# CBSE Class 9 Maths Chapter 1 Solution

Class 9 Maths Chapter 1 Solution “Number system”
“Number Systems” in Class 9 Maths typically covers the following topics:

Place value of digits in a number
Understanding of different number systems (binary, decimal, octal, and hexadecimal)
Conversion of numbers from one number system to another (e.g. decimal to binary, binary to decimal)
Addition, subtraction, multiplication, and division of numbers in different number systems
Understanding of negative binary numbers and two’s complement representation of negative numbers
It is important to understand the place value of digits and the concepts of different number systems to effectively perform operations and conversions between them.

# CBSE NCERT Solutions for Class 9 Mathematics Chapter 1

## 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?

Solution:

2. Find six rational numbers between 3 and 4.
Solution:

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

4. State whether the following statements are true or false. Give reasons for your
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.

Solution:

(i) True, because we can say that whole numbers are nothing but natural
numbers plus zero. Therefore, every natural number is a whole number but
every whole number is not a natural number. As 0 is not a natural number.

(ii) False, because integers include both positive and negative numbers. The
whole numbers include only positive numbers and negative numbers are not
whole numbers. Therefore, every integer is a not a whole number.

(iii) False, because rational numbers can also be in the form of fractions and
these fractional numbers are not whole numbers. For example, 2/2, 3/5 ,4/3. are
not whole numbers but they are rational numbers.

### Exercise: 1.2

###### (iii) False, because real numbers contain both rational and irrational numbers. Therefore, every real number cannot be 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.

Solution:

No, the square root of all positive numbers need not be irrational
For example, √4 = 2 and √9 = 3.
Here, 2 and 3 are rational.
Therefore, the square roots of all positive integers are not irrational.

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

solution:

### Exercise: 1.3

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

solution:1

question:2

question:3

solution:3

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:4

question:5

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:5
By performing the actual division operation, we see that, 1/17 =0.0588235294117647.
Therefore, the maximum number digits that can be in the repeating block of digits
in the above expansion are 16.

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 terminated decimal
representations (expansions). Can you guess what property q must satisfy?

solution:6

Let us look at some examples:
5/4= 1.25
21/8= 2.625
34/5= 6.8
From the above examples we may generally conclude that, terminating decimal
expansion will occur when denominator ‘q’ of rational number p /q are in the form,
2a × 5b, (2 to the power a and 5 to he power b ) where ‘a’ and ‘b’ are integers.

question:7

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

Solution:7

There are infinite numbers of non-terminating and non-recurring decimals. Further,
we observe that all irrational numbers are non-terminating. Some examples are,
(i) 0.645238456364…
(ii) √3 = 1.73205087…
(iii) 0.7235432436…

question:8

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

solution:8

By performing long division, the numbers can be represented as
5/7= 0.714285… and 9/11= 0.81818181…

Now, any three numbers between these numbers will satisfy the given question.
Therefore, the required numbers can be:
(i) 0.72534345029…
(ii) 0.7523028734 …
(iii) 0.77623402347 …

9. Classify the following numbers as rational or irrational:
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001

Solution:9

(i) √23 = 4.79583152331…
In the above expansion, the number is non-terminating and non-recurring,
therefore, it is an irrational number.

(ii) √225 = 15 =15/1
As the above number can be represented in p/q form, it is a rational number.

(iii) 0.3796
As the above number is terminating, it is a rational number.

(iv) 7.478478 … = 7.478
From the decimals it is a recurring number, but it is non-terminating.
Therefore, it is a rational number.

(v) 1.101001000100001
We see that the decimal expansion of this number is non-repeating and
non-recurring. Therefore, it is an irrational number.

### Exercise: 1.4

###### (v) To locate this, we divide the portion between 3.76 and 3.77 into 10 equal parts and hence locate 3.765

2. Visualize 4.�on the number line, up to 4 decimal places.

solution:
To visualize 4. ��on the number line, we must follow these steps:

(i) First, we see that 4.2 lies between 4 and 5.

(ii) Now, divide this portion into 10 equal parts.

(iii) Next, we locate 4.26. We observe that this lies between 4.2 and 4.3.

(iv) To get a more accurate visualization, we further divide this portion into 10
parts and locate it.

(v) Further, we visualize 4.262 and observe that it lies between 4.26 and
4.27.

(vi) To locate this, we divide the portion between 4.26 and 4.27 into 10 equal
parts and locate 4.262

(vii) We observe that 4.2626 lies between 4.262 and 4.263.

(viii) To find this, we divide the portion further into 10 parts and hence locate
4.2626.

### Exercise: 1.5

1. Classify the following numbers as rational or irrational:
(i) 2 − √5
(ii) �3 + √23�− √23
(iii) 2√7/7√7
(iv) 1/√2
(v) 2π

Solution:

(i) 2 − √5 = 2 − 2.2360679… = −0.2360679…
We see that the decimal expansion of this expression is non-terminating and
non-recurring. Therefore, it is an irrational number.

(ii) �3 + √23�− √23 = 3 =3/1

We see that the number can be represented in p/q form. Therefore, it is a
rational number.

(iii) 2√7/7√7 =2/7
As it can be represented in p/q form, therefore, it is a rational number.
(iv) 1/√2=1×√2/√2×√2= 0.7071067811 …

We see that the decimal expansion of the above expression is nonterminating and non-recurring. Therefore, it is an irrational number.

(v) 2π = 2(3.1415 … ) = 6.2830 …

The decimal expansion of this expression is non-terminating and nonrecurring. Therefore, it is an irrational number.

2. Simplify each of the following expressions:
(i) �3 + √3��2 + √2�
(ii) �3 + √3��3 − √3�²
(iii) �√5 + √2�
(iv) �√5 − √2��√5 + √2�
Solution:
(i) �3 + √3��2 + √2�= 3�2 + √2�+ √3�2 + √2�
= 6 + 3√2 + 2√3 + √6
(ii) �3 + √3��3 − √3�= (3)² − �√3�²
= 9 − 3 = 6
(iii) �√5 + √2�²= (5)²+ �√2�²+ 2�√5��√2�
= 5 + 2 + 2√10 = 7 + 2√10

(iv) �√5 − √2��√5 + √2�= �√5�² − �√2�²
= 5 − 2 = 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?

Solution:

There is no contradiction at all. When we measure a length with scale or any other instrument, we only obtain an approximate rational value which is rational. We never obtain an exact value. For this reason, we may not realize, that either c or d is irrational. Therefore, the fraction c/d is irrational. Hence, 𝜋 is irrational.

4. Represent √9.3 on the number line.

Solution:

To represent √9.3 on the number line, we first need to mark a line segment OB =9.3 on number line. Now, take BC of 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is √9.3.

5. Rationalise the denominators of the following:

question:2

question:3

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