Mathematics • Year 9 • Unit 1 • Lesson 9
Negative Indices, Evaluation
Build fluency with evaluating negative indices: flip fractions $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$, combine negative + positive indices, and finish every answer with positive indices only.
1. I do, fully worked example
Read every step. Each line has a short reason on the right so you can see why.
Problem. Evaluate $\left(\dfrac{2}{3}\right)^{-2}$.
Step 1, Spot the rule.
A fraction raised to a NEGATIVE index, flip the fraction and use the positive index.
Reason: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$.
Step 2, Flip the fraction, drop the minus.
$\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2$
Reason: top and bottom swap, and the negative sign in the index goes away.
Step 3, Apply power-of-a-quotient.
$\left(\dfrac{3}{2}\right)^2 = \dfrac{3^2}{2^2} = \dfrac{9}{4}$
Reason: square the top, square the bottom (L5 rule).
Step 4, Check the form.
$\dfrac{9}{4}$, positive, no negative indices left.
Reason: standard form needs positive indices only.
Answer: $\mathbf{\dfrac{9}{4}}$.
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. Simplify $a^{-3} \times a^5$, leaving your answer with a positive index.
Step 1, Spot the rule: same base on both sides of the $\times$, so use the __________________ rule and __________ the indices (even when one is negative).
Step 2, Add the indices:
$a^{-3} \times a^5 = a^{(\,\_\_\_\) + (\,\_\_\_\)}$
Step 3, Simplify the index:
$= a^{\_\_\_}$
Step 4, Final form:
$a^{-3} \times a^5 = \_\_\_\_\_\_$ (index is already positive, no conversion needed)
3. You do, independent practice
Show your working in the space under each problem. The first four are foundation (single-step evaluation). The middle two are standard (combine indices). The last two are extension (negative bases and final-form conversion).
Foundation, single-step evaluations
3.1 Evaluate $\left(\dfrac{1}{3}\right)^{-2}$. 1 mark
3.2 Evaluate $\left(\dfrac{1}{2}\right)^{-3}$. 1 mark
3.3 Evaluate $5^{-1}$ as a decimal. 1 mark
3.4 Evaluate $\left(\dfrac{3}{5}\right)^{-2}$. 1 mark
Standard, combine indices
3.5 Simplify $a^{-2} \times a^7$, leaving your answer with a positive index ($a \ne 0$). 2 marks
3.6 Simplify $b^{-4} \times b^{-2}$, leaving your answer with a positive index ($b \ne 0$). 2 marks
Extension, negative bases and final-form conversion
3.7 Evaluate $(-2)^{-3}$. 3 marks
3.8 Simplify $\dfrac{c^4}{c^{-1}}$, leaving your answer as a single power of $c$ with a positive index ($c \ne 0$). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (faded $a^{-3} \times a^5$)
Step 1: product rule; add the indices.
Step 2: $a^{(\mathbf{-3}) + (\mathbf{5})}$.
Step 3: $= a^{\mathbf{2}}$.
Step 4: $a^{-3} \times a^5 = \mathbf{a^2}$.
3.1, $\left(\dfrac{1}{3}\right)^{-2}$
Flip: $\left(\dfrac{1}{3}\right)^{-2} = \left(\dfrac{3}{1}\right)^2 = 3^2 = \mathbf{9}$.
3.2, $\left(\dfrac{1}{2}\right)^{-3}$
Flip: $\left(\dfrac{1}{2}\right)^{-3} = 2^3 = \mathbf{8}$.
3.3, $5^{-1}$ as a decimal
$5^{-1} = \dfrac{1}{5} = \mathbf{0.2}$.
3.4, $\left(\dfrac{3}{5}\right)^{-2}$
Flip: $\left(\dfrac{3}{5}\right)^{-2} = \left(\dfrac{5}{3}\right)^2 = \dfrac{5^2}{3^2} = \mathbf{\dfrac{25}{9}}$.
3.5, $a^{-2} \times a^7$
Product rule: $a^{-2+7} = \mathbf{a^5}$ (already positive index, no conversion needed).
3.6, $b^{-4} \times b^{-2}$
Product rule: $b^{-4 + (-2)} = b^{-6}$.
Convert to positive index: $b^{-6} = \mathbf{\dfrac{1}{b^6}}$.
3.7, $(-2)^{-3}$
Negative-index rule: $(-2)^{-3} = \dfrac{1}{(-2)^3}$.
Evaluate $(-2)^3 = -8$ (negative base, odd index, keeps the minus).
Combine: $\dfrac{1}{-8} = \mathbf{-\dfrac{1}{8}}$.
Note: the answer IS negative this time, because the base $-2$ is negative AND the index $3$ is odd.
3.8, $\dfrac{c^4}{c^{-1}}$
Quotient rule: $c^{4 - (-1)} = c^{4+1} = \mathbf{c^5}$.
"Subtracting a negative" is the same as adding.