Mathematics • Year 9 • Unit 1 • Lesson 8
Negative Indices — Introduction
Build fluency with the negative-index rule $a^{-n} = \dfrac{1}{a^n}$ (for $a \ne 0$). One worked example, one guided example with blanks, then graduated practice.
1. I do — fully worked example
Read every step. Each line has a short reason on the right so you can see why, not just what.
Problem. Write $2^{-3}$ as a fraction with a positive index, then evaluate.
Step 1 — Spot the negative index.
The index is $-3$ — that means "reciprocate". Use $a^{-n} = \dfrac{1}{a^n}$.
Reason: a negative index does NOT mean the answer is negative — it moves the base into the denominator.
Step 2 — Reciprocate the base.
$2^{-3} = \dfrac{1}{2^3}$
Reason: move the base to the denominator and drop the minus sign in the index.
Step 3 — Evaluate $2^3$.
$2^3 = 2 \times 2 \times 2 = 8$
Reason: we should evaluate any numerical power that's small enough to compute.
Step 4 — Final answer.
$2^{-3} = \dfrac{1}{8}$
Reason: a POSITIVE fraction — not $-8$. Negative INDEX, positive VALUE.
Answer: $\mathbf{\dfrac{1}{8}}$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Write $5^{-2}$ as a fraction with a positive index, then evaluate.
Step 1 — Spot the rule: the index is __________, so use the __________________-index rule $a^{-n} = \dfrac{1}{a^n}$.
Step 2 — Reciprocate the base:
$5^{-2} = \dfrac{1}{5^{\_\_\,}}$
Step 3 — Evaluate $5^2$:
$5^2 = 5 \times \_\_ = \_\_\_$
Step 4 — Final answer:
$5^{-2} = \dfrac{1}{\_\_\_}$
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (rewrite or evaluate one term). The middle two are standard (variables, restrictions). The last two are extension (quotient rule giving negative index).
Foundation — apply the rule directly
3.1 Rewrite $7^{-2}$ as a fraction with a positive index. Do not evaluate. 1 mark
3.2 Evaluate $3^{-2}$. 1 mark
3.3 Evaluate $4^{-1}$. 1 mark
3.4 Evaluate $2^{-4}$. 1 mark
Standard — variable bases and restrictions
3.5 Rewrite $x^{-4}$ with a positive index. State any restriction on $x$. 2 marks
3.6 Rewrite $y^{-6}$ with a positive index. 2 marks
Extension — quotient rule into a negative index
3.7 Use the quotient rule to simplify $\dfrac{a^2}{a^6}$ as a power of $a$ with a negative index, then rewrite that as a fraction with a positive index. 3 marks
3.8 A friend writes "$3^{-2} = 3 - 2 = 1$". In one or two sentences, say what mistake they have made and give the correct value of $3^{-2}$. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $5^{-2}$)
Step 1: index is $-2$ (or "negative two"); use the negative-index rule.
Step 2: $5^{-2} = \dfrac{1}{5^{\mathbf{2}}}$.
Step 3: $5^2 = 5 \times \mathbf{5} = \mathbf{25}$.
Step 4: $5^{-2} = \dfrac{1}{\mathbf{25}}$.
3.1 — $7^{-2}$
$a^{-n} = \dfrac{1}{a^n}$, so $7^{-2} = \mathbf{\dfrac{1}{7^2}}$.
3.2 — $3^{-2}$
$3^{-2} = \dfrac{1}{3^2} = \mathbf{\dfrac{1}{9}}$.
3.3 — $4^{-1}$
$4^{-1} = \dfrac{1}{4^1} = \mathbf{\dfrac{1}{4}}$.
3.4 — $2^{-4}$
$2^{-4} = \dfrac{1}{2^4} = \mathbf{\dfrac{1}{16}}$.
3.5 — $x^{-4}$
$x^{-4} = \mathbf{\dfrac{1}{x^4}}$, with $x \ne 0$ (otherwise the denominator is $0$, which is undefined).
3.6 — $y^{-6}$
$y^{-6} = \mathbf{\dfrac{1}{y^6}}$ (with $y \ne 0$).
3.7 — $\dfrac{a^2}{a^6}$
Quotient rule: $\dfrac{a^2}{a^6} = a^{2-6} = a^{-4}$.
Negative-index rule: $a^{-4} = \mathbf{\dfrac{1}{a^4}}$ (with $a \ne 0$).
Check by cancelling: $\dfrac{a^2}{a^6} = \dfrac{1}{a^4}$. Both methods agree.
3.8 — Friend's mistake on $3^{-2}$
The friend has treated the negative index as a SUBTRACTION ($3 - 2$). That's wrong — a negative index is the reciprocal of the positive index, not subtraction.
Correct: $3^{-2} = \dfrac{1}{3^2} = \mathbf{\dfrac{1}{9}}$.