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To get the terminating decimal expansion of a rational number $\dfrac{p}{q}$, if $q = {2^m}{5^n}$ then m and n must belong to?
(A) $Z$ (B) $N \cup \{ 0\} $ (C) $N$ (D) $R$

Last updated date: 14th Jul 2024
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Hint: For the given number to have a terminating decimal number as its value, the denominator must be a positive integer and it must not be a multiple of prime numbers other than 2 and 5.

According to the question, the given number is $\dfrac{p}{q}$.
For the number to be rational, q must be an integer. Therefore, m and n cannot have fractional values. They must be integers.
Further, the value of $\dfrac{p}{q}$ must be a terminating decimal. And the value of q is ${2^m}{5^n}$. From this we can conclude that m and n can have only positive values.
If m and n take negative values or zero, then we will not get the value of $\dfrac{p}{q}$ a decimal number.
So, m and n will have only positive integral values. And in the denominator we have q i.e. ${2^m}{5^n}$ which will always be a multiple of either 2 or 5 or both. Thus, the value of $\dfrac{p}{q}$ will always result in a terminating decimal number.
Positive integral values can also be referred to as natural numbers. Thus, the values of m and n must belong to natural numbers. Hence, option (C) is correct.

Note: If a denominator consists of 3 or a multiple of 3 or it consists of prime numbers other than 2 and 5 or a multiple of such prime numbers and these are not getting cancelled out from the numerator then we will never get a terminating decimal number as the value of our fraction. The result will always be a non-terminating decimal number.
For ex: $\dfrac{1}{3},\dfrac{4}{{13}},\dfrac{6}{7},\dfrac{{11}}{{17}}$