Fractional Indices [H]
Fractional Indices
The index laws from N7 extend naturally to fractional indices — powers written as fractions. The key insight is that a fractional index represents a root. Understanding why, rather than just memorising the rule, allows confident application even to unfamiliar expressions.
Why \( a^{1/n} = \sqrt[n]{a} \)
Apply the power-of-a-power law to \( a^{1/n} \) raised to the power \( n \):
\[ \left(a^{1/n}\right)^n = a^{\frac{1}{n} \times n} = a^1 = a \]So \( a^{1/n} \) is a number that, when raised to the power \( n \), gives \( a \). That is precisely the definition of the \( n \)-th root. Therefore \( a^{1/n} = \sqrt[n]{a} \).
The General Rule: \( a^{m/n} \)
For any fractional index \( \frac{m}{n} \) in its lowest terms:
\[ a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m} \]Both forms are equivalent. In examinations, the first — root first, then power — is strongly preferred because it keeps numbers smaller at every step.
Fractional Indices — Reference Table
| Index Form | Root Form | Evaluated Example |
|---|---|---|
| \( a^{1/2} \) | \( \sqrt{a} \) | \( 36^{1/2} = \sqrt{36} = 6 \) |
| \( a^{1/3} \) | \( \sqrt[3]{a} \) | \( 64^{1/3} = \sqrt[3]{64} = 4 \) |
| \( a^{1/n} \) | \( \sqrt[n]{a} \) | \( 32^{1/5} = \sqrt[5]{32} = 2 \) |
| \( a^{m/n} \) | \( \left(\sqrt[n]{a}\right)^m \) | \( 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4 \) |
| \( a^{-m/n} \) | \( \dfrac{1}{\left(\sqrt[n]{a}\right)^m} \) | \( 4^{-3/2} = \dfrac{1}{\left(\sqrt{4}\right)^3} = \dfrac{1}{8} \) |
Worked Examples
Evaluate \( 81^{1/4} \) and \( 1000^{1/3} \)
\[ 81^{1/4} = \sqrt[4]{81} \]Which number raised to the 4th power gives 81? Since \( 3^4 = 81 \): \( 81^{1/4} = 3 \).
\[ 1000^{1/3} = \sqrt[3]{1000} = 10 quad \text{since } 10^3 = 1000 \]Evaluate \( 27^{4/3} \) and \( 16^{3/4} \)
Root first, then power:
\[ 27^{4/3} = \left(\sqrt[3]{27}\right)^4 = 3^4 = 81 \] \[ 16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8 \]Both could be solved "power first, then root" but the numbers would be much larger at the intermediate step.
Evaluate \( 25^{-3/2} \) and \( 8^{-2/3} \)
The negative sign means take the reciprocal of the positive fractional index result:
\[ 25^{-3/2} = \frac{1}{25^{3/2}} = \frac{1}{\left(\sqrt{25}\right)^3} = \frac{1}{5^3} = \frac{1}{125} \] \[ 8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{\left(\sqrt[3]{8}\right)^2} = \frac{1}{2^2} = \frac{1}{4} \]Key Takeaways
- \( a^{1/n} = \sqrt[n]{a} \) — a unit fractional index is the \( n \)-th root, justified by the power-of-a-power law.
- \( a^{m/n} = \left(\sqrt[n]{a}\right)^m \) — take the root first, then apply the power. This keeps intermediate values small.
- \( a^{-m/n} = \frac{1}{a^{m/n}} \) — a negative fractional index combines the reciprocal rule with the fractional index rule.
- All index laws from integer indices continue to apply — combine them freely with fractional and negative indices.
- Check that the base is positive (for even roots) or handle negative bases carefully — odd roots of negative numbers are real and negative.