Courses
By rohit.pandey1
|
Updated on 19 Feb 2026, 13:43 IST
Looking for NCERT Solutions Class 9 Number System? You’re in the right place. At Infinity Learn, we provide detailed and easy-to-understand Class 9 Maths Chapter 1 Solutions designed as per the latest CBSE Class 9 Maths syllabus and exam guidelines.
The chapter Number System Class 9 NCERT lays the foundation for higher mathematics. Our expert-curated solutions help students clearly understand every concept covered in NCERT Class 9 Maths Chapter 1, including rational numbers, irrational numbers, real numbers, and laws of exponents.
The Class 9 Number Systems NCERT Solutions available on Infinity Learn are prepared by subject experts to ensure:
These solutions help students solve textbook questions accurately and confidently for board exams.
Students can download the Class 9 Maths Chapter 1 Solutions PDF for free and practice offline. The PDF includes:
This makes revision faster and more structured before school exams and future competitive exams.
Loading PDF...
Question 1: Is zero a rational number? Can you write it in the form pq where p and q are integers and q ≠ 0?
Answer:
Yes, zero is a rational number. It can be written in the form pq.
For example: 01, 02, 05 etc. are rational numbers where p and q are integers and q ≠ 0.
Question 2: Find six rational numbers between 3 and 4.
Answer:
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
To get six rational numbers between 3 and 4, the denominator must be 6 + 1 = 7
Here, 3 = (3 × 7) ÷ 7 = 21 ÷ 7 and 4 = (4 × 7) ÷ 7 = 287
So, the six rational can be obtained by changing numerator from 22 to 27.
Therefore, six rational numbers are: 227, 237, 247, 257, 267, 277
Question 3: Find five rational numbers between 35 and 45.
Answer:
By converting these numbers into decimal, we have
35 = 0.6 and 45 = 0.8
Hence, the five rational numbers between 35 and 45 are: 0.61, 0.62, 0.63, 0.64, 0.65.
Question 4: State whether the following statements are true or false. Give reasons for your answers.
Answer:
(1) True statement.
The collection of all natural numbers and 0 is called whole numbers.
(2) False statement.
Integers such as −1, −2 are non-whole numbers.
(3) False statement.
Rational numbers like 12, 15, 27, etc. are not a whole number.
Question 1: State whether the following statements are true or false. Justify your answers.
Answer:
(1) True statement, because all rational numbers and all irrational numbers form the group (collection) of real numbers.
(2) False statement, because no negative number can be the square root of any natural number.
(3) False statement, because rational numbers are also a part of real numbers.
Question 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.
Answer:
No, if we take a positive integer, say 4 its square root is 2, which is a rational number.
According to the Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In the figure:
OB2 = OA2 + AB2
=> OB2 = 12 + 12
=> OB2 = 2
=> OB = √2
Question 3: Show how √5 can be represented on the number line.
Answer:
Let us take the horizontal line XOX′ as the x-axis. Mark O as its origin such that it represents 0.
Cut off OA = 1 unit, AB = 1 unit.
=> OB = 2 units
Draw a perpendicular BC ⊥ OX
Cut off BC = 1 unit.
Since OBC is a right triangle.
OB2 + BC2 = OC2
=> 22 + 12 = OC2
=> OC2 = 4 + 1
=> OC2 = 5
=> OC = √5
With O as centre and OC as radius, draw an arc intersecting OX at D.
Since OC = OD
So, OD represents √5 on XOX′.
Question 1: Write the following in decimal form and say what kind of decimal expansion each has:
Answer:
(1) 36100 = 0.36
So, the decimal expansion of 36100 is terminating.
(2) 111 = 0.090909… = 0.09
Thus, the decimal expansion of 111 is non-terminating repeating.
(3) 418 = 4 + 18 = 338 = 4.125
Thus, the decimal expansion of 418 is terminating.
(4) 313 = 0.230769230769… = 0.230769
Thus, the decimal expansion of 313 is non-terminating repeating.
(5) 211 = 0.181818… = 0.18
Thus, the decimal expansion of 211 is non-terminating repeating.
(6) 329400 = 0.8225
Thus, the decimal expansion of 329400 is terminating.
Question 2: You know that 17 = 0.142857. Can you predict what the decimal expansions of 27, 37, 47, 57, 67 are, without actually doing the long division? If so, how?
Answer:
Given, 17 = 0.142857
Now,
27 = 2 × (17) = 2 × 0.142857 = 0.285714
37 = 3 × (17) = 3 × 0.142857 = 0.428571
47 = 4 × (17) = 4 × 0.142857 = 0.571428
57 = 5 × (17) = 5 × 0.142857 = 0.714285
67 = 6 × (17) = 6 × 0.142857 = 0.857142
Thus, without actually doing the long division we can predict the decimal expansions of the above given rational numbers.
Question 3: Express the following in the form pq, where p and q are integers and q ≠ 0.
