Rationalise the denominator and simplify it to find the value of 4 5 + 3 , given that 5 = 2 . 236 and 3 = 1 . 732

We have, 4 5 + 3 = 4 × ( 5 - 3 ) 5 + 3 × ( 5 - 3 ) = 4 × ( 5 - 3 ) 5 2 - 3 2 ∵ ( a + b ) ( a - b ) = a 2 - b 2 = 4 × ( 5 - 3 ) 5 - 3 = 4 × ( 5 - 3 ) 2 = 2 ( 5 - 3 ) = 2 ( 2 . 236 - 1 . 732 ) = 2 × 0 . 504 = 1 . 008 Therefore, the required answer is 1 . 008

Prove that the number 4 - 3 is irrational.

To prove 4 - 3 is an irrational number. Let 4 - 3 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers So, 4 - 3 = p q ⇒ 3 = 4 - p q ⇒ 3 = 4 q - p q Here, when p & q are integer then 4 q - p p is a rational number. It means 3 is also a rational number. This is a contradiction as 3 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence our assumption is false. Therefore, 4 - 3 is an irrational number.

Prove that the number 3 - 2 is irrational.

To prove 3 - 2 is an irrational number. Let 3 - 2 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers So, 3 - 2 = p q ⇒ 3 = p q + 2 On squaring both the sides, ⇒ 3 2 = p q + 2 2 ⇒ 3 = p 2 q 2 + 2 2 p q + 2 ⇒ 2 2 p q = - p 2 q 2 + 3 - 2 ⇒ 2 2 p q = 1 - p 2 q 2 ⇒ 2 = 1 2 × q p × 1 - p 2 q 2 ⇒ 2 = q 2 p - p 2 q 2 × q 2 p ⇒ 2 = q 2 p - p 2 q ⇒ 2 = q 2 - p 2 2 p q Here, when p & q are integer then q 2 - p 2 2 p q is a rational number. It means 2 is also a rational number. This is a contradiction as 2 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence our assumption is false. Therefore, 3 - 2 is an irrational number.

E M B I B E

ICSE MATHEMATICS CLASS 9>Pure Arithmetic>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE

B Nirmala Shastry Solutions for Exercise 4: MISCELLANEOUS EXERCISE

Author:B Nirmala Shastry

B Nirmala Shastry Mathematics Solutions for Exercise - B Nirmala Shastry Solutions for Exercise 4: MISCELLANEOUS EXERCISE

Attempt the practice questions from Exercise 4: MISCELLANEOUS EXERCISE with hints and solutions to strengthen your understanding. ICSE MATHEMATICS CLASS 9 solutions are prepared by Experienced Embibe Experts.

Questions from B Nirmala Shastry Solutions for Exercise 4: MISCELLANEOUS EXERCISE with Hints & Solutions

MEDIUM

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 7

$\sqrt{49 - 5} = 7 - \sqrt{5}$

MEDIUM

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 7

$\sqrt{9 + 7} = 3 + \sqrt{7}$

MEDIUM

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 7

$\sqrt{3} \sqrt{27} = 9$

MEDIUM

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 7

$(\sqrt{11} - \sqrt{7}) (\sqrt{11} + \sqrt{7}) = 4$

MEDIUM

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 7

$\frac{1}{3 - \sqrt{8}} = 3 + 2 \sqrt{2}$

HARD

9th ICSE

IMPORTANT

ICSE MATHEMATICS CLASS 9>Chapter 1 - Rational and Irrational Numbers>MISCELLANEOUS EXERCISE>Q 8

Rationalise the denominator and simplify it to find the value of $\frac{4}{\sqrt{5} + \sqrt{3}}$ , given that $\sqrt{5} = 2.236$ and $\sqrt{3} = 1.732$