Negative Indices
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Theory
A negative index means a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\), and any non-zero number to the power zero is \(1\). This Year 10A maths page covers evaluating with negative indices, simplifying algebraic terms with positive indices, writing sums of negative powers as a single fraction, and solving index equations whose answer is negative.
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: \(4^{-3}\).
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: \(0.2^{-2} = \left(\dfrac{1}{5}\right)^{-2} = 25\).
(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 \((x^2)^2 = x^4\).
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 \(\dfrac{1}{5x^3}\).
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: \(\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\).