Express the fraction as recurring decimal : 2 3

We express 2 3 as decimal. By using the long division method, Since 2 3 is a non-terminating decimal, because the remainder is not zero. ⇒ 2 3 = 0 . 66666 . . . . . . . = 0 . 6 ˙ Hence, 0 . 6 ˙ is the required decimal number.

Express the fraction as recurring decimal : 2 3

We express 2 3 as decimal. By using the long division method, Since 2 3 is a non-terminating decimal, because the remainder is not zero. ⇒ 2 3 = 0 . 66666 . . . . . . . = 0 . 6 ˙ Hence, 0 . 6 ˙ is the required decimal number.

Express the fraction as recurring decimal : 1 6

We express 1 6 as decimal. By using the long division method, Since 1 6 is a non-terminating decimal, because the remainder is not zero. ⇒ 1 6 = 0 . 16666 . . . . . . . = 0 . 1 6 ˙ Hence, 0 . 1 6 ˙ is the required decimal number.

Express the fraction as recurring decimal : 5 9

We express 5 9 as decimal. By using the long division method, Since 5 9 is a non-terminating decimal, because the remainder is not zero. ⇒ 5 9 = 0 . 5555 . . . . . . . = 0 . 5 ˙ Hence, 0 . 5 ˙ is the required decimal number.

Express the fraction as recurring decimal : 5 5 11

We first convert the mixed fraction into an improper fraction. 5 5 11 = 5 × 11 + 5 11 = 55 + 5 11 = 60 11 . We express 60 11 as decimal. By using the long division method, Since 60 11 is a non-terminating decimal, because the remainder is not zero. ⇒ 60 11 = 5 . 4545 . . . . . = 5 . 45 ¯ Hence, 5 . 45 ¯ is the required decimal number.

Express the fraction as recurring decimal : 6 7 15

We first convert the mixed fraction into an improper fraction. 6 7 15 = 6 × 15 + 7 15 = 90 + 7 15 = 97 15 We express 97 15 as decimal. By using the long division method, Since 97 15 is a non-terminating decimal, because the remainder is not zero. ⇒ 97 15 = 6 . 4666 . . . . . . . = 6 . 4 6 Hence, 6 . 4 6 is the required decimal number.