Answer:
(1) Let x = 0.6
=> x = 0.66666…………… …(1)
Multiply equation (1) by 10 on both sides, we get
10x = 6.6666….
=> 10x = 6 + 0.6666…
=> 10x = 6 + x [From equation (1)]
=> 10x − x = 6
=> 9x = 6
=> x = 69 = 23
(2) Let x = 0.47
=> x = 0.477777…………… …(1)
Multiply equation (1) by 10 on both sides, we get
10x = 4.7777………… …(2)
Multiply equation (2) by 10 on both sides, we get
=> 100x = 47 + 0.7777…
=> 100x = 43 + 4 + 0.7777…
=> 100x = 43 + 4.7777…
=> 100x = 43 + 10x [From equation (2)]
=> 100x − 10x = 43
=> 90x = 43
=> x = 4390
(3) Let x = 0.001
=> x = 0.001001001…………… …(1)
Multiply equation (1) by 1000 on both sides, we get
1000x = 1 + 0.001001…
=> 1000x = 1 + x [From equation (1)]
=> 1000x − x = 1
=> 999x = 1
=> x = 1999
Question 4: Express 0.99999 . . . in the form pq. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer:
Let x = 0.99999…………… …(1)
Multiply equation (1) by 10 on both sides, we get
10x = 9.99999….
=> 10x = 9 + 0.99999…
=> 10x = 9 + x [From equation (1)]
=> 10x − x = 9
=> 9x = 9
=> x = 99 = 1
The answer makes sense as 0.99999… is very close to 1. That’s why we can say that 0.99999… = 1
Question 5: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117? Perform the division to check your answer.
Answer:
Since, the number of entries in the repeating block of digits is less than the divisor. In 117 the divisor is 17.
So, the maximum number of digits in the repeating block is 16.
To perform the long division, we have
117 = 0.588235294117647
Thus, there are 16 digits in the repeating block in the decimal expansion of 117. Hence, our answer is verified.
Question 6: Look at several examples of rational numbers in the form pq (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:
Let some examples are:
25 = 0.4, 110 = 0.1, 32 = 1.5, 78 = 0.875
The denominator of all the rational numbers are in the form 2m × 5n where m and n are integers.
Question 7: Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
Three numbers whose decimal expansions are non-terminating non-recurring are:
√2 = 1.414213562….
√3 = 1.732050808….
√5 = 2.236067978….
Question 8: Find three different irrational numbers between the rational numbers 57 and 911.
Answer:
57 = 0.714285714285… = 0.714285
911 = 0.818181…… = 0.81
We know that there are infinitely many irrational numbers between two rational numbers.
So, the three irrational numbers are:
1. 0.72722722272222…..
2. 0.73733733373333…..
3. 0.74744744474444….
Question 9: Classify the following numbers as rational or irrational:
Answer:
(1) Since 23 is not a perfect square, so √23 is an irrational number.
(2) 225 = 15 × 15 = 152
So, 225 is a perfect square. Thus, √225 is a rational number.
(3) Since 0.3796 is a terminating decimal, so it is a rational number.
(4) 7.478478 = 7.478
Since it is a non-terminating and recurring (repeating) decimal, so it is a rational number.
(5) Since 1.101001000100001... is a non-terminating and non-repeating decimal number, so it is an irrational number.
Question 1: Visualize 3.765 on the number line, using successive magnification.
Answer:
1. First of all, we observe that 3.765 lies between 3 and 4. Divide this portion in 10 equal parts.
2. In the next step, we locate 3.765 between 3.7 and 3.8
3. To get more accurate visualization of representation, we divide this portion of number line into 10 equal parts and use a magnifying glass to visualize that 3.765 lies between 3.76 and 3.77.
4. Now to visualize 3.765 still more accurately, we divide the portion between 3.76 and 3.77 into 10 equal parts and locate 3.765.
Question 2: Visualize 4.26 on the number line up to 4 decimal places.
Answer:
1. First of all, we observe that 4.2626 (4.26) lies between 4 and 5. Divide this portion into 10 equal parts.
2. In the next step, we locate 4.2626 between 4.2 and 4.3
3. To get more accurate visualization of representation, we divide this portion of number line into 10 equal parts and use a magnifying glass to visualize that 4.2626 lies between 4.262 and 4.263.
4. Now to visualize 4.2626 still more accurately, we divide the portion between 4.262 and 4.263 into 10 equal parts and locate 4.2626.
Question 1: Classify the following numbers as rational or irrational:
Answer:
(1) 2 − √5
Since it is a difference of a rational and irrational number, so 2 − √5 is an irrational number.
(2) (3 + √23) − √23 = 3 + √23 − √23 = 3
Which is a rational number.
(3) 2√77√7 = (2 × √7) ÷ (7 × √7) = 27
Which is a rational number.
(4) 1√2
The quotient of rational and irrational is an irrational number. So, 1√2 is an irrational number.
