NCERT Solutions - Class 9 Mathematics

Chapter 1: Number Systems
Exercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
Exercise 1.5

Exercise 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Yes, zero is a rational number. We can write it as 0/1, 0/2, 0/3, and so on. In all these forms, p=0 and q is a non-zero integer.

2. Find six rational numbers between 3 and 4.

We can find rational numbers between 3 and 4 by converting them to fractions with a common denominator:

3 = 21/7 and 4 = 28/7

Six rational numbers between 3 and 4 are:

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

We can also write them as decimals: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

3. Find five rational numbers between 3/5 and 4/5.

To find rational numbers between 3/5 and 4/5, we can convert them to fractions with a larger denominator:

3/5 = 18/30 and 4/5 = 24/30

Five rational numbers between 3/5 and 4/5 are:

19/30, 20/30, 21/30, 22/30, 23/30

These can be simplified to: 19/30, 2/3, 7/10, 11/15, 23/30

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

True. Natural numbers are 1, 2, 3, ... and whole numbers are 0, 1, 2, 3, ... Every natural number is included in the set of whole numbers.

(ii) Every integer is a whole number.

False. Integers include negative numbers like -1, -2, -3, ..., while whole numbers are only 0, 1, 2, 3, ... Negative integers are not whole numbers.

(iii) Every rational number is a whole number.

False. Rational numbers include fractions like 1/2, 3/4, etc., which are not whole numbers. Whole numbers are only non-negative integers.

Exercise 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

True. Real numbers include both rational and irrational numbers. So every irrational number is a real number.

(ii) Every point on the number line is of the form √m, where m is a natural number.

False. Points on the number line can represent rational numbers, irrational numbers that are not square roots (like π, e), and other types of numbers. Not every point corresponds to √m where m is a natural number.

(iii) Every real number is an irrational number.

False. Real numbers include both rational and irrational numbers. Rational numbers like 2, 1/2, 0.75 are real numbers but not irrational.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

No, the square roots of all positive integers are not irrational. For example:

√1 = 1 (rational)

√4 = 2 (rational)

√9 = 3 (rational)

√16 = 4 (rational)

These are rational numbers. Only the square roots of perfect squares are rational; the square roots of other positive integers are irrational.

3. Show how √5 can be represented on the number line.

To represent √5 on the number line:

  1. Draw a number line and mark point O at 0 and point A at 2.
  2. At point A, draw a perpendicular line of length 1 unit and mark the endpoint as B.
  3. Join O to B. The length OB is √(2² + 1²) = √(4 + 1) = √5.
  4. Using a compass with center O and radius OB, draw an arc that intersects the number line at point C.
  5. Point C represents √5 on the number line.
4. Classroom activity (Constructing the 'square root spiral')

This is a hands-on activity to construct a square root spiral:

  1. Start with point O and draw a line segment OP₁ of unit length.
  2. Draw a line segment P₁P₂ perpendicular to OP₁ of unit length.
  3. Now draw a line segment P₂P₃ perpendicular to OP₂.
  4. Continue this process, each time drawing a line segment of unit length perpendicular to the previous hypotenuse.
  5. The lengths of the hypotenuses will be √2, √3, √4, etc.
  6. Joining these points creates a beautiful spiral called the square root spiral.

Exercise 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has:

(i) 36/100 = 0.36 (Terminating decimal)

(ii) 1/11 = 0.090909... = 0.09 (Non-terminating repeating decimal)

(iii) 4/8 = 1/2 = 0.5 (Terminating decimal)

(iv) 3/13 = 0.230769230769... = 0.230769 (Non-terminating repeating decimal)

(v) 2/11 = 0.181818... = 0.18 (Non-terminating repeating decimal)

(vi) 329/400 = 0.8225 (Terminating decimal)

2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

Yes, we can predict the decimal expansions by observing the pattern in 1/7 = 0.142857:

2/7 = 2 × 0.142857 = 0.285714

3/7 = 3 × 0.142857 = 0.428571

4/7 = 4 × 0.142857 = 0.571428

5/7 = 5 × 0.142857 = 0.714285

6/7 = 6 × 0.142857 = 0.857142

All these fractions have the same repeating block of digits (142857), just starting at different positions.

3. Express the following in the form p/q, where p and q are integers and q ≠ 0:

(i) 0.6 (with 6 repeating)

Let x = 0.666...

Then 10x = 6.666...

Subtracting: 10x - x = 6.666... - 0.666...

9x = 6 ⇒ x = 6/9 = 2/3

(ii) 0.47 (with 7 repeating)

Let x = 0.4777...

Then 10x = 4.777... and 100x = 47.777...

Subtracting: 100x - 10x = 47.777... - 4.777...

90x = 43 ⇒ x = 43/90

(iii) 0.001 (with 001 repeating)

Let x = 0.001001...

Then 1000x = 001.001001... = 1.001001...

Subtracting: 1000x - x = 1.001001... - 0.001001...

999x = 1 ⇒ x = 1/999

4. Express 0.99999 ... in the form p/q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Let x = 0.999...

Then 10x = 9.999...

Subtracting: 10x - x = 9.999... - 0.999...

