CBSE CBSE Class 9 NCERT Solution Mathematics Number System

CBSE CBS E Class Class 9th NCE NCERT RT Soluti Solution: on: Mathe Mathemati matics cs

Chapter 1 Number System Exer Ex erci cise se:: 1. 1.1 1 p zero a rati ration onal al numb number er?? Ca Can n you write rite it in the the for form , whe where re p and and q are are Ques Qu esti tion on 1: Is zero q integers and

q

�

0?

0

Yes, we ma may y rewr rewrit itee 0 as as Solution: Yes,

(where (where 0 and 1 are are integers integers and q=1 which which is not equal equal to

1

zero).

Question 2: Find six rational numbers between 3 and 4. infinitely y many rational rational between between 3 and 4, one one way to take them is Solution: There can be infinitel 3

21 7

and 4 

28 7

.

First rational number between 3 and 4 21 q1

= (rational nu number be between

21 7

and

21



7

<

28 7

7 2

28

49

7 = 7 =7 2 2 2

= 7

<

+

28 7

Second rational number between 3 and 4 21 q2





(rational nu number b et etween

21 7

and

21 7

<

7 2

91 28

= 7 <

7 2

+

7

2 = 91 2 28 

28 7

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CBSE CBS E Class Class 9th NCE NCERT RT Solutio Solution: n: Mathema Mathematic ticss

Third rational number between 3 and 4 7 q 3 = (rational nu number be between �

Similarly,

21 7



175 56



91 28



7 2

7

an and

2 21





28



28

49

7 = 7 = 105 2 2 28 7 105 28

=2

7 91

+





7 28 2 28 7 203 105 217 28 56



28



56



7

Hence, the the six rational rational numbers numbers that that are all lying lying between between 3 and 4 is

Question 3: Find five rational numbers between

Solution: Let

a



3 5

and

b



3 5

and

4 5

175 91 7 203 105 105 217 . , , , , , 56 28 2 56 28 56

.

4 5

A rational number between a and b =

a b 2

3

+

4

3 4 7 7 A ra rational nu number be b etween and = 5 5 = = 5 5 2 5 × 2 10 3 7 4 < < 5 10 5 3 7 + 3 7 5 10 = (6 + 7) = 13 Now, a rraational nu number be between and = 5 10 2 10 × 2 20 20 3 13 13 7 4

�

Similarly,

5



20



10



5

3 2 25 5 27 15 4 are rational rational numbers numbers between between and . , , 5 40 40 20 5

Hence, required rational numbers are

25 13 1 3 27 7 15 . , , , , 40 20 40 10 20


CBSE Class 9th NCERT Solution: Mathematics

Question 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. Solution: (i) True, because natural numbers are 1, 2, 3, 4,…… ...………, � and whole numbers are 0, 1, 2, 3, 4, 5, ……………………….., � . OR

The collections of whole numbers contain all the natural numbers. (ii) False ( negative integers are not included in the list of whole numbers.)

 1 6 10  (iii) False  , ,  are not whole numbers.  3 7 19 

Exercise: 1.2 Question 1: 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. Solution: (i) True (Since, Real number = Rational number + Irrational number) (ii) False (Since, no negative can be the square root of any natural number) Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


(iii) False (Since, rational numbers are also present in the set 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. Solution: No, the square roots of all positive integers are not irrational.

For Example,

9

�3

Here, ‘3` is a rational number.

Question 3: Show how Solution: We know that,

5 can

be represented on the number line.

5 � 4 1 2

�

2

2

1

Draw of right angled triangle OQP, such that OQ = 2 Units PQ = 1 Unit And

 OQP = 90°

Now, by using Pythagoras theorem, we have 2

2

OP � OQ  PQ �

�

OP � �

2

2

2

2

1

4 1 5

Now, take O as centre OP � 5 as radius, draw an arc, which intersects the line at point R. Hence, the point R represents 5 .



Question 4: Classroom activity (constructing the ‘square root spiral`). Solution: Take a large sheet of paper and construct the ‘square root spiral’ in the following

fashion. Start with a point O and draw a line segment P1 P2

perpendicular to

OP1

OP1

of unit length. Draw a line segment

of unit length (see figure).

