Lesson 8 ~25 min Unit 1 · Index Laws +85 XP

Negative Indices — Introduction

A negative index makes the base a reciprocal: $a^{-n} = \dfrac{1}{a^n}$ for $a \ne 0$.

Today's hook: Extend the pattern: $2^3 = 8, 2^2 = 4, 2^1 = 2, 2^0 = 1$… what must $2^{-1}$ equal? What about $2^{-2}$?

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Think First

warm-up

Continue the table: $2^3 = 8, 2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = ?, 2^{-2} = ?$. Each step divides by $2$. What values appear after $2^0$?

Record in your workbook.

The Big Idea

+5 XP

A negative index means "reciprocal of the positive index". It does NOT make the value negative.

$a^{-n} = \dfrac{1}{a^n}$ for $a \ne 0$. The negative sign in the index flips the base into the denominator. It does not change the sign of the value — it just makes a fraction.

$a^{-n} = \dfrac{1}{a^n}$, $a \ne 0$

Reciprocate

$2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$. Move to denominator, drop the minus.

Not negative

$2^{-3} = \tfrac{1}{8}$ (positive), NOT $-8$.

Pattern works

$2^1 = 2, 2^0 = 1, 2^{-1} = \tfrac{1}{2}, 2^{-2} = \tfrac{1}{4}$.

What You'll Master

objectives

Know

$a^{-n} = \dfrac{1}{a^n}$ for $a \ne 0$
Negative index $\to$ reciprocal
Pattern proof and quotient-rule proof

Understand

Why the descending pattern continues into fractions
How the quotient rule produces a negative index naturally
That a negative index does NOT flip the sign of the value

Can Do

Rewrite $a^{-n}$ as $\dfrac{1}{a^n}$
Continue a halving pattern past $a^0$
Convert between negative-index and fraction form

Words You Need

vocabulary

Negative indexAn exponent less than $0$, e.g. $a^{-3}$.

ReciprocalThe "flipped" version: reciprocal of $a^n$ is $\dfrac{1}{a^n}$.

$a^{-n} = \dfrac{1}{a^n}$The negative-index rule (for $a \ne 0$).

DenominatorThe bottom of a fraction — where $a^n$ lands.

Halving patternGoing $2^1 \to 2^0 \to 2^{-1}$ divides by $2$ each step.

Sign of valueNegative index does NOT make the value negative.

Spot the Trap

heads-up

Wrong: "$2^{-3} = -8$" — thinking the negative sign attaches to the value.

Right: $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$. The result is a positive fraction.

Wrong: "$a^{-n} = -a^n$" — flipping the sign of the value.

Right: $a^{-n} = \dfrac{1}{a^n}$. Move the power to the denominator and remove the minus.

Why It Works: The Pattern

+5 XP

Extending the descending halving pattern beyond $2^0$ gives fractions, not negatives.

$2^3 = 8 \to 2^2 = 4 \to 2^1 = 2 \to 2^0 = 1 \to 2^{-1} = \tfrac{1}{2} \to 2^{-2} = \tfrac{1}{4}$. Each step divides by $2$. So $2^{-1}$ must equal $\tfrac{1}{2}$ — the pattern continues smoothly.

$2^{-1} = \dfrac{1}{2}, \; 2^{-2} = \dfrac{1}{4}$

Why It Works: Quotient Rule

+5 XP

The quotient rule, applied when the bottom index is bigger, produces a negative index naturally.

$a^3 \div a^5 = a^{3-5} = a^{-2}$. But cancelling factors: $\dfrac{a^3}{a^5} = \dfrac{1}{a^2}$. So $a^{-2}$ must equal $\dfrac{1}{a^2}$. The two methods agree only if $a^{-n} = \dfrac{1}{a^n}$.

$a^{-2} = \dfrac{1}{a^2}$

Watch Me Solve It · 3 examples

Watch Me Solve It · Rewrite using the rule

+15 XP per step

PROBLEM

Write $2^{-3}$ as a fraction with positive index, then evaluate.

1
Apply the rule
$2^{-3} = \dfrac{1}{2^3}$
Move to denominator, drop the minus.
2
Evaluate $2^3$
$2^3 = 8$
3
Final
$\dfrac{1}{8}$
Note: positive value, not $-8$.

Answer$\dfrac{1}{8}$

Watch Me Solve It · Variable base

+15 XP per step

PROBLEM

Rewrite $x^{-4}$ with a positive index.

1
Apply the rule
$x^{-4} = \dfrac{1}{x^4}$
Same rule, variable base.
2
State restriction
$x \ne 0$
Cannot divide by $0$.
3
Final
$\dfrac{1}{x^4}$

Answer$\dfrac{1}{x^4}$

Watch Me Solve It · Via quotient rule

+15 XP per step

PROBLEM

Use the quotient rule on $\dfrac{a^2}{a^5}$ to show it equals $a^{-3}$, and rewrite with positive index.

1
Apply quotient rule
$\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}$
2
Apply negative-index rule
$a^{-3} = \dfrac{1}{a^3}$
3
Verify by cancellation
$\dfrac{a^2}{a^5} = \dfrac{1}{a^3}$ ✓
Both methods agree.

