# Revisiting Rational Numbers and the Decimal Expansion

**Theorem : **Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form , where p and q are coprime, and the prime factorization of q is in the form 2^{n}5^{m }, where n, m are non-negative integers.

**Theorem :** Let be a rational number, such that the prime factorization of is of the form ^{} , where are non-negative integers . Then has a decimal expansion which terminates.

Theorem :Â Let be a rational number, such that the prime factorization of is not of the form ^{} , where are non-negative integers . Then has a decimal expansion which is

non-terminating repeating.

Â

Which of the following rational numbers have the terminating decimal representation?

(i) Â Â Â (ii) (iii)

(iv) (v) (vi)

[Making use of the result that a rational number where and have no common factors will have a terminating representation if and only if the prime factors of are 2's or 5's or both.]

(i) The prime factor of 5 is 5. And hence the denominator can be written in the form of 2^{0 }x 5^{1} (i.e.) 1 x 5 = 5

Hence has a terminating decimal representation.

(ii) 20 = 4 x 5 = 2^{2} x 5.

The prime factors of 20 are both 2's and 5's. Hence has a terminating decimal.

(iii) The prime factor of 13 is 13. The denominator cannot be expressed in the form of 2^{n} x 2^{m} Hence has non-terminating decimal.

(iv) 40 = 2^{3} x 5.

The prime factors of 40 are both 2's and 5's. Hence has a terminating decimal.

(v) 125 = 5^{3}

The prime factor of 125 is 5's. Hence has a terminating decimal.

(vi) The prime factor of 7 is 7. Hence has a non-terminating decimal representation.

Â

Express as a decimal fraction.

0.109375

**
Â Â Â Â **600

Â Â Â

__576__

**Â Â Â Â Â Â**240

Â Â Â Â Â

__192__

**Â Â Â Â Â Â Â Â Â**480

**Â Â Â Â Â Â Â Â Â**

__448__

Â Â Â Â Â Â Â Â Â 320

Â Â Â Â Â Â Â Â Â

__320__

**Â Â Â Â Â Â Â Â Â Â Â Â Â**0

Therefore

**= 0.109375**

Â

Represent 0.57Â in the form of .

100= 57.57Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦(ii)

(ii) â€“ (i)

99 = 57

Therefore = .