Negative and Fractional Indices
Grade 6Negative indices — the reciprocal connection
A negative index means "take the reciprocal". That is it. a⁻ⁿ = 1 ÷ aⁿ. So 2⁻³ = 1 ÷ 8 = ⅛.
The reason this works follows from the division law: a² ÷ a⁵ = a^(2−5) = a⁻³. But written out long, a² ÷ a⁵ = 1 ÷ a³. So a⁻³ must equal 1/a³.
Fractional indices — roots and powers combined
A fractional index connects powers to roots:
a^(1/n) = the nth root of a
a^(m/n) = (nth root of a)^m
For a^(m/n): take the root first, then raise to the power. Roots make numbers smaller, so do that step first — it keeps the numbers manageable.
Worked example
EXAMPLE
Without a calculator, evaluate: (a) 8^(1/3) (b) 27^(2/3) (c) 4⁻²
(a) 8^(1/3) = cube root of 8 = 2.
(b) 27^(2/3): cube root of 27 = 3, then 3² = 9.
(c) 4⁻² = 1 ÷ 4² = 1 ÷ 16 = 1/16.
The zero index
Any base (except zero) raised to the power of 0 equals 1. So 7⁰ = 1, x⁰ = 1, and even (−5)⁰ = 1. This follows from the division law: a^n ÷ a^n = a⁰ = 1.
Try this one
Work out the value of each expression when \( x = 3 \).
(a) \( x^2 + 7 \) [1]
(b) \( 2x^3 - 10 \) [2]
Nova's hint:
Try to Solve it with Nova?
This is about substitution into expressions with indices.
Ask yourself: 'Do I need to apply the power to \( x \) before multiplying by the coefficient?'
For example, if \( x = 2 \), then \( 3x^2 = 3 \times 2^2 = 3 \times 4 = 12 \). Always deal with the power before multiplying.
