Class 9 Maths Chapter 1: Number Systems
Class 9th maths exercise 1.4
1. Classify as Rational or Irrational
(i) \(2 – \sqrt{5}\) → Irrational (Difference of rational and irrational is irrational)
(ii) \((3 + \sqrt{23}) – \sqrt{23} = 3\) → Rational (Simplifies to an integer)
(iii) \(\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}\) → Rational (Simplifies to a fraction)
(iv) \(\frac{1}{\sqrt{2}}\) → Irrational (Denominator is irrational)
(v) \(2\pi\) → Irrational (Product of rational and irrational is irrational)
2. Simplify the Expressions
(i) \((3 + \sqrt{3})(2 + \sqrt{2})\)
\(= 3 \cdot 2 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{3} \cdot \sqrt{2}\)
\(= 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}\)
(ii) \((3 + \sqrt{3})(3 – \sqrt{3})\)
\(= 3^2 – (\sqrt{3})^2\) (Using \((a + b)(a – b) = a^2 – b^2\))
\(= 9 – 3 = 6\)
(iii) \((\sqrt{5} + \sqrt{2})^2\)
\(= (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot \sqrt{2} + (\sqrt{2})^2\)
\(= 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}\)
(iv) \((\sqrt{5} – \sqrt{2})(\sqrt{5} + \sqrt{2})\)
\(= (\sqrt{5})^2 – (\sqrt{2})^2\) (Difference of squares)
\(= 5 – 2 = 3\)
3. Resolving the π Contradiction
While \(\pi = \frac{c}{d}\) appears rational, both \(c\) (circumference) and \(d\) (diameter) cannot be measured exactly in rational numbers. For a circle with rational diameter, the circumference becomes irrational, and vice versa. Thus, \(\pi\) remains irrational.
4. Represent \(\sqrt{9.3}\) on the Number Line
Step-by-Step Construction:
- First, approximate \(\sqrt{9.3}\):
Since \(9 < 9.3 < 16\), we know \(3 < \sqrt{9.3} < 4\).
More precisely, \(\sqrt{9.3} \approx 3.05\). - Draw a horizontal number line and mark integers 3 and 4.
- Divide the segment between 3 and 4 into 10 equal parts (each part = 0.1).
- Mark the point corresponding to approximately 3.05.
Visualization: 3 —-3.1—-3.2—-…—-√9.3 ≈ 3.05—-…—-4
5. Rationalise the Denominators
(i) \(\frac{1}{\sqrt{7}} = \frac{1 \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} = \frac{\sqrt{7}}{7}\)
(ii) \(\frac{1}{\sqrt{7} – \sqrt{6}} = \frac{1 \cdot (\sqrt{7} + \sqrt{6})}{(\sqrt{7} – \sqrt{6})(\sqrt{7} + \sqrt{6})} = \frac{\sqrt{7} + \sqrt{6}}{7 – 6} = \sqrt{7} + \sqrt{6}\)
(iii) \(\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{1 \cdot (\sqrt{5} – \sqrt{2})}{(\sqrt{5} + \sqrt{2})(\sqrt{5} – \sqrt{2})} = \frac{\sqrt{5} – \sqrt{2}}{5 – 2} = \frac{\sqrt{5} – \sqrt{2}}{3}\)
(iv) \(\frac{1}{\sqrt{7} – 2} = \frac{1 \cdot (\sqrt{7} + 2)}{(\sqrt{7} – 2)(\sqrt{7} + 2)} = \frac{\sqrt{7} + 2}{7 – 4} = \frac{\sqrt{7} + 2}{3}\)
Key Concepts
- Rationalising denominators removes irrational terms.
- \(\pi\) is irrational despite being a ratio of circumference to diameter.
- Simplification using algebraic identities like \((a + b)(a – b) = a^2 – b^2\).
Practice Questions
1. Simplify: \((\sqrt{3} + \sqrt{2})^2\)
\(= (\sqrt{3})^2 + 2 \cdot \sqrt{3} \cdot \sqrt{2} + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}\)
2. Rationalise: \(\frac{1}{\sqrt{5} – 1}\)
Multiply numerator and denominator by \(\sqrt{5} + 1\):
\(\frac{\sqrt{5} + 1}{(\sqrt{5})^2 – (1)^2} = \frac{\sqrt{5} + 1}{5 – 1} = \frac{\sqrt{5} + 1}{4}\)
3. Represent \(\sqrt{6.5}\) on the number line.
Approximate \(\sqrt{6.5} \approx 2.55\). Mark 2.55 between 2 and 3 on the number line.
4. Is \(\sqrt{3} + \sqrt{2}\) rational? Justify.
No, the sum of two irrational numbers is not necessarily rational.
5. Simplify: \((\sqrt{7} – \sqrt{5})(\sqrt{7} + \sqrt{5})\)
Using \((a – b)(a + b) = a^2 – b^2\):
\(= (\sqrt{7})^2 – (\sqrt{5})^2 = 7 – 5 = 2\)
