NCERT Solutions for Class 9 Math Chapter 1.3 NUMBER SYSTEM ALL SOLUTION

---

EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.4 EXERCISE 1.5 EXERCISE 1.6

---

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

(i)   `\36/100\ `

(ii)   `\1/11\ `

(iii)   `\4\frac{1}{8}\ `

(iv)   `\3/13\ `

(v)   `\2/11\ `

(vi)   `\329/400\ `

solution :

(i)   `\36/100\ `

   = 0.36 terminating.

(ii)   `\1/11\ `

   = 0.0909 non-terminating repeating.

(iii)   `\4\frac{1}{8}\ `

   = 4.125 Terminating.

(iv)   `\3/13\ `

   = 0.230769 non-terminating repeating.

(v)   `\2/11\ `

   = 0.18 non-terminating repeating.

(vi)   `\329/400\ `

   = 0.8225 Terminating.

2.    You know that `\1/7= 0.\overline{142857}\ ` you predict what the decimal expansions of `\2/7, 2/7, 3/7, 4/7, 5/7, 1/7, 6/7\ ` are, without actually doing the long division? If so, how?

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

solution :

   `\frac{1}{7} = 0.\overline{142857}\ `

∴ `\2 × \frac{1}{7}=2 × 0.\overline{142857}=0.\overline{285714}\ `

∴ `\3 × \frac{1}{7}=3 × 0.\overline{142857}=0.\overline{428571}\ `

∴ `\4 × \frac{1}{7}=4 × 0.\overline{142857}=0.\overline{571428}\ `

∴ `\5 × \frac{1}{7}=5 × 0.\overline{142857}=0.\overline{714285}\ `

∴ `\6 × \frac{1}{7}=6 × 0.\overline{142857}=0.\overline{857142}\ `

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

(i)    0.6

(ii)    0.47

(iii)    0.001

solution :

   (i)  0.6

   0.6 = 0.666....

   let `\x = 0.66...\ `.

   we will multiply both side by 10.

   So, `\10x = 6.66...\ `.

   Or, `\10x = 6 + 0.66...\ `.

   Or, `\10x = 6 + x\ `

   Or, `\10x - x = 6\ `

   Or, `\9x = 6\ `

   Or, `\x = \frac{6}{9}\ `

   So, `\x = \frac{2}{3}\ `

   (ii)  0.47

   0.47 = 0.4777....

   let `\x = 0.4777...\ `.

   we will multiply both side by 10.

   So, `\10x = 4.777...\ `.

   Or, `\10x = 4.3 + 0.477...\ `.

   Or, `\10x = 4.3 + x\ `

   Or, `\10x - x = 4.3\ `

   Or, `\9x = 4.3\ `

   Or, `\x = \frac{4.3}{9}\ `

   So, `\x = \frac{43}{90}\ `

   (iii)  0.001

   0.001 = 0.001001....

   let `\x = 0.001001...\ `.

   we will multiply both side by 1000.

   So, `\1000x = 1.001...\ `.

   Or, `\1000x = 1 + 0.001...\ `.

   Or, `\1000x = 1 + x\ `

   Or, `\1000x - x = 1\ `

   Or, `\999x = 1\ `

   Or, `\x = \frac{1}{999}\ `

   So, `\x =\frac{1}{999}\ `

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 :

   0.99999....

   let `\x = 0.99999...\ `.

   we will multiply both side by 10.

   So, `\10x = 9.9999...\ `.

   Or, `\10x = 9 + 0.9999...\ `.

   Or, `\10x = 9 + x\ `

   Or, `\10x - x = 9\ `

   Or, `\9x = 9\ `

   Or, `\x = \frac{9}{9}\ `

   Or, `\x = \frac{1}{1}\ `

   So, `\x = 1\ `

   The difference between 1 and 0.99999 is 0.00001 which is negligible.

   Hence, we can say that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

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 :

   `\1/17\ `

   Dividing 1 by 17

   `\1/17\ ` = 0.0588235294117647

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

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 1, 2, 3, 4, 5, 6, 7, 8, 9, 10....... Then the decimal expansion is terminating.

   For example :

   `\frac{1}{1}=1\ `, denominator q = 1 × 1

   `\frac{1}{2}=0.5\ `, denominator q = 2 × 1

   `\frac{1}{3}=0.\overline{3}\ `, denominator q = 3 × 1

   `\frac{1}{4}=0.25\ `, denominator q = 2 × 2

   `\frac{1}{5}=0.2\ `, denominator q = 5 × 1

   `\frac{1}{6}=0.\overline{16}\ `, denominator q = 2 × 3

   `\frac{1}{7}=0.\overline{142857}\ `, denominator q = 7 × 1

   `\frac{1}{8}=0.125\ `, denominator q = 2 × 2 × 2

   `\frac{1}{9}=0.\overline{1}\ `, denominator q = 3 × 3

   `\frac{1}{10}=0.1\ `, denominator q = 2 × 5

   We can say that the 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.

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

solution :

   We know that all irrational numbers are non-terminating non-recurring.

   Three numbers are,

   (i)    `\sqrt{5}=\ `2.2360679774997896964091736687313..................

   (ii)    `\sqrt{3}=\ `1.7320508075688772935274463415059..................

   (iii)    `\sqrt{7}=\ `2.6457513110645905905016157536393..................

8.    Find three different irrational numbers between the rational numbers `\frac{5}{7}\ ` and `\frac{9}{11}\ `.

solution :

   `\frac{5}{7}=0.\overline{714285}\ `

   `\frac{9}{11}=0.\overline{81}\ `

   Three different irrational numbers are,

   (i)   0.7507500750007500007500000...........

   (ii)    0.7707700770007700007700000...........

   (iii)    0.7907900790007900007900000...........

9.    Classify the following numbers as rational or irrational :

(i)    `\sqrt{23}\ `

(ii)    `\sqrt{255}\ `

(iii)     0.3796

(iv)    7.478478

(v)    1.101001000100001.....

solution :

   (i)   `\sqrt{23}\ `

   `\sqrt{23}=4.79583152331......\ `.

   Since the number is non-terminating non-recurring.

   Therefore, it is an irrational number.

   (ii)   `\sqrt{225}\ `

   `\sqrt{225}=15=\frac{15}{1}\ `

   Since the number can be represented in `\frac{p}{q}\ ` form.

   Therefore, it is a rational number.

   (iii)   0.3796

   0.3796 = 0.3796

   Since the number is terminating.

   Therefore, it is a rational number.

   (iv)   7.478478

   7.478478 = 7.478478

   Since the number is non-terminating but recurring.

   Therefore, it is a rational number.

   (v)   1.101001000100001.....

   1.101001000100001..... = 1.101001000100001.....

   Since the number is non-terminating non-recurring.

   Therefore, it is an irrational number.

---

EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.4 EXERCISE 1.5 EXERCISE 1.6