Answer$\dfrac{1}{a^3}$

Common Pitfalls

heads-up

Negative answer

$2^{-3} \ne -8$. A negative index produces a fraction, not a negative number.

Fix: Use $a^{-n} = \dfrac{1}{a^n}$ — result is always positive (when $a > 0$).

Subtracting the index

$3^{-2} \ne 3 - 2 = 1$. A negative index does not mean "subtract".

Fix: Reciprocate. $3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$.

Forgetting $a \ne 0$

$0^{-n} = \dfrac{1}{0^n} = \dfrac{1}{0}$ — division by $0$, undefined.

Fix: Always check the base is non-zero before using the rule.

Copy Into Your Books

The rule

$a^{-n} = \dfrac{1}{a^n}$
$a \ne 0$
Reciprocal, not negative

Pattern proof

$2^1=2 \to 2^0=1$
$\to 2^{-1} = \tfrac{1}{2}$
$\to 2^{-2} = \tfrac{1}{4}$

Quotient proof

$\dfrac{a^3}{a^5} = a^{-2}$
Also $= \dfrac{1}{a^2}$
$\therefore a^{-2} = \dfrac{1}{a^2}$

Examples

$5^{-1} = \tfrac{1}{5}$
$3^{-2} = \tfrac{1}{9}$
$x^{-4} = \dfrac{1}{x^4}$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Negative-index drill

4 problems

Four drills to convert negative indices to fractions.

1 Rewrite $5^{-2}$ as a fraction.
$\dfrac{1}{5^2}$.$\dfrac{1}{25}$
2 Rewrite $y^{-6}$ with positive index.
Move to bottom.$\dfrac{1}{y^6}$
3 Evaluate $2^{-4}$.
$\dfrac{1}{2^4} = \dfrac{1}{16}$.$\dfrac{1}{16}$
4 Use quotient rule: simplify $\dfrac{a^2}{a^6}$ as a negative index, then as a fraction.
$a^{2-6} = a^{-4}$.$a^{-4} = \dfrac{1}{a^4}$

Complete in your workbook.

Quick Check · 5 questions

Evaluate $2^{-1}$.

+10 XP

Write $x^{-5}$ with a positive index.

+10 XP

Evaluate $3^{-2}$.

+10 XP

Simplify $\dfrac{a^3}{a^7}$ with a positive index ($a \ne 0$).

+10 XP

Evaluate $4^{-1}$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyEasy3 MARKS

Q6. Rewrite each with a positive index: (a) $7^{-2}$, (b) $a^{-5}$, (c) $10^{-3}$.

Answer in your workbook.

UnderstandMedium3 MARKS

Q7. Complete the table for powers of $3$: $3^3, 3^2, 3^1, 3^0, 3^{-1}, 3^{-2}$. State the pattern that connects each value to the next.

Answer in your workbook.

ReasonHard3 MARKS

Q8. Using both the quotient rule and direct cancellation, show that $\dfrac{a^2}{a^5} = \dfrac{1}{a^3}$. Hence justify $a^{-3} = \dfrac{1}{a^3}$.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $\dfrac{1}{2}$.

2. C — $\dfrac{1}{x^5}$.

3. A — $\dfrac{1}{9}$.

4. D — $\dfrac{1}{a^4}$.

5. B — $\dfrac{1}{4}$.

Show Your Working Model Answers

Q6 (3 marks): (a) $7^{-2} = \dfrac{1}{7^2} = \dfrac{1}{49}$ [1]; (b) $a^{-5} = \dfrac{1}{a^5}$, $a \ne 0$ [1]; (c) $10^{-3} = \dfrac{1}{10^3} = \dfrac{1}{1000}$ [1].

Q7 (3 marks): $3^3 = 27, 3^2 = 9, 3^1 = 3, 3^0 = 1, 3^{-1} = \tfrac{1}{3}, 3^{-2} = \tfrac{1}{9}$ [2]. Each value is the previous one divided by $3$ — a consistent halving (here, "thirding") pattern continuing through fractions [1].

Q8 (3 marks): Quotient rule: $\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}$ [1]. Direct cancellation: $\dfrac{a \cdot a}{a \cdot a \cdot a \cdot a \cdot a} = \dfrac{1}{a^3}$ [1]. Since both methods must give the same result, $a^{-3} = \dfrac{1}{a^3}$ [1].

Stretch Challenge · +25 XP, +10 coins

Negative-Index Detective

If $x^{-n} = \dfrac{1}{64}$ and $x = 2$, find the value of $n$.

Reveal solution

$2^{-n} = \dfrac{1}{2^n} = \dfrac{1}{64}$, so $2^n = 64 = 2^6$, giving $n = 6$.

Quick Review

Negative-index rule

$a^{-n} = \dfrac{1}{a^n}$

Restriction

$a \ne 0$

Pattern

$2^0, 2^{-1}, 2^{-2} = 1, \tfrac{1}{2}, \tfrac{1}{4}$

Not negative

Value stays positive (for $a>0$)

Quotient origin

$\dfrac{a^3}{a^5} = a^{-2}$

Move to bottom

Drop the minus sign

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