Locate 2 on the number line.

At O draw a unit square O A B C on number line with each side 1 unit. By Pythagoras theorem O B = 1 2 + 1 2 = 2 We have seen that O B = 2 Using a compass with centre O and radius OB, draw an arc on the right side to O intersecting the number line at the point P . Now P corresponds to 2 on the number line.

Locate 3 on the number line.

Given, 3 Take a point O corresponding to zero. Consider a point A so that OA = 1 unit. Construct a perpendicular at A and name it as AZ . Draw an arc such that AB = 1 unit. Applying Pythagoras Theorem on △ AOB , Therefore, OB 2 = OA 2 + AB 2 ⇒ OB 2 = 1 2 + 1 2 ⇒ OB 2 = 1 + 1 ⇒ OB = 2 Therefore, OB = 2 Considering O as the centre and OB = 2 as the radius, construct an arc which cuts the number line at the point C . Thus, OB = OC = 2 Draw a perpendicular at C and name it as CY . Draw an arc of radius 1 unit which cut CY at E . Applying Pythagoras Theorem on △ OCE , we get, OE 2 = OC 2 + CE 2 ⇒ OE 2 = ( 2 ) 2 + 1 2 ⇒ OE 2 = 2 + 1 ⇒ OE 2 = 3 Therefore, OE = 3 Considering O as the centre and OE = 3 as the radius, construct an arc which cuts the number line at the point D . Thus, OD = OE = 3 Hence, 3 is represented by point D on the number line.

Find the decimal expansion of 10 3 , 7 8 and 1 7 .

Decimal expansion of 10 3 can be given as, 3 . 3 3 3 . . . 3 1 0 9 1 0 9 1 0 9 1 0 9 1 Which can be written as 3 . 3 ¯ . Hence, decimal expansion of 10 3 is 3 . 3 ¯ . Decimal expansion of 7 8 can be given as, 0 . 8 7 5 8 7 . 0 6 4 6 0 5 6 4 0 4 0 0 Hence, decimal expansion of 7 8 is 0 . 875 Decimal expansion of 1 7 can be given as, 0 . 1 4 2 8 5 7 7 1 . 0 7 3 0 2 8 2 0 1 4 6 0 5 6 4 0 3 5 5 0 4 9 1 it can be also written as 0 . 14285714 . . . . Hence, decimal expansion of 1 7 is 0 . 142857 ¯

Show that 0 . 333 . . . = 0 . 3 can be expressed in the form p q , where p and q are integers and q ≠ 0 .

Let, x = 0 . 33333 . . . . . . . . . . . = 0 . 3 . . . . . ( i ) Multiplying on both sides by ' 10 ' we get, 10 x = 3 . 333333 . . . . . . . . . . . = 3 . 3 . . . . . ( ii ) Subtract ( i ) from ( ii ) we get, 10 x - x = 3 . 333 . . . . - 0 . 333 . . . . . 9 x = 3 . 00000 . . . . . = 3 x = 3 9 = 1 3 Hence, 0 . 33333 . . . . . . . . = 1 3

Show that 1 . 272727 … = 1 . 27 ¯ can be expressed in the form p q , where p and q are integers and q ≠ 0 .

We have to express 1 . 27 in the form of p q , where p and q are integers and q ≠ 0 . Let x = 1 . 27 ¯ . . . . . . . . . . . . i Multiply Eq. ( i ) by 100 as there are two recurring digits, 100 x = 127 . 27 . . . . . . . . . . . . ii Subtracting ( i ) from ( ii ) , we get, 100 x - x = 127 . 27 - 1 . 27 ⇒ 99 x = 126 ⇒ x = 126 99 ⇒ x = 14 11 Hence, 1 . 27 can be expressed as 14 11 .

Show that 0 . 2353535 … = 0 . 2 35 ¯ can be expressed in the form p q , where p and q are integers and q ≠ 0 .

Given: Number 0 . 235353535 . . . . . = 0 . 2 35 ¯ To show: it can be expressed in the form of p q , where p and q are integers and q ≠ 0 . Let x = 0 . 2353535 . . . . . . . . . . i Multiply by 10 both side in ( i ) , we get, 10 x = 2 . 353535 . . . . . . . . . . ii Multiply by 1000 both side in ( i ) , we get, 1000 x = 235 . 353535 . . . . . . . . . iii Subtract ( ii ) from ( iii ) , we get, 1000 x - 10 x = 235 . 353535 . . . . - 2 . 353535 . . . ⇒ 990 x = 233 ⇒ x = 233 990 Thus, 0 . 23535 . . . = 233 990 , where p = 233 and q = 990 . Hence, proved.

Find an irrational number between 1 7 and 2 7 .

We saw that 1 7 = 0 142857 ¯ . So, you can easily calculate 2 7 = 0 . 285714 ¯ To find an irrational number between 1 7 and 2 7 , we find a number which is non-terminating non-recurring lying between them. 1 7 = 0.14285714 . . . . 2 7 = 0.28571428 . . . . There can be a infinite number of irrational numbers between the given two numbers. Irrational numbers between these two numbers are :- 0.1505005000500005 … 0.202002000200002 …