**Download NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.3 – Number System**

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

**(i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

**(vi) **

**Solution:**

(i) = 0.36 , it is terminating decimal

We can check by performing long division as follows:

(ii) = , it is non terminating repeating decimal

(iii) = = 4.125 , it is a terminating decimal.

(iv) = , it is non terminating repeating decimal.

(v) = 2 x

for we have solved earlier in part (ii) where =

Hence = 2 x = , it is non terminating repeating decimal.

(vi) = 0.8225 , it is a terminating decimal.

**(2). You know that = 0.142857 . 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 carefully.]**

**Solution:**

While performing long division for

We can easily observe that each one from the numerator of the given fractions appear in the remainder of the above operation which has to be carried out further by dividing them by 7 . And so just by using this observation we can predict the required decimal expansions as follows:

=

=

=

=

=

**(3). Express the following in the form , where p and q are integers and q ≠ 0.**

**(i) **

**(i) **

**(i) **

**Solution:**

(i) Follow the steps below :

Step (I) Assume X = ———- (1)

Step (II) Since there is only one repeating block , multiply X by 10

We get

10X = ———- (2)

Step (III) Write as

= 6 + ————– (3)

Step (IV) Using equation (1) and (2) , equation (3) can be written as

10X = 6 + X ————– (4)

Now,

Solving for X from equation (4) , we get

9X = 6

Therefore,

X = =

Hence we obtain

=

(ii) Follow the steps below :

Step (I) Assume X = ———- (1)

Step (II) Since there is only one repeating block , multiply X by 10

We get

10X = ———- (2)

Step (III) Write as

= 4.3 + ————– (3)

Step (IV) Using equation (1) and (2) , equation (3) can be written as

10X = 4.3 + X ————– (4)

Now,

Solving for X from equation (4) , we get

9X = 4.3

Therefore,

X = =

Hence we obtain

=

(iii) Follow the steps below :

Step (I) Assume X = ————- (1)

Step (II) Since there are three repeating block , multiply X by 1000

We get

1000X = —————— (2)

Step (III) Write as

= 1 + ————– (3)

Step (IV) Using equation (1) and (2) , equation (3) can be written as

1000X = 1 + X ————- (4)

Now,

Solving for X from equation (4) , we get

999X = 1

Therefore,

X =

Hence we obtain

=

**(4). Express 0.99999…. in the form . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.**

**Solution:**

(i) Follow the steps below to write 0.9999…. in the form :

Step (I) Assume X = ———– (1)

Step (II) Since there is only one repeating block , multiply X by 10

We get

10X = ——————- (2)

Step (III) Write as

= 9 + ——— (3)

Step (IV) Using equation (1) and (2) , equation (3) can be written as

10X = 9 + X —————— (4)

Now,

Solving for X from equation (4) , we get

9X = 9

Therefore,

X = = 1

Hence we obtain

= 1

**(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 digitts in the repeating block must be less than the denominator i.e. 17. So the mximum number of digits in the repeating block in this case is 16.

Perform the long division for as follows:

Thus we obtained

= , which has exactly 16 digits in the repeating block.

**(6). Look at several examples of rational numbers in the form (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 can easily observe that in such rational numbers q , apart from 1 and itself ; is either divided by 2 or 5 or both.

So we can say that the prime factorization of q must have only powers of 2 or 5 or both apart from 1 and the number itself.

**(7). Write three numbers whose decimal expansions are non-terminating non-recurring.**

**Solution:**

There are numerous number of such examples , three of them are as follows:

0.1011011101111011111……..

0.0100100010000100000……..

0.2122122212222122222……..

**(8). Find three different irrational numbers between the rational numbers and **

**.**

**Solution:**

We can calculate

=

and

=

Out of infinite possibilities three are as follows :

0.72722722272222……

0.7277277727777……..

0.80800800080000……

**(9). Classify the following numbers as rational or irrational :**

**(i) √23 **

**(ii) √225 **

**(iii) 0.3796**

**(iv) 7.478478… **

**(v) 1.101001000100001…**

**Solution:**

(i) √23 : Irrational

(ii) √225 = 15 : Rational

(iii)0.3796 : Rational

(iv)7.478478… : Rational

(v)1.101001000100001… : Irrational.

**NCERT Solutions For Class 9 ****Maths Chapter 1 – Number System**

**NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.1****NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.2****NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.4****NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.5****NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.6**

**Download NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.3 – Number System**