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CBSE Class 9 Mathematics Question 8 of 9

Number Systems — Question 8

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Question 5

Rationalize the denominators of the following:

(i) 17\dfrac{1}{\sqrt{7}}

(ii) 17−6\dfrac{1}{\sqrt{7} - \sqrt{6} }

(iii)15+2\dfrac{1}{\sqrt{5} + \sqrt{2} }

(iv)17−2\dfrac{1}{\sqrt{7} - 2 }

Answer

(i)17\dfrac{1}{\sqrt{7}}

Multiply by (7)(\sqrt{7}) in numerator and denominator

= 17\dfrac{1}{\sqrt{7}} x 77\dfrac{\sqrt{7}}{\sqrt{7}}

= 77\dfrac{\sqrt{7}}{7}

Hence, 17\dfrac{1}{\sqrt{7}} = 77\dfrac{\sqrt{7}}{7}

(ii) 17−6\dfrac{1}{\sqrt{7} - \sqrt{6}}

Multiply by 17+6\dfrac{1}{\sqrt{7} + \sqrt{6}} in numerator and denominator

= 17−6\dfrac{1}{\sqrt{7} - \sqrt{6} } x 7+67+6\dfrac{\sqrt{7}+ \sqrt{6}}{\sqrt{7} + \sqrt{6}}

= 7+6(7)2−(6)2\dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (6)^2 }

= 7+67−6\dfrac{\sqrt{7} + \sqrt{6} }{7 - 6}

= 7+61\dfrac{\sqrt{7} + \sqrt{6} }{1}

= (7+6)(\sqrt{7} + \sqrt{6})

=Hence, 17−6\dfrac{1}{\sqrt{7} - \sqrt{6}} = (7+6)(\sqrt{7} + \sqrt{6})

(iii)15+2\dfrac{1}{\sqrt{5} + \sqrt{2}}

Multiply by 5+2{\sqrt{5} + \sqrt{2}} in numerator and denominator

= 15+2\dfrac{1}{\sqrt{5} + \sqrt{2}} x 5−25−2\dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}

= 5−2(5)2−(2)2\dfrac{\sqrt{5} - \sqrt{2}} {(\sqrt{5})^2 - (\sqrt{2})^2 }

= 5−25−2\dfrac{\sqrt{5} - \sqrt{2}} {5 - 2}

= 5−23\dfrac{\sqrt{5} - \sqrt{2}} {3}

=Hence, 15+2\dfrac{1}{\sqrt{5} + \sqrt{2}} = 5−23\dfrac{\sqrt{5} - \sqrt{2}} {3}

(iv)17−2\dfrac{1}{\sqrt{7} - 2}

Multiply by 7+2{\sqrt{7} + {2}} in numerator and denominator

= 17−2\dfrac{1}{\sqrt{7} - 2} X 7+27+2\dfrac{\sqrt{7} + 2 }{\sqrt{7} + 2 }

= 7+2(7)2−(2)2\dfrac{\sqrt{7} + 2} {(\sqrt{7})^2 - (2)^2 }

= 7+27−4\dfrac{\sqrt{7} + 2}{7 - 4}

= 7+23\dfrac{\sqrt{7} + 2}{3}

= Hence, 17−2\dfrac{1}{\sqrt{7} - 2 } = 7+23\dfrac{\sqrt{7} + 2}{3}

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Mathematics | Number SystemsWeb Content • Interactive Notes

Number Systems — Interactive Study Guide

Master the real number system, irrational numbers, surds, rationalisation, and exponent laws.

The Number Hierarchy

Think of numbers as nested boxes: Natural numbers are inside Whole numbers, which are inside Integers, which are inside Rational numbers, which are inside Real numbers.

Key Insight: Between any two rational numbers, there are infinitely many irrational numbers, and vice versa. The number line is “dense” with both types!

Identifying Rational vs Irrational

NumberTypeReason
√4Rational√4 = 2 (perfect square)
√7Irrational7 is not a perfect square
0.333...RationalRecurring decimal = 1/3
0.10100100010...IrrationalNon-terminating, non-recurring
πIrrationalNon-terminating, non-recurring
22/7RationalIt is p/q form (just an approximation of π)

Rationalisation — Quick Method

To rationalise a denominator with surds, multiply top and bottom by the conjugate:

  • Conjugate of (a + √b) is (a − √b)
  • Conjugate of (√a − √b) is (√a + √b)

The denominator becomes rational because (a+b)(a−b) = a² − b².

Quick Self-Check

  1. Is √(16/9) rational or irrational? (Rational: = 4/3)
  2. Simplify: √50 + √18 (= 5√2 + 3√2 = 8√2)
  3. Rationalise: 1/(√3 + 1) (= (√3 − 1)/2)
  4. Find: 81/3 (= 2)

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