Mathematics • Year 10 • Unit 1 • Lesson 10
Zero and Negative Indices
Build fluency with two extensions of the index laws from Lesson 10: the zero-index rule a⁰ = 1 (for a ≠ 0) and the negative-index rule a⁻ⁿ = 1/aⁿ. Convert between negative and positive index forms with confidence.
1. I do — fully worked example
Read every line. Each step has a short reason.
Problem. Simplify (x⁻³ × x⁵) ÷ x⁻², leaving your answer with positive indices.
Step 1 — Tackle the numerator using Index Law 1 (add).
x⁻³ × x⁵ = x⁻³⁺⁵ = x²
Reason: same base x, multiply → ADD the indices (including the negative one).
Step 2 — Divide using Index Law 2 (subtract).
x² ÷ x⁻² = x²⁻⁽⁻²⁾ = x²⁺² = x⁴
Reason: subtracting a negative is the same as adding. Be very careful with the sign.
Step 3 — Check the answer has positive indices.
x⁴ already has a positive index — nothing more to do.
Reason: had the result been x⁻⁴, we would have rewritten as 1/x⁴ using the negative-index rule.
Answer: x⁴.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. Evaluate 5⁰ + 2⁻³, giving your answer as a fraction.
Step 1 — Apply the zero-index rule:
5⁰ = ____ (because any non-zero number to the power of zero is ____)
Step 2 — Apply the negative-index rule a⁻ⁿ = 1/aⁿ:
2⁻³ = 1 / 2____ = 1 / ____
Step 3 — Add the two values:
5⁰ + 2⁻³ = ____ + (1 / ____) = ____ + ____ = ______
Step 4 — Write as a single fraction:
Final answer = ______
3. You do — independent practice
Show your working. The first four are foundation (single rule). The middle two are standard (combine 2 rules). The last two are extension (multiple negative indices).
Foundation — single rule
3.1 Evaluate 9⁰. 1 mark
3.2 Evaluate 7⁰ + 4⁰. 1 mark
3.3 Evaluate 2⁻⁴ as a fraction. 1 mark
3.4 Rewrite x⁻⁵ with a positive index. 1 mark
Standard — combine two ideas
3.5 Simplify x⁴ × x⁻⁶, leaving your answer with a positive index. 2 marks
3.6 Simplify y³ ÷ y⁷, leaving your answer with a positive index. 2 marks
Extension — push your thinking
3.7 Simplify (x⁻³ × x⁵) ÷ x⁻², leaving your answer with positive indices. 3 marks
3.8 Evaluate 3⁻² × 3⁵ ÷ 3³, giving your answer as a single number. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (5⁰ + 2⁻³)
Step 1: 5⁰ = 1 (because any non-zero number to the power of zero is 1).
Step 2: 2⁻³ = 1 / 2³ = 1 / 8.
Step 3: 5⁰ + 2⁻³ = 1 + (1 / 8) = 8/8 + 1/8 = 9/8.
Step 4: Final answer = 9/8 (or 1⅛).
3.1 — 9⁰
Any non-zero number to the power of zero is 1. So 9⁰ = 1.
3.2 — 7⁰ + 4⁰
Both equal 1. So 1 + 1 = 2.
3.3 — 2⁻⁴
2⁻⁴ = 1 / 2⁴ = 1/16.
3.4 — x⁻⁵ with positive index
x⁻⁵ = 1/x⁵.
3.5 — x⁴ × x⁻⁶
x⁴⁺⁽⁻⁶⁾ = x⁻² = 1/x² (using a⁻ⁿ = 1/aⁿ).
3.6 — y³ ÷ y⁷
y³⁻⁷ = y⁻⁴ = 1/y⁴.
3.7 — (x⁻³ × x⁵) ÷ x⁻²
Numerator: x⁻³⁺⁵ = x².
Divide: x² ÷ x⁻² = x²⁻⁽⁻²⁾ = x⁴.
Answer: x⁴.
3.8 — 3⁻² × 3⁵ ÷ 3³
Combine indices: 3(−2 + 5 − 3) = 3⁰ = 1.
The zero-index rule is the punchline here.