9x = 9 ⇒ x = 1

So 0.999... = 1

This might seem surprising at first, but it makes sense because 0.999... is infinitely close to 1. In mathematics, 0.999... is exactly equal to 1, not just approximately.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

The maximum number of digits in the repeating block of 1/17 can be at most 16. This is because when we divide 1 by 17, the remainders can be from 1 to 16. Once we get a remainder of 0 or a remainder we've seen before, the decimal either terminates or starts repeating.

Performing the division: 1/17 = 0.0588235294117647 (repeating)

The repeating block has 16 digits, which is the maximum possible.

6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

For a rational number p/q (in simplest form) to have a terminating decimal expansion, the denominator q must be of the form 2^m × 5^n, where m and n are non-negative integers.

Examples:

1/2 = 0.5 (q = 2 = 2¹ × 5⁰)

3/4 = 0.75 (q = 4 = 2² × 5⁰)

7/8 = 0.875 (q = 8 = 2³ × 5⁰)

13/125 = 0.104 (q = 125 = 2⁰ × 5³)

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Three numbers with non-terminating non-recurring decimal expansions are:

1. √2 = 1.4142135623...

2. π = 3.1415926535...

3. e = 2.7182818284...

These are irrational numbers.

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

First, let's find the decimal approximations:

5/7 ≈ 0.7142857143...

9/11 ≈ 0.8181818182...

Three irrational numbers between them could be:

1. 0.72072007200072000072... (pattern of increasing zeros)

2. 0.75075007500075000075... (pattern of increasing zeros)

3. 0.808008000800008... (pattern of increasing zeros)

These are irrational because they are non-terminating and non-repeating.

9. Classify the following numbers as rational or irrational:

(i) √23 - Irrational (23 is not a perfect square)

(ii) √225 = 15 - Rational (225 is a perfect square)

(iii) 0.3796 - Rational (terminating decimal)

(iv) 7.478478... = 7.478 - Rational (repeating decimal)

(v) 1.101001000100001... - Irrational (non-terminating, non-repeating pattern)

Exercise 1.4

1. Classify the following numbers as rational or irrational:

(i) 2 - √5 - Irrational (difference of rational and irrational)

(ii) 3 + √23 - √23 = 3 - Rational

(iii) (2√7)/(7√7) = 2/7 - Rational

(iv) 1/√2 - Irrational

(v) 2π - Irrational (product of rational and irrational)

2. Simplify each of the following expressions:

(i) 3 + √3(2 + √2) = 3 + 2√3 + √6

(ii) 3 + √3(3 - √3) = 3 + 3√3 - 3 = 3√3

(iii) (√5 + √2)² = 5 + 2√10 + 2 = 7 + 2√10

(iv) (√5 - √2)(√5 + √2) = 5 - 2 = 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

There is no contradiction. The definition π = c/d doesn't imply that both c and d are integers. In fact, for any circle, if the diameter is rational, the circumference will be irrational, and vice versa. So π is still irrational despite being expressed as a ratio.

4. Represent √9.3 on the number line.

To represent √9.3 on the number line:

  1. Mark point A at 9.3 on the number line.
  2. Extend the number line to the left of 0 and mark point B at -1.
  3. Find the midpoint M of AB. M = (9.3 + (-1))/2 = 8.3/2 = 4.15.
  4. With M as center and MA as radius, draw a semicircle.
  5. At point O (0), draw a perpendicular that intersects the semicircle at point C.
  6. OC = √9.3.
  7. With O as center and OC as radius, draw an arc that intersects the number line at point D.
  8. Point D represents √9.3 on the number line.
5. Rationalise the denominators of the following:

(i) 1/√7 = (1/√7) × (√7/√7) = √7/7

(ii) 1/(√7 - √6) = 1/(√7 - √6) × (√7 + √6)/(√7 + √6) = (√7 + √6)/(7 - 6) = √7 + √6

(iii) 1/(√5 + √2) = 1/(√5 + √2) × (√5 - √2)/(√5 - √2) = (√5 - √2)/(5 - 2) = (√5 - √2)/3

(iv) 1/(√7 - 2) = 1/(√7 - 2) × (√7 + 2)/(√7 + 2) = (√7 + 2)/(7 - 4) = (√7 + 2)/3

Exercise 1.5

1. Find:

(i) 64^(3/2) = (√64)³ = 8³ = 512

(ii) 32^(1/5) = ⁵√32 = 2

(iii) 125^(1/3) = ³√125 = 5

2. Find:

(i) 9^(3/2) = (√9)³ = 3³ = 27

(ii) 32^(1/5) = ⁵√32 = 2

(iii) 16^(1/4) = ⁴√16 = 2

(iv) 125^(-1/3) = 1/(125^(1/3)) = 1/³√125 = 1/5

3. Simplify:

(i) 2^(3/2) × 2^(1/2) = 2^(3/2 + 1/2) = 2² = 4

(ii) (1/3)⁷ = 1/3⁷ = 1/2187

(iii) 11^(1/2) / 11^(1/4) = 11^(1/2 - 1/4) = 11^(1/4)

(iv) 7^(1/2) × 8^(1/2) = (7 × 8)^(1/2) = 56^(1/2) = √56 = 2√14

NCERT Solutions for Class 9 Mathematics - Number Systems