A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form $\dfrac{p}{10^4}$, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

A rational number whose terminating decimal expansion has its last non-zero digit at the 4^th decimal place can be expressed as :

⇒ $\dfrac{p}{10^4}$

where p is an integer.

If p were divisible by 10, then the last non-zero digit would have occurred at the 3^rd decimal place (or earlier), contradicting our assumption. Hence, p must not be divisible by 10.

Now, 10⁴ = 2⁴ × 5⁴.

Since p is not divisible by 10 = 2 × 5, p does not contain both 2 and 5 as factors. So, p contains at most powers of 2 only, or powers of 5 only (or neither), but not both.

When we reduce $\dfrac{p}{2^4 \times 5^4}$ to lowest form :

Case 1 : If p is odd (no factor of 2), then no 2's in numerator can cancel with 2's in denominator. So, the lowest-form denominator still contains 2⁴.

Case 2 : If p is not divisible by 5 (no factor of 5), then no 5's can cancel. So, the lowest-form denominator still contains 5⁴.

Examples :

⇒ 0.0625 = $\dfrac{625}{10^4} = \dfrac{1}{16} = \dfrac{1}{2^4}$. Here, p = 625 = 5⁴, and lowest-form denominator is 2⁴.

⇒ 0.0008 = $\dfrac{8}{10^4} = \dfrac{1}{1250} = \dfrac{1}{2 \times 5^4}$. Here, p = 8 = 2³, and lowest-form denominator is divisible by 5⁴.

So, the denominator in lowest form must be divisible by 2⁴ or 5⁴ (at least one of them), but not necessarily both.

Hence, such a number can be written as $\dfrac{p}{10^4}$ where p is not divisible by 10. The denominator in lowest form must be divisible by 2⁴ or 5⁴ (or both), but it is not necessary for it to be divisible by both.