NCERT Solutions for Class 9 Math's Chapter 1 Number System

Question

NCERT Solution for Class 9 Math Chapter 1 Real Number is one of the most important chapters of class 9. Our solution covers all the important topics related to Real Numbers with a detailed explanation that aims to help students to grasp the concepts with complete clarity. These solutions are precise and it is according to the latest syllabus provided by the CBSE. Our NCERT solutions are the best way to assist the students in their CBSE exam preparation and as well as for competitive exams like JEE Mains, JEE Advance or any other similar exams. These solutions are explained in the step-wise method.

NCERT solutions for class 9 Math's chapter 1 number systems begin with an introduction to the number system and the various types of numbers within it. Natural numbers, whole numbers, integers, rational numbers, irrational numbers, and so on are all types of numbers in the number system. The NCERT solutions class 9 Math's chapter 1 covers all of the fundamentals of the number system, which will aid in the formation of the fundamental foundation of mathematics. These NCERT Math's Solutions assist students in solving problems adroitly and efficiently for board exams. They also concentrate on formulating Math solutions that are simple for students to understand. The NCERT Solutions for Class 9 aim to provide students with detailed and step-by-step explanations for all of the answers to the questions in this Chapter's exercises.

According to the exam, these topics are important. Sarthaks experts recommend that learners go through all of the topics in NCERT Solutions Class 9 Math's to gain a complete understanding of this chapter. Every year, a large number of questions from this chapter are asked, and it carries a significant weightage in the CBSE board examination. Our solutions must be used for better understanding, answering questions, revising, completing assignments, and doing homework. Students can also find answers to NCERT intext questions, exercises, and questions from the back of the chapter. These solutions not only provide the necessary solutions but also provide you with a thorough understanding of all the associated concepts.

Answer 1

NCERT Solutions for Class 9 Math's Chapter 1 Number System

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

Answer:

We know that, a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as , 0/1, 0/2, 0/3 ..

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

2. Find six rational numbers between 3 and 4.

Answer:

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6 + 1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

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

Answer:

There are infinite rational numbers between 3/5 and 4/5.

To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5

with 5+1=6 (or any number greater than 5)

i.e., (3/5) × (6/6) = 18/30

and, (4/5) × (6/6) = 24/30

The numbers in between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5.

Hence,19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/5.

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

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Answer:

(i) Every natural number is a whole number.

True

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers = 1,2,3,4…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number.

(ii) Every integer is a whole number.

False

Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers.

i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…}

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

Every whole number is an integer; however, every integer is not a whole number.

(iii) Every rational number is a whole number.

False

Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.

i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers = 0,1,2,3….

Hence, we can say that integers includes whole numbers as well as negative numbers.

Every whole numbers are rational, however, every rational numbers are not whole numbers.

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

(i) 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.

(iii) Every real number is an irrational number.

Answer:

(i) Every irrational number is a real number.

True

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, 0.102…

Every irrational number is a real number, however, every real numbers are not irrational numbers.

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

False

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 = 3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

(iii) Every real number is an irrational number.

False

The statement is false, the real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, 0.102…

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5 +√2, 6.23146…. , 0.101001001000….

Every irrational number is a real number, however, every real number is not irrational.

6. 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.

Answer:

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

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

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

Answer:

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB² + BC² = CA²

2² + 1² = CA² = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose center was A.

Thus, √5 is represented on the number line as shown in the figure.

Answer 2

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

(i) \(\frac{36}{100}\)

(ii) \(\frac1{11}\)

(iii) \(4\frac18\)

(iv) \(\frac 3{13}\)

(v) \(\frac2{11}\)

(vi) \(\frac{329}{400}\)

Answer:

(i) \(\frac{36}{100}\) 

= 0.36 (Terminating)

(ii) \(\frac1{11}\)

= 0.0909... = \(0.\overline{09}\) (non terminating and repeating)

(iii) \(4\frac18\)

= 4.125 (Terminating)

(iv) \(\frac 3{13}\) 

(v) \(\frac2{11}\) 

(non terminating and repeating)

(vi) \(\frac{329}{400}\)

= 0.8225 (Terminating)

9. 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?

[Hint: Study the remainders while finding the value of 1/7 carefully.]

Answer:

10. Express the following in the form p/q, where p and q are integers and q \(\ne\) 0.

(i) \(0.\bar 6\) 

(ii) \(0.\overline {47}\) 

(iii) \(0. \overline{001}\)

Answer:

(i) \(0.\bar 6\) 

\(0.\bar6 = 0.666..\)

Assume that x = 0.666…

Then,10x = 6.666…

10x = 6 + x

9x = 6

x = 2/3

(ii) \(0.\overline {47}\) 

\(0.\overline {47} = 0.4777..\) 

= (4/10)+(0.777/10)

Assume that x = 0.777…

Then, 10x = 7.777…

10x = 7 + x

x = 7/9

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

(iii) \(0. \overline{001}\) 

\(0. \overline{001} = 0.001001..\)

Assume that  x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 1

x = 1/999

11. 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.

Answer:

Assume that x = 0.9999…..   Eq. (a)

Multiplying both sides by 10,

10x = 9.9999….   Eq. (b)

Eq.(b) – Eq.(a), we get

10x = 9.9999

–x = -0.9999…

9x = 9

x = 1

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

12. 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.

