A rational number in its lowest form has denominator . How many decimal places will its decimal expansion have? Explain your answer.

Question

Homi · Accepted Answer

Understand the Question

A rational number $\dfrac{p}{q}$ in simplest form has a terminating decimal expansion if the prime factorisation of its denominator $q$ is of the form $2^m \times 5^n$ .
The number of decimal places after which the decimal expansion terminates is equal to the maximum of the powers of $2$ and $5$ , which is $\max(m, n)$ .
Here, the denominator is $2^3 \times 5^1$ , so we determine the highest exponent between the prime factors $2$ and $5$ .

Step 1 · Analyse the Denominator and Express as a Power of 10

Given denominator of the rational number in lowest form: $\text{Denominator} = 2^3 \times 5^1$

To make the powers of $2$ and $5$ equal, multiply the numerator and denominator by $5^2$ :

\begin{aligned} \text{Denominator} &= 2^3 \times 5^1 \times 5^2 \\[0.6em] &= 2^3 \times 5^3 \\[0.6em] &= (2 \times 5)^3 \\[0.6em] &= 10^3 \end{aligned}

Since the denominator is $10^3$ , dividing by $10^3$ shifts the decimal point $3$ places to the left.

Alternatively, for a denominator of the form $2^m \times 5^n$ , the number of decimal places is: $\max(m, n) = \max(3, 1) = 3$

Answer

$3\text{ decimal places}$

Common Mistakes

Adding Exponents: Adding the powers of $2$ and $5$ (e.g., $3 + 1 = 4$ ) instead of taking the maximum exponent $\max(3, 1) = 3$ .
Selecting the Smaller Exponent: Incorrectly choosing the lower power ( $1$ decimal place) instead of the highest power required to form complete tens ( $10^3$ ).

Question 12