Mathematics • Year 9 • Unit 1 • Lesson 7

The Zero Index

Build fluency with $a^0 = 1$ (for $a \ne 0$) — see one worked example, fill in a guided example, then practise on your own with numbers, variables and brackets.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every step. The reason on the right tells you why the line is allowed.

Problem. Evaluate $4 \times 7^0 - 2$.

Step 1 — Spot the zero index.

There is a $7^0$ in the expression — that part needs the zero-index rule first.

Reason: order of operations — deal with the power before multiplying or subtracting.

Step 2 — Apply $a^0 = 1$.

$7^0 = 1$ (because $7 \ne 0$).

Reason: any non-zero base raised to $0$ equals $1$ — not $0$.

Step 3 — Substitute back.

$4 \times 7^0 - 2 = 4 \times 1 - 2$

Reason: replace the $7^0$ with $1$ and leave the rest of the expression alone.

Step 4 — Multiply, then subtract.

$4 \times 1 - 2 = 4 - 2 = 2$

Reason: BIDMAS — multiplication before subtraction.

Answer: $\mathbf{2}$.

Stuck? Revisit lesson § "Spot the Trap" — $7^0 = 1$, NOT $0$ and NOT $7$.

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. Evaluate $5 \times 9^0 + 8^0$.

Step 1 — Identify the zero-index parts: there are __________ powers of $0$ in this expression. Both bases are non-zero, so we can apply the __________________-index rule.

Step 2 — Apply $a^0 = 1$ to each:

$9^0 = \_\_\_\_$ and $8^0 = \_\_\_\_$

Step 3 — Substitute back:

$5 \times 9^0 + 8^0 = 5 \times \_\_\_\_ + \_\_\_\_$

Step 4 — Multiply, then add:

$= \_\_\_\_ + \_\_\_\_ = \_\_\_\_$

Step 5 — Final answer:

$5 \times 9^0 + 8^0 = \_\_\_\_$

Stuck? Revisit lesson § "Watch Me Solve It · Numerical base" — same idea as the $5^0 + 12^0$ example.

3. You do — independent practice

Show working in the space under each problem. The first four are foundation (pure zero-index). The middle two are standard (combine with arithmetic). The last two are extension (brackets and reasoning).

Foundation — apply the rule directly

3.1 Evaluate $9^0$. 1 mark

3.2 Evaluate $(-4)^0$. 1 mark

3.3 Simplify $(5y)^0$, assuming $y \ne 0$. 1 mark

3.4 Evaluate $(1.5)^0$. 1 mark

Standard — combine with arithmetic

3.5 Evaluate $6 \times 8^0$. 2 marks

3.6 Evaluate $10^0 + 3 \times 2^0 - 1$. 2 marks

Extension — brackets and reasoning

3.7 Simplify $(3x)^0$ and $3x^0$ (assume $x \ne 0$). Are they the same? Explain in one sentence. 3 marks

3.8 Use the quotient rule to show that $a^5 \div a^5 = 1$ (for $a \ne 0$). Then explain why this forces $a^0 = 1$. 2 marks

Stuck on 3.7? Look at where the brackets are. $(3x)^0$ means the WHOLE $3x$ is raised to $0$. $3x^0$ means only $x$ is raised to $0$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (faded $5 \times 9^0 + 8^0$)

Step 1: two powers of $0$; both bases non-zero, so apply the zero-index rule.
Step 2: $9^0 = \mathbf{1}$ and $8^0 = \mathbf{1}$.
Step 3: $5 \times 9^0 + 8^0 = 5 \times \mathbf{1} + \mathbf{1}$.
Step 4: $= \mathbf{5} + \mathbf{1} = \mathbf{6}$.
Step 5: $5 \times 9^0 + 8^0 = \mathbf{6}$.

3.1 — $9^0$

Any non-zero base raised to $0$ equals $1$. So $9^0 = \mathbf{1}$.

3.2 — $(-4)^0$

The base is $-4$, which is non-zero. So $(-4)^0 = \mathbf{1}$. (The negative sign doesn't matter — only that the base is non-zero.)

3.3 — $(5y)^0$

The brackets show the WHOLE base is $5y$. With $y \ne 0$, the base is non-zero, so $(5y)^0 = \mathbf{1}$.

3.4 — $(1.5)^0$

$1.5$ is non-zero, so $(1.5)^0 = \mathbf{1}$. The rule doesn't care if the base is a whole number or a decimal.

3.5 — $6 \times 8^0$

$8^0 = 1$, so $6 \times 8^0 = 6 \times 1 = \mathbf{6}$. (Common slip: writing $0$ because of the $0$ index. The index is $0$ but the VALUE is $1$.)

3.6 — $10^0 + 3 \times 2^0 - 1$

$10^0 = 1$ and $2^0 = 1$. So $1 + 3 \times 1 - 1 = 1 + 3 - 1 = \mathbf{3}$.

3.7 — $(3x)^0$ vs $3x^0$

$(3x)^0 = \mathbf{1}$ — the WHOLE bracket $3x$ is raised to $0$, and any non-zero base to $0$ is $1$.
$3x^0 = 3 \times x^0 = 3 \times 1 = \mathbf{3}$ — only the $x$ has the $0$ index; the $3$ has an implied index of $1$.
They are NOT the same. The brackets decide what the index applies to.

3.8 — Quotient-rule proof of $a^0 = 1$

$\dfrac{a^5}{a^5}$ is something divided by itself, so it equals $\mathbf{1}$ (provided $a \ne 0$).
By the quotient rule, $\dfrac{a^5}{a^5} = a^{5-5} = a^0$.
Since both expressions equal the same thing, $a^0 = \mathbf{1}$. The two rules force the zero-index rule on us.