Question

Without actual division, classify the decimal expansion of following numbers as terminating or non-terminating and recurring.
A) $\dfrac{7}{{16}}$
B) $\dfrac{{13}}{{150}}$
C) $\dfrac{{ - 11}}{{75}}$
D) $\dfrac{{17}}{{200}}$

Answer 1

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

Complete step by step solution:
 let $x = \dfrac{p}{q}$ be a rational number, such that the prime factorization of q is the form of ${2^n}{5^m}$. Where, n,m are non negative integers. Then, x has a decimal expansion which terminates.
Step2 (i) Given $\dfrac{7}{{16}}$ is a form of $\dfrac{p}{q}$ where, p =7 and q= 16.
The prime factorization of q is given as
16 = 2× 2× 2× 2 = ${2^4}$
Therefore, given rational number $\dfrac{7}{{16}}$ is written as $\dfrac{7}{{{2^4}{5^0}}}$
According to lemma , the factorization of q i.e. 16 is the form of ${2^4}{5^0}$. Where n = 4 and m=0 is non negative integers. So, given $\dfrac{7}{{16}}$is terminating.
(ii) given $\dfrac{{13}}{{150}}$is form of $\dfrac{p}{q}$ where p = 13 and q = 150
The prime factorization of q i.e. 150 is written as
150= 2× 3× 5× 5 = ${2^1}{3^1}{5^2}$
Therefore, given number $\dfrac{{13}}{{150}}$ is written as $\dfrac{{13}}{{150}}$=$\dfrac{{13}}{{{2^1}{3^1}{5^2}}}$
According to lemma, the factorization of q i.e. 150 is not of the form of ${2^n}{5^m}$. Where n and m are non negative integers.
So, $\dfrac{{13}}{{150}}$ is non terminating recurring.
(iii) Given $\dfrac{{ - 11}}{{75}}$ is a form of $\dfrac{p}{q}$where p = -11 and q = 75.
The prime factorization of q i.e. 75 = 3× 5× 5 = ${3^1}{5^2}$
Therefore, given number $\dfrac{{ - 11}}{{75}}$ is written as $\dfrac{{ - 11}}{{{3^1}{5^2}}}$
According to lemma, the factorization of q i.e. 75 is not of the form ${2^n}{5^m}$. Where n and m are non negative integers.
So, $\dfrac{{ - 11}}{{75}}$is non terminating recurring.
(iv) Given $\dfrac{{17}}{{200}}$ is a form of $\dfrac{p}{q}$where p = 17 and q = 200.
The prime factorization of q i.e. 200 = 2×2× 2× 5× 5 = ${2^3}{5^2}$
Therefore, given number $\dfrac{{17}}{{200}}$ is written as $\dfrac{{17}}{{{2^3}{5^2}}}$
According to lemma, the factorization of q i.e. 200 is the form of ${2^n}{5^m}$. Where n and m are non-negative integers.

So, $\dfrac{{17}}{{200}}$ is terminating.

Note:
let x = p/q be a rational number, such that the prime factorization of q is not of the form ${2^n}{5^m}$, where n ,m are non negative integers. Then, x has a decimal expansion , which is non terminating repeating (recurring)