The decimal expansion of $\dfrac{p}{q}$ will be terminating precisely when the prime factors of $q$ are only 2, only 5 or both 2 and 5. Can you explain why?

The decimal expansion of $\dfrac{p}{q}$ terminates if and only if the denominator q (in lowest terms) has only 2 and/or 5 as its prime factors.

Reason :

Our number system is base-10, and 10 = 2 × 5. So, any denominator that is a power of 10 produces a terminating decimal.

If q has only the prime factors 2 and/or 5, we can multiply both the numerator and the denominator by a suitable number to make the denominator a power of 10. This gives a terminating decimal.

Example 1 :

⇒ $\dfrac{3}{20} = \dfrac{3}{2^2 \times 5} = \dfrac{3 \times 5}{2^2 \times 5 \times 5} = \dfrac{15}{100} = 0.15$ (terminating).

Example 2 :

⇒ $\dfrac{7}{8} = \dfrac{7}{2^3} = \dfrac{7 \times 5^3}{2^3 \times 5^3} = \dfrac{875}{1000} = 0.875$ (terminating).

But, if q has any prime factor other than 2 or 5 (such as 3, 7, 11, etc.), no multiplication can make the denominator a power of 10. So, the long division will never give a remainder of 0, and the decimal expansion will be non-terminating.

Example 3 :

⇒ $\dfrac{1}{3} = 0.3333\ldots$ (non-terminating, repeating). Here 3 is a prime factor other than 2 or 5.

Hence, the decimal expansion of $\dfrac{p}{q}$ terminates only when the denominator q (in lowest form) has 2 and/or 5 as its only prime factors, because only then can it be expressed as a fraction with a power of 10 in the denominator.