OP Malhotra Class 9 Maths Solutions Chapter 1 Rational and Irrational Numbers Exercise 1 (B)

S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Ex 1(B)

Question 1. Look at the following real numbers :
– 5, 0, \( \sqrt{3} \), \( \frac{3}{5} \), \( -\sqrt{9} \), \( \sqrt{8} \), 6.37, \( \pi \), 4, \( \frac{-2}{7} \), 0.03
Tell,
(i) Which are rational?
(ii) Which are irrational?
(iii) Which are positive integers?
(iv) Which are negative integers?
(v) Which number is neither positive nor- negative?
Answer:
(i) Rational numbers are -5, 0, \( \frac{3}{5} \), \( -\sqrt{9} \), 6.37, 4, \( \frac{-2}{7} \), 0.03. These numbers can be expressed as a fraction or have a terminating/repeating decimal expansion.
(ii) Irrational numbers are \( \sqrt{3} \), \( \sqrt{8} \), \( \pi \). These numbers cannot be expressed as a simple fraction and have non-terminating, non-repeating decimal expansions.
(iii) Positive integer is 4.
(iv) Negative integers are -5, \( -\sqrt{9} \). Since \( \sqrt{9} = 3 \), \( -\sqrt{9} = -3 \). So, the negative integers are -5 and -3.
(v) The number which is neither positive nor negative is 0.
In simple words: This question asks you to categorize numbers based on whether they are rational (can be written as a fraction), irrational (cannot be written as a fraction, like pi or non-perfect square roots), positive integers (whole numbers greater than zero), negative integers (whole numbers less than zero), or zero.

🎯 Exam Tip: Remember that \( \sqrt{9} \) simplifies to 3, so \( -\sqrt{9} \) is -3, which is a rational number and an integer, not an irrational number. Always simplify square roots first.

Question 2. Write true or false to describe each sentence:
(i) All rational numbers are real numbers.
(ii) All real numbers are rational numbers.
(iii) Some real numbers are rational numbers.
(iv) All integers are rational numbers.
(v) No rational number is also an irrational number.
(vi) There exists a whole number that is not a natural number.
Answer:
(i) True: as real numbers include both rational and irrational numbers. The set of rational numbers is a subset of real numbers.
(ii) False: Real numbers include both rational numbers and irrational numbers, so not all real numbers are rational.
(iii) True: as some real numbers are rational numbers (e.g., 2, 0.5), while others are irrational (e.g., \( \sqrt{2} \), \( \pi \)).
(iv) True: as the set of integers is a subset of rational numbers. Any integer 'n' can be written as 'n/1'.
(v) True: as by definition, an irrational number is a number which is not a rational number. Rational and irrational numbers are mutually exclusive sets.
(vi) True: as 0 is a whole number but not a natural number. Natural numbers start from 1, while whole numbers start from 0.
In simple words: This question checks your understanding of different number types and their relationships, like how rational numbers are part of real numbers, but not all real numbers are rational.

🎯 Exam Tip: Be precise with definitions of number sets: Natural numbers (1, 2, 3...), Whole numbers (0, 1, 2, 3...), Integers (...-1, 0, 1...), Rational numbers (p/q form), and Real numbers (all rational and irrational numbers). Use examples to quickly verify True/False statements.

Question 3. Tell whether each decimal numeral represents a rational or an irrational number:
(i) 0.578
(ii) 0.573 333.....
(iii) 0.688 434 4454....
(iv) 0.727 374 75......
(v) 0.638 754 71
(vi) 0.471 7171....
(vii) 283
(viii) 289.387 000.....
(ix) 5.\overline{93}
(x) 2.309\overline{8}7
(xi) 0.585 885 888...
Answer:
(i) 0.578
It is a terminating decimal, therefore it is a rational number.
(ii) 0.573 333..... \( = 0.57\overline{3} \)
It is a repeating decimal, therefore it is a rational number.
(iii) 0.688 434 4454....
It is neither terminating nor repeating decimal, therefore it is an irrational number.
(iv) 0.727 374 75...
It is neither terminating nor repeating decimal, therefore it is an irrational number.
(v) 0.638 754 71
It is neither terminating nor repeating decimal, therefore it is an irrational number.
(vi) 0.471 7171.... \( = 0.4\overline{71} \)
It is a repeating decimal, therefore it is a rational number.
(vii) 283
It is a terminating decimal (can be written as 283.0), therefore it is a rational number.
(viii) 289.387 000.....
It is a terminating decimal (since the zeros repeat indefinitely, it effectively ends), therefore it is a rational number.
(ix) 5.\overline{93}
It is a repeating decimal, therefore it is a rational number.
(x) 2.309\overline{8}7
It is a terminating decimal (the notation \( \overline{8}7 \) in this context is implied to mean it's a fixed part, making the whole number terminating), therefore it is a rational number.
(xi) 0.585 885 888...
It is a non-repeating decimal and non-terminating, therefore it is an irrational number.
In simple words: Rational numbers have decimals that stop or repeat a pattern, like 0.5 or 0.333... Irrational numbers have decimals that go on forever without any repeating pattern, like \( \sqrt{2} \) or \( \pi \).

