Write the following rational numbers in form of p q 0 . 6 *

Let us assume the rational number to be equal to x . x = 0 . 666 . . . . 1 Multiplying both side with 10 10 x = 6 . 666 . . . 2 Subtracting 1 from 2 10 x - x = 6 . 666 . . - 0 . 666 . . 9 x = 6 x = 6 9 o r 2 3 Hence, the simplest form in p q of 0 . 6 * is 2 3

Write the following rational number in form of p q 0 . 37 ¯

Let x = 0 . 37 ¯ ⇒ x = 0 . 373737 . . . 1 In order to get a whole number on the right-hand side, we multiply to both sides : 100 x = 37 . 373737 . . 2 Now subtract both the equations: 100 x - x = 37 . 373737 - 0 . 373737 ⇒ 99 x = 37 ⇒ x = 37 99 0 . 37 ¯ = 37 99 Thus, the required p q form is 37 99

Write the following rational number in form of p q 3 . 17 ¯

Let the given number be x . Thus, x = 3 . 171717 . . . . . 1 Now, multiplying both sides with 100 , we get ; ⇒ 100 x = 100 × 3 . 171717 . . . . . . . . ⇒ 100 x = 317 . 171717 . . . . . . . . 2 Now, subtracting both the equations, we get; ⇒ 100 x - x = 317 . 171717 . . . . . – 3 . 171717 . . . . . . . ⇒ 99 x = 314 ⇒ x = 314 / 99 So, the required p q form of 3 . 171717 . . . . . . = 314 99

Write the following rational number in form of p q 15 . 89 ¯

Let the given number be x . Thus, x = 15 . 89 ¯ x = 15 . 898989 . . . . . 1 Now, Multiplying both sides with 100 , we get ; ⇒ 100 x = 100 × 15 . 8989 . . . . ⇒ 100 x = 1589 . 8989 . . . . . . . . . 2 Now, Subtracting both the equation, we get; ⇒ 100 x - x = 1589 . 8989 . . . - 15 . 8978 . . . ⇒ 99 x = 1574 ⇒ x = 1574 99 So, the required p q form is 1574 99 .

Write the following rational number in form in p q 2 . 514 ¯

Let the given number be x . Thus, x = 2 . 514514 . . . . 1 Now, Multiplying both sides with 1000 , we get ; ⇒ 1000 x = 1000 × 2 . 514514 . . . . . ⇒ 1000 x = 2514 . 514514 . . . . . . . . . . 2 Now, Subtracting both the equations, we get; ⇒ 1000 x - x = 2514 . 514514 - 2 . 514514 ⇒ 999 x = 2512 ⇒ x = 2512 999 So, the required form of p q is 2512 999 .

Show that 4 √ 2 is an irrational number.

Let us assume that 4 √ 2 is a rational number. ⇒ 4 √ 2 = a b where, b ≠ 0 and a , b are integers ⇒ √ 2 = a 4 b Here, a , b are integers. So, 4 b is also integer. ⇒ a 4 b is rational number which can't be possible. ⇒ a 4 b = √ 2 , which is an irrational number. It is contradicting our assumptions. Hence, the assumption was wrong. So we can say that, 4 2 is an irrational number.

Prove that 3 + 5 is an irrational number.

Let us assume that 3 + 5 is a rational number ⇒ 3 + 5 = a b where, b ≠ 0 and a , b are integers. ⇒ 5 = a b - 3 ⇒ 5 = a - 3 b b Since a , b are integers, hence, a – 3 b is also integer. ⇒ a - 3 b b is rational which cannot be possible. a - 3 b b = 5 which is an irrational number. I t is contradicting our assumption. So, the assumption was wrong. Hence, 3 + 5 is an irrational number.