Class 9 NCERT Solutions- Chapter 1 Number System - Exercise 1.3

Chapter 1 of the Class 9 NCERT Mathematics textbook, titled "Number System," introduces students to various types of numbers, including real numbers, irrational numbers, and the representation of numbers on the number line. Exercise 1.3 focuses on the representation of irrational numbers on the number line and helps students understand how to approximate and locate these numbers using geometrical methods.

Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.3

This section provides detailed solutions for Exercise 1.3 from Chapter 1 of the Class 9 NCERT Mathematics textbook. The solutions guide students through the process of representing irrational numbers on the number line and understanding their properties.

NCERT Solutions for Class 9 Maths, Chapter 1, Number Systems, Exercise 1.3, meticulously crafted by our subject experts, facilitating effortless learning for students. These solutions serve as a valuable reference for students tackling exercise problems. Exercise 1.3 delves into the realm of real numbers and their decimal expansions, aiding in a comprehensive understanding of the topic.

1. Explore the NCERT Solutions for Exercise 1.3 in Chapter 1 of Class 9 Maths, focusing on Number Systems.

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

(i) 36/100 

Solution:

In the given question, we get

👁 Image

Here, the remainder becomes zero.

Hence, decimal expansion becomes terminating.

36/100 = 0.36

(ii) 1/11 

Solution:

In the given question, we get

👁 Image

Here, the remainder never becomes zero and remainders repeat after a certain stage.

Hence, decimal expansion is non-terminating recurring

1/11 = 

(iii) 

Solution:

Here, 

In the given question, we get

👁 Image

Here, the remainder becomes zero.

Hence, decimal expansion becomes terminating.

 = 4.125

(iv) 3/13 

Solution:

In the given question, we get

👁 Image

Here, the remainder never becomes zero and remainders repeat after a certain stage.

Hence, decimal expansion is non-terminating recurring

3/13 = 

(v) 2/11 

Solution:

In the given question, we get

👁 Image

Here, the remainder never becomes zero and remainders repeat after a certain stage.

Hence, decimal expansion is non-terminating recurring

2/11 =

(vi) 329/400 

Solution:

In the given question, we get

👁 Image

Here, the remainder becomes zero.

Hence, decimal expansion becomes terminating.

329/400 = 0.8225

Question 2. You know that = Can you predict what the decimal expansions of are, without actually doing the long division? If so, how?

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

Solution:

As it is given,

So, 

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

(i) 

Solution:

= 0.66666......

Lets's take, x = 0.66666......

10x = 6.666....

So, 

10x - x = (6.6666.....) - (0.66666........)

9x = 6

x = 6/9 

x = 2/3 

Hence, x is in the form p/q, here p and q are integers and q ≠ 0.

(ii) 

Solution:

 = 0.4777777......

Lets's take, x = 0.4777777......

10x = 4.77777.......

So,

10x - x = (4.77777.......) - (0.4777777......)

9x = 4.3

9x = 43/10 

x = 43/90

Hence, x is in the form p/q, here p and q are integers and q ≠ 0.

(iii) 

Solution:

 = 0.001001001......

Lets's take, x = 0.001001001......

1000x = 1.001001001......

So,

1000x - x = (1.001001001......) - (0.001001001......)

999x = 1

x = 1/999

Hence, x is in the form p/q, here p and q are integers and q ≠ 0.

Question 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.

Solution:

Lets's take, x = 0.99999......

10x = 9.99999....

So,

10x - x = (9.99999.....) - (0.99999........)

9x = 9

x = 1

As, 0.9999..... just goes on, then at some point of time there is no gap between 1 and 0.9999....

We can observe that, 0.999 is too much near 1, hence, 1 is justified as the answer.

Hence, x is in the form p/q, where p and q are integers and q ≠ 0.

Question 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.

Solution:

In the given question,

👁 Image

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

Here, the remainder never becomes zero and remainders repeat after a certain stage.

Hence, decimal expansion is non-terminating recurring

1/17 = 

Question 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?

Solution:

We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. 

Let's take some example,

1/2 = 0. 5, denominator q = 2¹

7/8 = 0. 875, denominator q =2³

4/5 = 0. 8, denominator q = 5¹

So, we conclude that 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.

In the form of 2^m × 5ⁿ, where n, m are natural numbers.

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

Solution:

As we know that all irrational numbers are non-terminating non-recurring. 

So,

√5 = 2.23606798.......

√27 =5.19615242......

√41 = 6.40312424.....

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

Solution:

As, decimal expansion of

5/7 = 

9/11 = 

Hence, three different irrational numbers between them can be as follows,

0.73073007300073000073…

0.75075007300075000075…

0.76076007600076000076…

Question 9. Classify the following numbers as rational or irrational :

(i) √23 

Solution:

√23 = 4.79583152......

As the number is non-terminating non-recurring.

It is an irrational number.

(ii) √225

Solution:

√225 = 15 = 15/1 

As the number can be represented in p/q form, where q ≠ 0.

It is a rational number.

(iii) 0.3796

Solution:

As, the number 0.3796, is terminating.

It is a rational number.

(iv) 7.478478...

Solution:

As, the number 7.478478, is non-terminating but recurring.

It is a rational number.

(v) 1.101001000100001...

Solution:

As, the number 1.101001000100001..., is non-terminating but recurring.

It is a rational number.

Class 9 Maths Number System:

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