🎯 Exam Tip: The key to classifying decimals is to observe if they terminate (end) or if they repeat a block of digits. If neither, they are irrational. Numbers like 283 can be written as 283.0, making them terminating decimals and thus rational.

Question 4. List three distinct irrational numbers.
Answer: We know that an irrational number has a non-terminating and non-repeating decimal expansion. Three distinct irrational numbers are \( \pi \), \( \sqrt{5} \), \( \sqrt{6} \), \( \sqrt{7} \) etc. For example, \( \sqrt{2} \), \( \sqrt{3} \), and \( \pi \) are common irrational numbers. The square roots of non-perfect squares are good examples.
In simple words: Irrational numbers are numbers whose decimal forms never end and never repeat, like \( \pi \) or the square root of any number that isn't a perfect square.

🎯 Exam Tip: Common examples of irrational numbers are \( \pi \) and square roots of prime numbers (e.g., \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \)). Any non-terminating, non-repeating decimal also works.

Question 5. Show that (i) \( \sqrt{3} \), (ii) \( \sqrt{5} \) are not rational numbers.
Answer:
(i) Let \( \sqrt{3} \) be a rational number.
Let \( \sqrt{3} = \frac{p}{q} \) where p and q are integers, have no common factors other than 1, and \( q \neq 0 \).
Squaring both sides,
\( 3 = \frac{p^2}{q^2} \)
\( \implies p^2 = 3q^2 \) ... (i)
Since \( 3q^2 \) is a multiple of 3, \( p^2 \) must also be a multiple of 3.
\( \implies p \) is a multiple of 3.
Let \( p = 3m \) for some integer m.
Squaring both sides,
\( p^2 = (3m)^2 \)
\( \implies p^2 = 9m^2 \) ... (iii)
From (i) and (iii), we have:
\( 3q^2 = 9m^2 \)
\( q^2 = 3m^2 \)
Since \( 3m^2 \) is a multiple of 3, \( q^2 \) must also be a multiple of 3.
\( \implies q \) is a multiple of 3 ... (iv)
From (ii) and (iv), both p and q are multiples of 3. This contradicts our initial assumption that p and q have no common factors other than 1.
Hence, \( \sqrt{3} \) is not a rational number.
Hence proved.
(ii) Let \( \sqrt{5} \) be a rational number.
Let \( \sqrt{5} = \frac{p}{q} \) where p and q are integers, have no common factors other than 1, and \( q \neq 0 \).
Squaring both sides,
\( 5 = \frac{p^2}{q^2} \)
\( \implies p^2 = 5q^2 \) ... (i)
Since \( 5q^2 \) is a multiple of 5, \( p^2 \) must also be a multiple of 5.
\( \implies p \) is a multiple of 5.
Let \( p = 5m \) for some integer m.
Squaring both sides,
\( p^2 = (5m)^2 \)
\( \implies p^2 = 25m^2 \) ... (iii)
From (i) and (iii), we have:
\( 5q^2 = 25m^2 \)
\( q^2 = 5m^2 \)
Since \( 5m^2 \) is a multiple of 5, \( q^2 \) must also be a multiple of 5.
\( \implies q \) is a multiple of 5 ... (iv)
From (ii) and (iv), both p and q are multiples of 5. This contradicts our initial assumption that p and q have no common factors other than 1.
Hence, \( \sqrt{5} \) is not a rational number.
Hence proved.
In simple words: To prove a number like \( \sqrt{3} \) or \( \sqrt{5} \) is irrational, you assume it IS rational (can be written as a fraction p/q), then show that this assumption leads to a contradiction, meaning it must be irrational.

🎯 Exam Tip: This is a classic proof by contradiction. The structure is key: Assume rational, set equal to p/q (coprime), square both sides, show p and q share a common factor, conclude contradiction, and state it's irrational. Use precise language for each step.

Question 6. Is \( \sqrt{100} + \sqrt{36} \) the same as \( \sqrt{100+36} \)? Give reasons.
Answer: No, \( \sqrt{100} + \sqrt{36} \) is not the same as \( \sqrt{100+36} \). The operation of square root does not distribute over addition.
Let's evaluate both expressions:
\( \sqrt{100} + \sqrt{36} = \sqrt{(10)^2} + \sqrt{(6)^2} = 10 + 6 = 16 \)
And
\( \sqrt{100+36} = \sqrt{136} \approx 11.66 \) (approx)
It is clear that \( \sqrt{100} + \sqrt{36} \) and \( \sqrt{100+36} \) are not the same because \( 16 \neq 11.66 \).
In simple words: You can't just add numbers inside a square root if they are separate square roots. You have to calculate each square root first, or add the numbers first, depending on how the problem is written.

🎯 Exam Tip: Remember that \( \sqrt{a} + \sqrt{b} \neq \sqrt{a+b} \) and \( \sqrt{a} - \sqrt{b} \neq \sqrt{a-b} \). Square roots can only be combined under multiplication or division (e.g., \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \)). Always perform operations inside the square root before taking the root, or take individual roots before combining if they are separate terms.