Now, draw a line segment

P2 P3

perpendicular to OP2 . Then draw a line segment

perpendicular to OP3 . Continuing in this manner, you can get the line segment

Pn �1 Pn

P3 P4

by drawing

a line segment of unit length perpendicular to OPn �1 . In This manner, you will have created the points

P2 , P3 ,........., Pn ,......, and

joined them to create a beautiful spiral depicting

2, 3, 4,......

Exercise: 1.3 Question 1: Write the following in decimal form and say what kind of decimal expansion each

has (i)

36 100



1

(ii)

11

(iii)

(iv)

(v)

(vi)

4

1 8

3 13 2 11

329 400

Solution:

(i) Clearly,

36 100

can be written as 0.36 36

�

100

= 0.36

(Terminating decimal)

(ii) Dividing 1 by 11, we get 0.0909 11) 100 99 100 99 1 1

�

11

(iii) We have,

1 4 8





0.090909...  0.09

4  8 1 8

(Non-terminating repeating)

33 

8

Dividing 33 by 8, we get Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


�

33 8



4.125



0.230769

(Terminating)

(iv) We have, 3/13

Dividing 3 by 13, we get

�

3 13

(Non- terminating repeating)

(v) We have, 2/11




2

�

11



(Non-terminating repeating)

0.181818...  0.18

(vi) We have, 320/400


329/400 = 0.8225

�

Question 2: You know that

1 7



(Terminating)

0.142857 . Can you predict what the decimal expansions of

2 3 4 5 6 , , , , are, without actually doing the long division. If so; how? 7 7 7 7 7

[Hint Study the remainders while finding the value

Solution: We have,

1 7



1 7

of carefully.]

0.142857



2 7 �



2



2

1 7

2  0.142857

�

0.285714

�

3

7 3 7 

7 � 3  0.142857

3 7



0.428571



4

4 7 



4 7





5



7

1 7

5  0.142857



0.714285



6



6 

4  0.142857



6 7

7

0.571428

5 7

1



5 7

1



1 7

6  0.142857

0.857142

Question3: Express the following in the form

p

where p and q are integers and q  0 .

q

(i) 0.6 (ii) 0.47 (iii) 0.001 Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


Solution:

(i) Let

x

=

0.6 =

0.666…

… (i)

Multiplying Eq. (i) by 10, we get 10 x = 6.666…

… (ii)

On subtracting Eq. (ii) from Eq. (i), we get (10 x – x ) = (6.666…) – (0.666…) 9 x = 6

�

(ii) Let

x

= 6/9

x

= 2/3

x

= 0.47 = 0.4777….

… (iii)

Multiplying Eq. (iii) by 10, we get 10 x = 4.777…

… (iv)

Multiplying Eq. (iv) by 10, we get

… (v)

100 x = 47.777 On subtracting Eq. (v) from Eq. (iv), we get (100 x – 10 x ) = (47.777…) – (4.777…)

90 x = 43

(iii) Let

� x

x

=

43 90

= 0.001 = 0.001001001…

… (vi)

Multiplying Eq. (vi) by (1000), we get

1000 x = 1.001001001…

… (vii)

On subtracting Eq. (vii) by Eq. (vi), we get (1000 x – x ) = (1.001001001….) – (0.001001001….) Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


999 x = 1 �

x

=

1 999

Question 4: Express 0.99999… in the form

p

. Are you surprised by your answer? With your

q

teacher and classmates discuss why the answer makes sense. Solution: Let

x

= 0.99999…

… (i)

Multiplying Eq. (i) by 10, we get 10 x = 9.99999…

… (ii) On subtracting Eq. (ii) by Eq. (i), we get (10 x – x ) = (9.99999…) – (0.99999…) 9 x = 9 �

x

x

=

9 9

=1

Question 5: What can the maximum number of digits be in the

repeating block of digits in the decimal expansion of? Perform the division to check your answer. Solution: The maximum number of digits in the repeating block of

digits in the decimal expansion of

1 17

is 17 – 1 = 16

See the figure beside, We have, Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


Thus,

1 17

�

0.0588235294117647.... , a block of 16 digits is repeated.

