10.1 Introduction | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 3 min read
10.1 Introduction – this guide gives you a concise, exam-ready overview of 10.1 Introduction from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
10.2 Powers with Negative Exponents
This section extends the concept of exponents to include negative powers. While positive exponents represent repeated multiplication, negative exponents represent repeated division or the reciprocal of the positive power. For example, 10² = 10 × 10 = 100, 10¹ = 10, and 10⁰ = 1. Following this pattern, 10⁻¹ is defined as 1 ÷ 10 = 1/10, and similarly, 10⁻² = 1 ÷ 10 ÷ 10 = 1/100 = 1/10². This shows that a negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent. The section provides examples with other bases such as 3⁻¹ = 1/3, 3⁻² = 1/3², and so on. It generalizes that for any non-zero integer a and positive integer m, a⁻ᵐ = 1/aᵐ. This means a negative exponent is the multiplicative inverse of the positive power. The section also illustrates how decimal numbers can be expressed in expanded form using both positive and negative exponents, for example, 1425.36 = 1 × 10³ + 4 × 10² + 2 × 10¹ + 5 × 10⁰ + 3 × 10⁻¹ + 6 × 10⁻². This understanding is fundamental for working with scientific notation and algebraic expressions involving powers.
📊 Diagram: Figures illustrating powers with negative exponents, including the pattern for 3⁻¹, 3⁻², 3⁻³; Expanded form of decimal numbers using exponents.
🧪 Activity: Try finding multiplicative inverses of given powers with negative exponents; expand numbers like 1025.63 using exponents.
🔗 Connection: Prepares for the next section on Laws of Exponents, including operations involving negative powers.
Frequently asked questions
1. Evaluate. (i) $ 3^{-2} $ (ii) $ (-4)^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-5} $
Solution: (i) $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$
(ii) $(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$
(iii) $\left( \frac{1}{2} \right)^{-5} = 2^5 = 32$
2. Simplify and express the result in power notation with positive exponent. (i) $ (-4)^5 \div (-4)^8 $ (ii) $ \left( \frac{1}{2^3} \right)^2 $ (iii) $ (-3)^4 \times \left( \frac{5}{3} \right)^4 $ (iv) $ (3^{-7} \div 3^{-10}) \times 3^{-5} $ (v) $ 2^{-3} \times (-7)^{-3} $
Solution: (i) $(-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}$
(ii) $\left( \frac{1}{2^3} \right)^2 = \left( 2^{-3} \right)^2 = 2^{-6} = \frac{1}{2^6} = \frac{1}{64}$
(iii) $(-3)^4 \times \left( \frac{5}{3} \right)^4 = \left( -3 \times \frac{5}{3} \right)^4 = (-5)^4 = 5^4 = 625$
(iv) $(3^{-7} \div 3^{-10}) \times 3^{-5} = 3^{-7 - (-10)} \times 3^{-5} = 3^{3} \times 3^{-5} = 3^{3-5} = 3^{-2} = \frac{1}{9}$
(v) $2^{-3} \times (-7)^{-3} = \frac{1}
3. Find the value of. (i) $ (3^0 + 4^{-1}) \times 2^2 $ (ii) $ (2^{-1} \times 4^{-1}) \div 2^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-2} + \left( \frac{1}{3} \right)^{-2} + \left( \frac{1}{4} \right)^{-2} $ (iv) $ (3^{-1} + 4^{-1} + 5^{-1})^0 $ (v) $ \left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^2 $
Solution: (i) $3^0 = 1$, $4^{-1} = \frac{1}{4}$, so $(1 + \frac{1}{4}) \times 2^2 = \frac{5}{4} \times 4 = 5$
(ii) $(2^{-1} \times 4^{-1}) \div 2^{-2} = \left( \frac{1}{2} \times \frac{1}{4} \right) \div \frac{1}{2^2} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{1}{2}$
(iii) $\left( \frac{1}{2} \right)^{-2} = 2^2 = 4$, $\left( \frac{1}{3} \right)^{-2} = 3^2 = 9$, $\left( \frac{1}{4} \right)^{-2} = 4^2 = 16$, sum = $4 + 9 + 16 = 29$
(iv) Any number to the power 0 is 1, so $(3^
4. Evaluate (i) $\frac{8^{-1} \times 5^3}{2^{-4}}$ (ii) $(5^{-1} \times 2^{-1}) \times 6^{-1}$
Solution: (i) $\frac{8^{-1} \times 5^3}{2^{-4}} = \frac{\frac{1}{8} \times 125}{\frac{1}{2^{-4}}} = \frac{\frac{125}{8}}{2^{-4}} = \frac{125}{8} \times 2^{4} = \frac{125}{8} \times 16 = 125 \times 2 = 250$
(ii) $(5^{-1} \times 2^{-1}) \times 6^{-1} = \frac{1}{5} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{60}$
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