Negative Indices & Reciprocals

Unit 2 • Lesson 02 of 05

Negative Indices & Reciprocals

Grade 11 Mathematics • G.C.E. O/L Sri Lanka Local Syllabus

Learning Objectives

  • Understand the meaning of a negative index as the reciprocal of a positive power.
  • Rewrite algebraic expressions with negative indices using positive indices only.
  • Evaluate numerical expressions and fractions raised to negative powers.

Negative Index Rules

Using the quotient law, \(a^0 \div a^n = a^{0-n} = a^{-n}\). Since \(a^0 = 1\), we obtain \(\frac{1}{a^n} = a^{-n}\).

Negative Power to Positive
\[a^{-n} = \frac{1}{a^n} \quad (a \neq 0)\]
Denominator Negative Power
\[\frac{1}{a^{-n}} = a^n \quad (a \neq 0)\]
Fractional Base with Negative Exponent
\[\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n = \frac{b^n}{a^n}\]

Step-by-Step Worked Examples

Example 1: Evaluate \(\left(\frac{2}{3}\right)^{-3} \times 2^{-2}\)
  1. Invert the fraction \(\frac{2}{3}\) to change \(-3\) to \(+3\), and write \(2^{-2}\) as \(\frac{1}{2^2}\): \[\left(\frac{3}{2}\right)^3 \times \frac{1}{2^2}\]
  2. Expand the powers and multiply: \[\frac{27}{8} \times \frac{1}{4} = \frac{27}{32}\]
Example 2: Express with positive indices only: \(\frac{3x^{-4}y^2}{6x^2y^{-3}}\)
  1. Simplify the numerical coefficient \(\frac{3}{6} = \frac{1}{2}\) and shift terms with negative exponents across the fraction bar: \[\frac{y^2 \cdot y^3}{2 \cdot x^2 \cdot x^4}\]
  2. Apply the product law in the numerator and denominator: \[\frac{y^{2+3}}{2x^{2+4}} = \frac{y^5}{2x^6}\]

Practice Drill

1. Write with positive indices only: \(\frac{4x^{-2}y^3}{2x^3y^{-4}}\)
Show Step-by-Step Answer

Step 1: Divide coefficients (\(\frac{4}{2} = 2\)) and shift negative powers across the fraction line:
\(\frac{2y^3 \cdot y^4}{x^3 \cdot x^2}\)

Step 2: Apply the product law:
\(\mathbf{\frac{2y^7}{x^5}}\)

2. Find the value of: \(5^{-2} + \left(\frac{1}{5}\right)^{-1}\)
Show Step-by-Step Answer

Step 1: Convert each term to positive index form:
\(5^{-2} = \frac{1}{5^2} = \frac{1}{25}\) and \(\left(\frac{1}{5}\right)^{-1} = 5^1 = 5\)

Step 2: Add the values:
\(\frac{1}{25} + 5 = \mathbf{5\frac{1}{25}}\) (or \(\frac{126}{25}\))

3. Evaluate: \((2^{-1} \times 4^{-1})^{-2}\)
Show Step-by-Step Answer

Step 1: Simplify inside the bracket first:
\(2^{-1} \times 4^{-1} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\)

Step 2: Apply the outer power \(-2\):
\(\left(\frac{1}{8}\right)^{-2} = 8^2 = \mathbf{64}\)

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