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: Consider many rational numbers in the form

p q

� q  0 ,

where

p

and

q

are integers

with no common factor other that 1 and having terminating decimal representations. 1 1 5 36 7 19 29 etc. , , , , , , 2 4 8 25 125 20 16

Let the various such rational numbers be

In all cases, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10. 1 2 1 4 5 8

�

�

�

= 36 25

125 20 29 16

25

�

1 25 4  25

5 10

�

0.5

�

25 100

�

0.25

5 125 8 125 625 1000

�

7 19

1 5

�

0.625

36  4 7 8

125  8

19  5 20  5

144

�

25  4

�

�

�

100 �

�

29  6 25 16  6 25

�

56 100

95 100 �

�

1.44

�

0.056

0.95

18125 10000

�

1.8125

From the above, we find that the decimal expansion of above numbers are terminating. Along with we see that the denominators of above numbers are in the form 2m  5n , where

m

and

n

are



natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.

Question 7: Write three numbers whose decimal expansions are non-terminating non-recurring. Solution: 0.74074007400074000074… 0.6650665006650006650000… 0.70700700070000…

Question 8: Find three different irrational numbers between the rational

numbers

5 7

and

9 11

.

Solution: To find irrational numbers, firstly we shall divide 5 by 7 and 9 by

11, So, 5

Thus,

7

0.714285... � 0.714285

9

Thus, 

�

11

�

0.8181... � 0.81

The required numbers are 0.73073007300073000073... 0.7650765007650007650000... 0.80800800080000...

Question 9: Classify the following numbers as rational or irrational (i) (ii)

23 225



(iii) 0.3796 (iv) 7.478478... (v) 1.101001000100001... Solution:

(i) Irrational (Since, it is not a perfect square.) (ii) Rational (Since, it is a whole number) (iii) 0.3796 = Rational (terminating) (iv) 7.478478… = 7.478 = Rational (non-terminating repeating.) (v) 1.101001000100001… = Irrational (non-terminating non-repeating.)

Exercise: 1.4 Question 1: Visualize 3.765 on the number line, using successive magnification. Solution: We know that, 3.765 lies between 3 and 4. So, let us divide the part of the number line between and 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7 and 3.8 Figure 1. Now, we imagine to divide this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and soon. To see this clearly, we magnify this as shown in Figure 2.

Again 3.765 lie between 3.76 and 3.77 Figure 2. So, let us focus on this portion of the number line Figure 3 and imagine dividing it again into ten equal parts Figure 3. Here, we can visualize th that 3.761 is the first mark and 3.765 is the 5 mark in these subdivisions. We call this process of visualization of representation of number on the number line through a magnifying glass as the process of successive magnification. So, we get seen that it is possible by sufficient successive magnification of visualize the position (or representation) of a real number with a terminating decimal expansion on the number line.



Question 2: Visualize 4.26 on the number line, upto 4 Decimal places. Solution: We adopt process by successive magnification and successively decrease the length of

the portion of the number line in which 4.26 is located. Since 4.26 is located between 4 and 5 and is divided into 10 equal parts Figure 1. In further, we locate Figure 2.

4.26

between 4.2 and 4.3

To get more accurate visualization of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualize that 4.26 lies between 4.26 and 4.27. To visualize 4.26

more clearly we divide again between 4.26 and 4.27 into 10 equal parts and visualize the

representation of

4.26

between 4.262 and 4.263 Figure 3.

Now, for a much better visualization between 4.262 and 4.263 is again divided into 10 equal parts Figure 4. Notice that

4.26

is located closer to 4.263 then to 4.262 at 4.2627.



�

Remark We can adopt the process endlessly in this manner and simultaneously

imagining the decrease in the length of the number line in which 4.26 is located.

Exercise: 1.5 Question 1: Classify the following numbers as rational or irrational. (i) 2 � 5 (ii) (3  23) � 23



(iii)

(iv)

2 7 7 7 1 2

(v) 2π Solution:

(i) Irrational (Since, 2 is a rational number and

5 is an irrational number)

Therefore, 2 � 5 is an irrational number. Since, the difference of a rational number and an irrational number is irrational. (ii) 3 + 23 � 23 (iii)

(iv)

2 7 7 7 1 2



2 7



3 (Rational)

(Rational)

(Irrational since,

Therefore,

1 2

1 � 0 is

a rational number and

2



0 is an irrational number)

is an irrational number.

