Find the decimal expansion of 1 7 .

The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. We know that we can find the decimal expansion of two digits by finding their decimal value. Hence, 7 ) 1 . 0 ( 0 . 142857 . . . . . 7 30 28 20 14 60 56 40 35 50 49 1 Therefore, the decimal expansion of 1 7 is 0 . 142857 ¯ Thus the decimal expansion is "Non-terminating but repeating".

Find the decimal expansion of 1 7 .

The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. We know that we can find the decimal expansion of two digits by finding their decimal value. Hence, 7 ) 1 . 0 ( 0 . 142857 . . . . . 7 30 28 20 14 60 56 40 35 50 49 1 Therefore, the decimal expansion of 1 7 is 0 . 142857 ¯ Thus the decimal expansion is "Non-terminating but repeating".

Find the decimal expansion of 7 8 .

The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. Therefore, The decimal expansion of 7 8 is 0 . 875 .

Find the decimal expansion of 20 3 .

The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. Hence, 20 3 = 6 . 6 - The digit or the decimal expansion of 20 3 is non-terminating because there is a presence of same number again and again.

Express 0.3333 . . . . in the form of p q q ≠ 0 .

Let x = 0.3333 … . → 1 Since, there is one repeating digit. ∴ we multiply both sides by 10 . 10 x = 0.333 … × 10 Clearly, 10 x = 3.333 … → 2 Simplify, 1 - 2 ⇒ 10 x - x = 3 .333 … - 0 .333 … 9 x = 3 x = 3 9 x = 1 3 Therefore, 0 . 3333 . . . . = 1 3

Express 1 . 27 - in the form p q q ≠ 0 in the simplest form.

Let say, x = p q where q ≠ 0 Take, x = 1.2727 … . . . . . 1 Multiply the above equation by 100 on both sides we get, 100 x = 127.2727 … . . . . . 2 2 - 1 , 100 x - x = 127.2727 . . . . - 1.272727 … 99 x = 126 x = 126 99 On simplify, x = p q = 14 11

Express 0.2 35 in the p q form q ≠ 0 .

Let, x = 0.2 35 - ⇒ x = 0.23535353 … (1) Multiply both sides by 10 , ⇒ 10 x = 2.353535 … (2) Multiply both sides by 100 , ⇒ 1000 x = 235.353535 … (3) On subtracting (2) from (3). 1000 x - 10 x = 235 - 2 x = 233 990 Therefore, 0.2 35 - = 233 990

Find an irrational number between 1 7 and 2 7 .

To find an irrational number between 1 7 and 2 7 . 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 …