Question

If $ x = \dfrac{p}{q} $ be 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 terminating.
A. True
B. False
C. Neither
D. Either

Answer 1

Hint: We can find the solution of the given problem by using a method of contradiction. In this, if we take ‘q’ as prime factor of $ {2^n}{5^m} $ and find a solution using this and if using this we get an answer as true then for actual problem our answer will be false as we are doing it for contradiction.

Complete step-by-step answer:
Here, we have $ x = \dfrac{p}{q}, $ where $ p $ and $ q $ are integers.
But \[q\] is not in the form of $ {2^n} \times {5^m} $ where \[n,m\] are non-negative integers.
To find the required solution we assume that the denominator of fraction ‘q’ has a prime factorization of the form $ {2^n}{5^m} $ .
Therefore, a given fraction $ \dfrac{p}{q} $ is written as $ \dfrac{p}{{{2^n}{5^m}}} $ .
Then from above we have:
  $ \dfrac{p}{q} = \dfrac{p}{{{2^n}{{.5}^m}}} $
Or
 $
  \dfrac{p}{q} = \dfrac{{{2^m}.p}}{{{2^n}{{.2}^m}{{.5}^m}}} \\
   \Rightarrow \dfrac{p}{q} = \dfrac{p}{{{2^{n - m}}.{{(2.5)}^m}}} \\
   \Rightarrow \dfrac{p}{q} = \dfrac{p}{{{2^{n - m}}{{\left( {10} \right)}^m}}} \\
 $
In above we see that the denominator of the right hand side fraction term is a product of some power or $ 2\,\,and\,\,10 $ .
And we know that dividing any number with either by $ 10,\,\,100({10^2}),\,\,1000({10^3}),\,\,\,\,\,\,or\,\,we\,\,write\,as\,\,{(10)^m} $ result is always a terminating decimal expression.
Also we know that every number on division by either $ 2,\,4({2^2}),\,\,8({2^3}),..........or\,\,by\,\,({2^{any\,\,number}}) $ result so obtain is always a terminating decimal expression.
Hence, from above we see that the rational number in which denominator having a prime factor in the form of $ {2^n} \times {5^m} $ is always a terminating rational number.
Therefore, from the above assumption we can say that if q has no prime factor of the form $ {2^n} \times {5^m} $ is always a non-terminating rational number.
So, the correct answer is “Option B”.

Note: We know that the denominator of the fraction is not of the form $ {2^n} \times {5^m} $ . Means it will have some other prime factor say either $ 3 $ or some multiple of $ 3 $ and we know that such a fractions are non-terminating rational numbers and hence we can clearly see if denominator of a fraction has factor of the form $ {2^n} \times {5^m} $ then it will be a terminating or if not then non-terminating.