(5) 2π = 2 × π = Product of a rational and an irrational (which is an irrational number)
So, 2π is an irrational number.
Question 2: Simplify each of the following expressions:
Answer:
(1) (3 + √3)(2 + √2) = 3(2 + √2) + √3(2 + √2)
= 3 × 2 + 3 × √2 + √3 × 2 + √3 × √2
= 6 + 3√2 + 2√3 + √6
(2) (3 + √3)(3 − √3) = 3(3 − √3) + √3(3 − √3)
= 3 × 3 − 3 × √3 + √3 × 3 − √3 × √3
= 9 − 3√3 + 3√3 − 3
= 6
(3) (√5 + √2)2 = (√5 + √2)(√5 + √2)
= √5(√5 + √2) + √2(√5 + √2)
= √5 × √5 + √5 × √2 + √2 × √5 + √2 × √2
= 5 + √10 + √10 + 2
= 7 + 2√10
(4) (√5 − √2)(√5 + √2) = √5(√5 + √2) − √2(√5 + √2)
= √5 × √5 + √5 × √2 − √2 × √5 − √2 × √2
= 5 + √10 − √10 − 2
= 3
Question 3: Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = cd. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Answer:
When we measure the length of a line with a scale or with any other device, we only get an approximate rational value, i.e., c and d both are irrational.
=> cd is irrational and hence π is irrational.
Thus, there is no contradiction in saying that π is irrational.
Question 4:
Represent √9.3 on the number line.
Answer:
To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C such that BC = 1. Draw the perpendicular bisector of AC which intersects AC at O. Taking O as center and OA as radius, draw a semi-circle which intersects D at the perpendicular at B. Now taking O as center and OD as radius, draw an arc which intersects AC produced at E. Hence, OE = √9.3.
Question 5: Rationalize the denominators of the following:
Answer:
(1) 1√7 = 1√7 × √7√7 = (1 × √7) ÷ (√7 × √7) = √77
(2) 1(√7 − √6) = 1(√7 − √6) × (√7 + √6)(√7 + √6)
= (√7 + √6) ÷ {(√7)2 − (√6)2}
= (√7 + √6) ÷ (7 − 6)
= (√7 + √6)
(3) 1(√5 + √2) = 1(√5 + √2) × (√5 − √2)(√5 − √2)
= (√5 − √2) ÷ {(√5)2 − (√2)2}
= (√5 − √2) ÷ (5 − 2)
= (√5 − √2)3
(4) 1(√7 − 2) = 1(√7 − 2) × (√7 + 2)(√7 + 2)
= (√7 + 2) ÷ {(√7)2 − 22}
= (√7 + 2) ÷ (7 − 4)
= (√7 + 2)3
Question 1: Find:
Answer:
(1) 641/2 = (82)1/2 = 82 × 1/2 = 81 = 8
(2) 321/5 = (25)1/5 = 25 × 1/5 = 21 = 2
(3) 1251/3 = (53)1/3 = 53 × 1/3 = 51 = 5
Question 2: Find:
Answer:
(1) 93/2 = (32)3/2 = 32 × 3/2 = 33 = 27
(2) 322/5 = (25)2/5 = 25 × 2/5 = 22 = 4
(3) 163/4 = (24)3/4 = 24 × 3/4 = 23 = 8
(4) 125−1/3 = (53)−1/3 = 53 × (−1/3) = 5−1 = 15
Question 3: Simplify:
Answer:
(1) 22/3 × 21/5 = 2(2/3 + 1/5) = 2(10 + 3)/15 = 213/15
(2) (133)7 = (3−3)7 = 3−3 × 7 = 3−21
(3) 111/2111/4 = 111/2 × 11−1/4 = 111/2 − 1/4 = 11(2−1)/4 = 111/4
(4) 71/2 × 81/2 = (7 × 8)1/2 = 561/2
No courses found
The chapter Number System Class 9 NCERT covers rational numbers, irrational numbers, real numbers, representation of real numbers on the number line, laws of exponents, and decimal expansions. These concepts form the foundation for Class 10 Mathematics and competitive exams.
Students can download the NCERT Solutions Class 9 Number System PDF from Infinity Learn, it provides chapter-wise, step-by-step solutions aligned with the latest CBSE syllabus. Always ensure the solutions cover all exercises from 1.1 to 1.6.
Class 9 Maths Chapter 1 Solutions are usually sufficient for school exams because most questions are directly based on NCERT exercises. Practicing all in-text and exercise problems thoroughly helps in scoring high marks.
To understand irrational numbers, focus on their definition (non-terminating, non-repeating decimals) and practice representing them on the number line. Referring to detailed Class 9 Number Systems NCERT Solutions with step-wise explanations makes the concept clearer.
The CBSE Class 9 Maths Number System chapter builds the base for topics like polynomials, quadratic equations, and real numbers in Class 10. A strong understanding helps students perform better in board exams and competitive tests like JEE and NTSE.