Chapter 1: Real Numbers (Exercise 1.5)

Admin 06 Aug, 2026 8 बार पढ़ा गया

Question 1:

Write the simplest form (simplest surd) of the following:

Solution:

$\sqrt{108}$:
Prime factorisation of $108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^2 \times 3$
$\sqrt{108} = \sqrt{6^2 \times 3} = 6\sqrt{3}$
$\sqrt[3]{128}$:
Prime factorisation of $128 = 2^7 = 2^3 \times 2^3 \times 2$
$\sqrt[3]{128} = \sqrt[3]{2^3} \times \sqrt[3]{2^3} \times \sqrt[3]{2} = 2 \times 2 \times \sqrt[3]{2} = 4\sqrt[3]{2}$
$\sqrt[4]{243}$:
Prime factorisation of $243 = 3^5 = 3^4 \times 3$
$\sqrt[4]{243} = \sqrt[4]{3^4 \times 3} = 3\sqrt[4]{3}$

Question 2:

Write the Rationalising Factor (R.F.) for each of the following surds:

Solution:

$3\sqrt{5}$:
Multiplying by $\sqrt{5}$ gives $3\sqrt{5} \times \sqrt{5} = 3 \times 5 = 15$ (Rational).
- Rationalising Factor: $\sqrt{5}$
$\sqrt{13} - \sqrt{7}$:
Using $(a-b)(a+b) = a^2 - b^2$ : $(\sqrt{13} - \sqrt{7})(\sqrt{13} + \sqrt{7}) = 13 - 7 = 6$ (Rational).
- Rationalising Factor: $\sqrt{13} + \sqrt{7}$
$5 + 2\sqrt{3}$:
Using $(a+b)(a-b) = a^2 - b^2$ : $(5 + 2\sqrt{3})(5 - 2\sqrt{3}) = 25 - 12 = 13$ (Rational).
- Rationalising Factor: $5 - 2\sqrt{3}$

Question 3:

Simplify the following surd expressions:

Solution:

$5\sqrt{3} + 2\sqrt{27} - \sqrt{75}$:
Simplifying each term:
- $2\sqrt{27} = 2\sqrt{9 \times 3} = 2(3\sqrt{3}) = 6\sqrt{3}$
- $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
Substituting back:
$= 5\sqrt{3} + 6\sqrt{3} - 5\sqrt{3}$
$= (5 + 6 - 5)\sqrt{3} = 6\sqrt{3}$
$2\sqrt[3]{4} + 5\sqrt[3]{4} - 3\sqrt[3]{4}$:
Taking $\sqrt[3]{4}$ common:
$= (2 + 5 - 3)\sqrt[3]{4} = 4\sqrt[3]{4}$

Question 4:

Compare the surds $\sqrt{3}$ and $\sqrt[3]{5}$ (Find which one is greater).

Solution:

Expressing in exponent form:

\sqrt{3} = 3^{\frac{1}{2}}, \quad \sqrt[3]{5} = 5^{\frac{1}{3}}

LCM of denominators of powers ( $2$ and $3$ ) is $6$ .

Convert exponents to have common denominator $6$ :

3^{\frac{1}{2}} = 3^{\frac{3}{6}} = (3^3)^{\frac{1}{6}} = (27)^{\frac{1}{6}} = \sqrt[6]{27}

5^{\frac{1}{3}} = 5^{\frac{2}{6}} = (5^2)^{\frac{1}{6}} = (25)^{\frac{1}{6}} = \sqrt[6]{25}

Since $27 > 25$ , we have $\sqrt[6]{27} > \sqrt[6]{25}$ .

Therefore, $\sqrt{3} > \sqrt[3]{5}$.

1. Types of Numbers

Natural Numbers ( $\mathbb{N}$ ): $\{1, 2, 3, 4, ...\}$
Whole Numbers ( $\mathbb{W}$ ): $\{0, 1, 2, 3, 4, ...\}$
Integers ( $\mathbb{Z}$ ): $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$
Rational Numbers ( $\mathbb{Q}$ ): Numbers in $\frac{p}{q}$ form ( $p, q \in \mathbb{Z}, q \neq 0$ ).
Irrational Numbers ( $\mathbb{Q}'$ ): Non-terminating, non-recurring decimals (e.g., $\sqrt{2}, \sqrt{3}, \pi$ ).
Real Numbers ( $\mathbb{R}$ ): Combination of all Rational and Irrational numbers ( $\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}'$ ).

2. Key Operations on Radicals / Surds

3. Laws of Exponents for Real Numbers