\(\sqrt[5]{32}=32^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^{ ̶5̶\times\frac{1}{ ̶5̶}}=2\)

\(\sqrt { 2 } \) =1.414213...
\(\sqrt { 3} \) =1.73205....
\(\frac { 7 }{ 9 } \) =0..7777
\(\frac { 6 }{ 7 } \) =0.857142
\(\frac { 23 }{ 6 } \) =3.8333...
\(\frac { 48 }{ 5 } \) =9.6
(i) Arrange the ten real numbers in the given boxes in ascending order.

(ii) Arrange the same numbers in the boxes given below in descending order.

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

Two irrational numbers between the given two rational numbers are 0.12201100111 ..... and 0.12301100111 .......

\(\sqrt{2}=1.414\)
\(\sqrt{3}=1.732\)
The two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.514 and 1.632

\({6\over 7}=0.\overline {857142}\)
\({12\over 13}=0.\overline{923076}\)

The two irrational numbers are  0.866142 ...and 0.903076.......

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

The two rational numbers are 2.2362 and 2.2363 (It has many answers)

Represent \(\sqrt{6.5}\) on a number line.
Steps of construction:
1. Draw a line and mark a point A and B such that AB = 6.5 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 = \(\sqrt{6.5}\) which can be marked in the number line as the value of BE = BD = \(\sqrt{6.5}\).

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 = \(\sqrt{3}\)