Mathematics • Year 9 • Unit 1 • Lesson 7
Zero Indices in the Real World
Use $a^0 = 1$ inside calculator shortcuts, scoring formulas, tournaments and patterns. Then explain — in your own words — why a power of $0$ gives $1$, not $0$.
1. Word problems
Each problem uses the zero-index rule $a^0 = 1$ from Lesson 7. Show working — a single final answer with no working only earns half marks.
1.1 — Calculator check. Mia is checking a friend's calculator. She types $\mathbf{6 \times 12^0 + 3}$ and gets $9$ on the screen. Her friend says it should be $3$, claiming "the zero power makes the whole thing zero".
(a) Who is right? Show the full calculation.
(b) Where does her friend's reasoning go wrong? 3 marks
1.2 — Game score formula. An indie game gives a player score of $S = 100 \times L^0 + 25B$, where $L$ is the level number and $B$ is the number of bosses defeated. A new player has cleared no bosses ($B = 0$) but is on Level $7$.
(a) Show that $L^0 = 1$ no matter what the level number is (as long as $L \ne 0$).
(b) What is the new player's score? 3 marks
1.3 — Halving down to zero. Olivia writes powers of $2$ to remember the rule: $2^4 = 16, 2^3 = 8, 2^2 = 4, 2^1 = 2$. Each time she goes down one in the index, the value halves.
(a) Following Olivia's pattern, what value does $2^0$ have to be? Show the halving.
(b) Why does this argument force $a^0 = 1$ for any non-zero base, not just $a = 2$? 3 marks
1.4 — Tournament that goes nowhere. An esports league round has $2^5 = 32$ players. After every round, exactly half are eliminated. After $5$ rounds, there is $1$ winner. Using the quotient rule, this can be written as $\dfrac{2^5}{2^5}$.
(a) Use the quotient rule from Lesson 4 to show that $\dfrac{2^5}{2^5} = 2^0$.
(b) Use real-world common sense to show the same fraction must equal $1$ (only one winner). Hence explain what value $2^0$ must take. 3 marks
1.5 — Recipe scaling factor. A bakery uses a "scaling multiplier" to adjust the size of a batch. If they keep the batch unchanged, the multiplier is $k = R^0$, where $R$ is the recipe number ($R \ne 0$). Today the head chef writes "$k = R^0 = 1$, so the batch stays the same".
(a) Explain in one sentence why the chef's note is correct, no matter which recipe number $R$ is used.
(b) The trainee writes "$k = 1 \times R^0 + 0$" instead. Simplify the trainee's expression to show it gives the same answer. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate insists that "$5^0 = 0$, because zero of anything is zero". In your own words, explain (i) which lesson rule they have got wrong, (ii) why the rule $a^0 = 1$ has to be true (use either the halving pattern or the quotient rule), and (iii) one quick sanity check the student could run to spot their mistake.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Calculator check
(a) $12^0 = 1$. So $6 \times 12^0 + 3 = 6 \times 1 + 3 = 6 + 3 = \mathbf{9}$. Mia's calculator is right.
(b) The friend has confused "the index is $0$" with "the value is $0$". The index $0$ means the value is $1$, so $6 \times 12^0$ is $6 \times 1 = 6$, not $0$.
1.2 — Game score formula
(a) For any non-zero level $L$, the zero-index rule says $L^0 = 1$. So whether $L = 1, 7$ or $99$, $L^0 = 1$.
(b) $S = 100 \times 7^0 + 25 \times 0 = 100 \times 1 + 0 = \mathbf{100}$ points.
The $L^0$ in the formula is basically a "constant $100$" no matter which level the player is on.
1.3 — Halving down to zero
(a) Continuing: $2^1 = 2 \to 2^0 = 2 \div 2 = \mathbf{1}$.
(b) Replace "halving" with "dividing by the base $a$". From $a^1 = a$, divide by $a$: $a^0 = a \div a = 1$ (provided $a \ne 0$). The same logic works for any non-zero base.
1.4 — Tournament that goes nowhere
(a) Quotient rule: $\dfrac{2^5}{2^5} = 2^{5-5} = 2^0$.
(b) After $5$ halving rounds, $32$ players become $1$ winner. So $\dfrac{2^5}{2^5} = \dfrac{32}{32} = 1$. Since the SAME fraction equals both $2^0$ and $1$, we must have $\mathbf{2^0 = 1}$.
1.5 — Recipe scaling factor
(a) For any non-zero recipe number $R$, the rule $a^0 = 1$ applies — so $R^0 = 1$ regardless of which recipe.
(b) $k = 1 \times R^0 + 0 = 1 \times 1 + 0 = \mathbf{1}$. Same answer as the head chef.
2.1 — Explain your thinking (sample response)
The classmate has the zero-index rule wrong. The rule says $a^0 = 1$ for any non-zero $a$, NOT $0$. The mistake is confusing "the index is $0$" with "the value is $0$".
We can see the rule is forced on us by the descending pattern: $5^3 = 125 \to 5^2 = 25 \to 5^1 = 5$, and each step divides by $5$. To stay on the pattern, $5^0$ must equal $5 \div 5 = 1$. The quotient rule says the same thing: $\dfrac{5^n}{5^n} = 5^0$, but anything non-zero divided by itself is $1$, so $5^0 = 1$.
A quick sanity check: type $5 \wedge 0$ into any calculator — it gives $1$. Or compute $\dfrac{5^3}{5^3} = \dfrac{125}{125} = 1$, and that's $5^0$ by the quotient rule.
Marking: 1 mark for naming the zero-index rule and identifying the value confusion; 1 for a valid proof (pattern OR quotient); 1 for the correct value $5^0 = 1$; 1 for a clean sanity-check suggestion.