Answer:

1/17

Dividing 1 by 17:

There are 16 digits in the repeating block of the decimal expansion of 1/17.

13. 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?

Answer:

We observe that when q is 2, 4, 5, 8, 10… 

Then the decimal expansion is terminating. 

For example:

1/2 = 0. 5, denominator q = 2¹

7/8 = 0. 875, denominator q =2³

4/5 = 0. 8, denominator q = 5¹

We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

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

Answer:

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

1. √3 = 1.732050807568
2. √26 =5.099019513592
3. √101 = 10.04987562112

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

Answer:

Three different irrational numbers are:

1. 0.73073007300073000073…
2. 0.75075007300075000075…
3. 0.76076007600076000076…

Answer 3

16. Classify the following numbers as rational or irrational according to their type:

(i) √23

(ii) √225

(iii) 0.3796

(iv) 7.478478

(v) 1.101001000100001…

Answer:

(i) √23

√23 = 4.79583152331…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii) √225

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

17. Visualise 3.765 on the number line, using successive magnification.

Answer:

18. Visualise \(4.\overline{26}\) on the number line, up to 4 decimal places.

Answer:

\(4.\overline{26} = 4.26262626....\) 

\(4.\overline{26}\) up to 4 decimal places = 4.2626

19. Classify the following numbers as rational or irrational:

(i) 2 –√5

(ii) (3 +√23)- √23

(iii) 2√7/7√7

(iv) 1/√2

(v) 2

Answer:

(i) 2 –√5

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2 - √5 = 2 - 2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 – √5 is an irrational number.

(ii) (3 +√23)- √23

(3 + √23) – √23 = 3 + √23 – √23

= 3

= 3/1

Since the number 3/1 is in p/q form, (3 + √23) - √23 is rational.

(iii) 2√7/7√7

2√7/7√7 = ( 2/7) × (√7/√7)

We know that (√7/√7) = 1

Hence, (2/7) × (√7/√7) = (2/7) × 1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2 × √2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2

We know that, the value of = 3.1415

Hence, 2 = 2 × 3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

20. Simplify each of the following expressions:

(i) (3 + √3)(2 + √2)

(ii) (3 + √3)(3 - √3)

(iii) (√5 + √2)²

(iv) (√5 - √2)(√5 + √2)

Answer:

(i) (3 + √3)(2 + √2)

(3 + √3)(2 + √2)

Opening the brackets, we get, (3 × 2) + (3 × √2) + (√3 × 2) + (√3 × √2)

= 6 + 3√2 + 2√3 + √6

(ii) (3 + √3)(3 - √3)

(3 + √3)(3 - √3 ) = 3² - (√3)² = 9 - 3

= 6

(iii) (√5 + √2)²

(√5 + √2)² = √5² + (2 × √5 × √2) + √2²

= 5 + 2 × √10 + 2 

= 7 + 2√10

(iv) (√5 - √2)(√5 + √2)

(√5 - √2)(√5 + √2) = (√5² - √2²)

= 5 - 2 

= 3

Answer 4

21. 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?

Answer:

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

22. Represent \(\sqrt{9.3}\) on the number line.

Answer:

Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.

Find mid point of AC and mark it as O.

Draw a semicircle taking O as centre and AO as radius. Draw BD ⊥ AC.

Draw an arc taking B as centre and BD as radius meeting AC produced at E such that BE = BD = \(\sqrt{9.3}\) units.

23. Rationalize the denominators of the following:

(i) \(\frac1{\sqrt 7}\)

(ii) \(\frac1{\sqrt7 - \sqrt 6}\)

(iii) \(\frac1{\sqrt5 + \sqrt 2}\)

(iv) \(\frac1{\sqrt7 - 2}\)

Answer:

24. Find:

(i) 64^1/2

(ii) 32^1/5

(iii) 125^1/3

Answer:

(i) 64^1/2

64^1/2 = (8 × 8)^1/2

= (8²)^½

= 8¹    [⸪2 × 1/2 = 2/2 = 1]

= 8

(ii) 32^1/5

32^1/5 = (2⁵)^1/5

= (2⁵)^⅕

= 2¹    [⸪5 × 1/5 = 1]

= 2

(iii) 125^1/3

(125)^1/3 = (5 × 5 × 5)^1/3

= (5³)^⅓

= 5¹ (3 × 1/3 = 3/3 = 1)

= 5

25. Find:

(i) 9^3/2

(ii) 32^2/5

(iii)16^3/4

(iv) 125^-1/3

Answer:

(i) 9^3/2

9^3/2 = (3 × 3)^3/2

= (3²)^3/2

= 3³   [⸪ 2 × 3/2 = 3]

= 27

(ii) 32^2/5

32^2/5 = (2 × 2 × 2 × 2 × 2)^2/5

= (2⁵)^2/5

= 2²     [⸪5 × 2/5 = 2]

= 4

(iii)16^3/4

16^3/4 = (2 × 2 × 2 × 2)^3/4

= (2⁴)^3/4

= 2³   [⸪ 4 × 3/4 = 3]

= 8

(iv) 125^-1/3

125^-1/3 = (5 × 5 × 5)^-1/3

= (5³)^-1/3

= 5^-1   [⸪ 3 × -1/3 = -1]

= 1/5