Question

If $\dfrac{p}{q}$ is a rational number $\left( q\ne 0 \right)$ , what is the condition on q so that the decimal representation of $\dfrac{p}{q}$ is terminating?

Answer 1

Hint: Recall the definition of a rational number and focus on the point that a rational number is terminating if and only if the denominator has only 2 and 5 as its prime factors, i.e., for a rational number $\dfrac{p}{q}$ to be terminating in its decimal form the denominator must be of the form ${{2}^{a}}\times {{5}^{b}}$ , where a and b are positive integers.

Complete step-by-step solution -
Before moving to the options, let us talk about the definitions of rational numbers followed by irrational numbers.
So, rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers and $q \ne 0$ . In other words, we can say that the numbers which are either terminating or recurring when converted to decimal form are called rational numbers. All the integers fall under this category.
Now moving to the solution. It is given that $\dfrac{p}{q}$ is a rational number $\left( q\ne 0 \right)$ . Now for a number to be terminating, we know that for a rational number $\dfrac{p}{q}$ to be terminating in its decimal form the denominator must be of the form ${{2}^{a}}\times {{5}^{b}}$ , where a and b are positive integers. Also, q is the denominator of the rational number and both p and q are integers.
So, we can conclude that $q={{2}^{a}}\times {{5}^{b}}$ , such that both a and b are positive integers.

Note: Be careful about the point that p and q both are integers, and an additional necessary condition for a number $\dfrac{p}{q}$ to be rational is both p and q must be co-primes. It is also an important thing to mention that for $q={{2}^{a}}\times {{5}^{b}}$ , a and b are positive integers, as if we take a and b to be negative then q would be a fraction violating the condition of a rational number. For example: if we take a=-1 and b=-2, the value of q comes out to be $q={{2}^{a}}\times {{5}^{b}}={{2}^{-1}}\times {{5}^{-2}}=\dfrac{1}{2\times {{5}^{2}}}=\dfrac{1}{50}$ , which is not an integer and for the number to be rational q needs to be an integer.