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.
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: $\mathbf{2}$.
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:
$5 \times 9^0 + 8^0 = \_\_\_\_$
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
How did this worksheet feel?
What I'll revisit before next class:
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.