Mathematics • Year 9 • Unit 1 • Lesson 8
Negative Indices in the Real World
Use the negative-index rule $a^{-n} = \dfrac{1}{a^n}$ in everyday contexts: file sizes, batteries fading, photo zoom, fundraising drives, and a halving pattern. Then explain — in your own words — why a negative index makes a fraction, not a negative number.
1. Word problems
Each problem uses the rule from Lesson 8: $a^{-n} = \dfrac{1}{a^n}$. Show your working — a single final answer with no working only earns half marks.
1.1 — File sizes that halve. A video editing app saves a draft, then saves smaller previews at $\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dfrac{1}{16}$ of the original file size. The size of the $n$-th preview (compared to the original) can be written as $2^{-n}$.
(a) Show that $2^{-1}, 2^{-2}, 2^{-3}, 2^{-4}$ give the four fractions above.
(b) Write the size of the $5$th preview as a fraction with positive index, and evaluate it. 3 marks
1.2 — Battery fades. An old phone battery's effective capacity is modelled as $\dfrac{1}{2^h}$ of full, where $h$ is the number of hours since unplugging. After $0, 1, 2, 3$ hours, the formula gives the fractions $\dfrac{1}{1}, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}$ — that is, $2^{-h}$ for $h = 0, 1, 2, 3$.
(a) Use $a^0 = 1$ to confirm the $h = 0$ value matches "full battery".
(b) Use the negative-index rule to evaluate the battery at $h = 4$. Give your answer as a fraction. 3 marks
1.3 — Photo zoom out. Each time Ari taps "zoom out" on a photo, the image becomes $\dfrac{1}{3}$ as wide. After $n$ taps, the photo is $3^{-n}$ as wide as the original.
(a) After $2$ taps, write the photo's width as a fraction with positive index.
(b) After $4$ taps, evaluate the width as a fraction. 3 marks
1.4 — Quotient rule on shrinking donations. A school fundraiser tracks total donations $D$ after $5$ weeks and after $7$ weeks. The ratio of week-$5$ donations to week-$7$ donations is $\dfrac{a^5}{a^7}$, where $a$ is the weekly growth factor (greater than $1$, so totals grow week-on-week).
(a) Use the quotient rule to simplify $\dfrac{a^5}{a^7}$ as a single power of $a$ with a negative index.
(b) Rewrite that with a positive index. Briefly explain why your answer is a fraction less than $1$ (use the fact $a > 1$). 3 marks
1.5 — Pattern down past zero. Ben writes the powers of $3$ going DOWN: $3^3 = 27, 3^2 = 9, 3^1 = 3$.
(a) What value should come next ($3^0$)? Why?
(b) Continuing the pattern, what value should $3^{-1}$ have? Express it both as a fraction with positive index and as a decimal. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate writes "$2^{-3} = -8$". In your own words, explain (i) what mistake they have made about the meaning of a negative index, (ii) which rule from Lesson 8 they have forgotten, and (iii) what the correct value of $2^{-3}$ is, with one sentence justifying the sign of the answer. Refer to the words "reciprocal" or "denominator" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — File sizes that halve
(a) $2^{-1} = \dfrac{1}{2}$, $2^{-2} = \dfrac{1}{4}$, $2^{-3} = \dfrac{1}{8}$, $2^{-4} = \dfrac{1}{16}$ ✓ — exactly the listed fractions.
(b) $2^{-5} = \dfrac{1}{2^5} = \mathbf{\dfrac{1}{32}}$ of the original size.
1.2 — Battery fades
(a) $2^{-0} = 2^0 = \mathbf{1}$ — i.e., full battery. ✓
(b) $2^{-4} = \dfrac{1}{2^4} = \mathbf{\dfrac{1}{16}}$ of full capacity at $h = 4$.
1.3 — Photo zoom out
(a) $3^{-2} = \mathbf{\dfrac{1}{3^2}} = \dfrac{1}{9}$ of original width after $2$ taps.
(b) $3^{-4} = \dfrac{1}{3^4} = \mathbf{\dfrac{1}{81}}$ of original width after $4$ taps.
1.4 — Quotient rule on shrinking donations
(a) Quotient rule: $\dfrac{a^5}{a^7} = a^{5-7} = \mathbf{a^{-2}}$.
(b) Negative-index rule: $a^{-2} = \mathbf{\dfrac{1}{a^2}}$.
Since $a > 1$, we have $a^2 > 1$, so $\dfrac{1}{a^2} < 1$. The week-$5$ total really is a fraction of the week-$7$ total, as expected.
1.5 — Pattern down past zero
(a) $3^0 = \mathbf{1}$ — following the pattern, each step divides by $3$, so $3^1 \div 3 = 1$. The zero-index rule from Lesson 7 says this directly.
(b) Continuing the pattern, $3^{-1} = 1 \div 3 = \mathbf{\dfrac{1}{3}}$ as a fraction, or $\mathbf{0.\overline{3}}$ (about $0.33$) as a decimal.
2.1 — Explain your thinking (sample response)
The classmate has assumed that the negative sign in the index attaches to the VALUE — that "$-3$ in the index means the answer is negative". That's wrong. A negative index means take the reciprocal: move the base to the denominator and drop the minus sign.
The rule they have forgotten is the negative-index rule: $a^{-n} = \dfrac{1}{a^n}$, for $a \ne 0$.
Applied to $2^{-3}$: $2^{-3} = \dfrac{1}{2^3} = \mathbf{\dfrac{1}{8}}$. The answer is a positive fraction, not $-8$. The sign is positive because $2^3 = 8$ is positive, so its reciprocal $\dfrac{1}{8}$ is also positive — the minus is in the INDEX, not in the value.
Marking: 1 mark for naming the negative-index rule; 1 for "reciprocal" or "denominator" used correctly; 1 for the correct value $\tfrac{1}{8}$; 1 for a clear, full-sentence explanation of the sign.