
Divide the numbers. Then subtract the indices, even though one is negative.
| \(8 \div 2\) | \(=\) | \(4\) |
| \(y^4 \div y^{-3}\) | \(=\) | \(y^{4-(-3)}\) |
| \(\) | \(=\) | \(y^{4+3} = y^7\) |
Subtracting a negative is the same as adding.
| \(\dfrac{8y^4}{2y^{-3}}\) | \(=\) | \(\boldsymbol{4y^7}\) |
Another way to see it: \(\dfrac{1}{y^{-3}} = y^3\), so the \(y^3\) moves up to the top and joins the \(y^4\).
Common pitfalls
Frequently asked questions
What does a negative index mean?
A negative index means take the reciprocal: \(a^{-n} = \dfrac{1}{a^n}\), so \(3^{-2} = \dfrac{1}{9}\).
Does a negative power make a number negative?
No. A negative power gives a fraction, not a negative number. \(5^{-2} = \dfrac{1}{25}\), which is positive.
How do you get rid of a negative index?
Move the part the index touches to the other side of the fraction line and make the index positive: \(3x^{-4} = \dfrac{3}{x^4}\).
What happens to a fraction with a negative power?
It turns upside down and the index becomes positive: \(\left(\dfrac{3}{4}\right)^{-2} = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}\).
Why is a to the power negative n equal to 1 over a to the n?
\(\dfrac{a^2}{a^5} = a^{-3}\) by the division law, and cancelling gives \(\dfrac{1}{a^3}\). Both are the same, so \(a^{-3} = \dfrac{1}{a^3}\).