(i) \(\sqrt { 10 } \) is an irrational number ( since 10 is not a perfect square number).
(ii) \(\sqrt { 49 } =7=\frac { 7 }{ 1 } \) a rational number(since 49 is a perfect square number).
(iii) 0.025 is a rational number (since it is a terminating decimal).
(iv) \(0.7\overline { 6 } \) = 0.7666…. is a rational number ( since it is a non – terminating and recurring decimal expansion).
(v) 2.505500555…. is an irrational number (since it is a non – terminating and non–recurring decimal).
(vi) \(\frac { \sqrt { 2 } }{ 2 } =\frac { \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \) is an irrational number ( since 2 is not a perfect square number).

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Relational between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 5 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
= \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
Rational numbers between\(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \)
=\(\frac { 1 }{ 2 } \left( \frac { 10+7 }{ 20 } \right) \)
=\(\frac { 17 }{ 40 } \)
Rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \) =\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \times \frac { 17 }{ 40 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 20+17 }{ 40 } \right) \)
=\(\frac { 37 }{ 80 } \)
Thus the rational numbers are \(\frac { 7 }{ 20 } ,\frac { 17 }{ 40 } and\frac { 37 }{ 80 } \)

Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
Now the rational numbers between \(-\frac { 202 }{ 24 } and\frac { 15 }{ 24 } are-\frac { 19 }{ 24 } ,-\frac { 18 }{ 24 } ,-\frac { 17 }{ 24 } ,....,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 22 }{ 24 } ,...,\frac { 14 }{ 24 } \)
we can take any seven of them \(\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,\frac { 4 }{ 24 } ,\frac { 5 }{ 24 } ,\frac { 6 }{ 24 } ,\frac { 7 }{ 24 } \)

\(1\over 11\) = 0.0909 ......... = \(0.\overline{09}\)
\(\therefore \frac{1}{33}=\frac{1}{3}\times \frac{1}{11}\)
 \(=\frac{1}{3}\times 0.0909.......\)
= 0.0303........ = \(0.\overline{03}\)
Also \(\frac{71}{33}=2\frac{5}{33}=2+\frac{5}{33}\)
\(=2+5\times \frac{1}{33}\)
= 2 + 5 \(\times\) \(0.\overline {03}\)
= 2 + (5 \(\times\) 0.030303 .......)
= 2 + 0.151515 .......
= 2 + \(0.\overline {15}\)
= \(2.\overline {15}\)

\(3.4\bar { 5 } \) = 3.45454545...
 = 3.4545 (correct to 4 decimal places)
The number lies between 3 and 4

(i) Divide the distance between 4 and 5 into 10 equal intervals.
(ii) Mark the point 4.8 which is second from the left of 5 and eighth from the right of 4
(iii) 4.86 lies between 4.8 and 4.9. Divide the distance into 10 equal intervals.
(iv) Mark the point 4.86 which is fourth from the left of 4.9 and sixth from the right of 4.8
(v) 4.863 lies between 4.86 and 4.87. Divide the distance into 10 equal intervals.
(vi) Mark point 4.863 which is seventh from the left of 4.87 and third from the right of 4.86.

0.3010011000111.. .. and 0.3020020002 .......
Two irrational numbers between 0.3010011000111.... and 0.3020020002.... are 0.3010111000111 and 0.3012311000111.....

Represent \(\sqrt{4.7}\) on a number line.
Steps of construction:
1. Draw a line and mark a point A and B such that AB = 4.7 cm.
2. Mark a point C on this line such that BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of AC and let it be "O".
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{4.7}\) , which can be marked in the number line as the value of BE = BD = \(\sqrt{4.7}\).

Represent \(\sqrt{3}\) on a number line.
Steps of construction:
1. Draw a line and mark a point A and B such that AB = 3 cm.
2. Mark a point C on this line such that BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of Ac and let it be "O".
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{3}\) which can be marked in the nu~ber line as the value of BE = BD = \(\sqrt{3}\)