Since, the quotient of a non-zero rational number with an irrational number is irrational.

(v)

2π

(Irrational, since

Therefore,

2π is

2 is a rational number and π i s an irrational number)

an irrational number.

Since the product of a non-zero rational number with an irrational number is an irrational.

Question 2: Simplify each of the following expressions: (i) (3  3) (2  2) Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


(ii) (3 � 3) (3  3) (iii) ( 5 � 2)

2

(iv) ( 5  2) ( 5 � 2) Solution:

(i) (3 � 3) (2 � 2)

(ii) (3 � 3) (3  3)

�



3(2 � 2) � 3 2 � 2



6 �3 2 �2 3 � 6



2

3



[�(a � b)

2

�  5



a

2

� 3

2

2

�

�

b

2

]

[�(a  b) (a � b)  a

2

b

2

]

2

2

2ab � b ]



5 � 2 10 � 2



7 � 2 10 2

π

2

2 5  2 � ( 2)

(iv) ( 5  2) ( 5 � 2) 

Question 3: Recall,

[�(a � b) (a  b)  a

93  6



(iii) ( 5 � 2)2





2

�  �  5



2



5 2  3

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: Actually

c d

22 �

7

which is an approximate value of π .



Question 4: Represent 9.3 on the number line. Solution: Firstly we draw AB = 9.3 units. Now, from B, mark a distance of 1 unit. Let this point

be C . Let O be the mid-point of AC . Now, draw a semi-circle with centre O and radius OA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Therefore, The distance BD = 9.3 Draw an arc with centre

B

and radius BD which intersects the number line at point

E ,

then the

point E represents 9.3 .

Question 5: Rationalize the denominator of the following: (i)

(ii)

(iii)

(iv)

1 7

1 7

�

6

1 5

2

1 7

�2

Solution: Get Study Material, Solved Question Papers, Syllabus, Sample Papers, Expert’s video, Online Test And much more.…


(i)

1

7

�

7

(ii)

7 1

7

=

(Multiplying and Dividing by

7 7 6

�

7 6

7 6



7 6

( 7)

2



( 6)

2



7 6

7

)

7 6



76

(Multiplying and dividing by 7  6 ) (iii)

1

5 2

�

5+ 2

5 2



5 2

( 5) 5 �

(iv)

1

�

7 2

7 2



2





( 2)

2

5 �

(Multiplying and dividing by 5  2 )

2



2

3

72



72

5

2

( 7)

2



7 2



74

2



(Multiplying and Dividing by 7  2 )

2

7 2 3

Exercise: 1.6 Question 1: Find: 1

(i) 64 2 1

(ii) 32 5 1

(iii) 1253 Solution:

(i) 64

1

1

2

2



(8 � 8)

2�



8

1 2



8

m

(Since, (a )

n

a

mn

)



1

1

(ii) 32  (2 � 2 � 2 � 2 � 2)  2 5

5

1

1

3�

(iii) 125  (5 � 5 � 5)  5 3

3

5�

1 5

2

1 3

5

Question 2: Find 3

(i) 9 2 2

(ii) 32 5 3

(iii) 16 4 

(iv) 125

1 3

Solution: 3

3 2 2

(i) 9  (3 )  33  27 2

2

2 5 5

(ii) 32  (2 )  22  4 5

3

3 4 4

(iii) 16  (2 )  23  8 4



(iv) 125

1 3

 (5 ) 3



1 3

 51 

1 5

Question 3: Simplify: 2

1

(i) 2 .2 5 3

1 (ii)  3  3 

7



1

(iii)

112 1

114 1

1

(iv) 7 .8 2 2

Solution: 2

1

2 1

(i) 2 .2  2 3

5

�

�

a b a b (Since, x .x  x )

3 5

10�3

2

15

13

2

15

7

1 1 1 (ii)  3   37  21  321 3 3  3 1

(iii)

112 1

1



 11 11 2

1 4

1 1



 11

2 4

a b ab (Since, ( x )  x )

2 1

 11

114

4

(Since,

x a x

b

 x a b )

1

 114 1

1

1

1

(iv) 7 .8  (7.8)  (56) 2 2

2

2

(Since, x . y  ( xy) ) a

a

a

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CBSE CBSE Class 9 NCERT Solution Mathematics Number System

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