A fraction terminates iff its denominator has only 2s and 5s as prime factors. Proof and examples.

Why Does 17/8 Terminate but 7/6 Does Not? — Decimal Expansions

Question

Which of the following fractions have terminating decimal expansions, and which do not? Justify your answer.

\frac{17}{8} \qquad \frac{7}{6}

Also, find the actual decimal expansion of the terminating fraction.

Solution — Step by Step

A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal if and only if the prime factorization of $q$ contains only the primes 2 and 5 — no other prime factor.

This is the key theorem from NCERT Chapter 1. Everything we do below is just applying this test.

Factorize the denominator: $8 = 2^3$ .

Only 2s — no 5s, no 3s, no 7s, nothing else. So $\frac{17}{8}$ terminates.

Factorize the denominator: $6 = 2 \times 3$ .

There’s a 3 in the factorization. Since 3 is neither 2 nor 5, the condition fails. So $\frac{7}{6}$ does not terminate — it’s a recurring decimal.

We need to convert $\frac{17}{8}$ into a decimal. The cleanest method: multiply numerator and denominator to make the denominator a power of 10.

\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{1000} = \frac{2125}{1000} = \mathbf{2.125}

For confirmation: $7 \div 6 = 1.1666\ldots = 1.1\overline{6}$ . The 6 repeats forever. This is exactly what the theorem predicted.

Why This Works

Every decimal we write is really a fraction with denominator $10^n$ for some $n$ — like $2.125 = \frac{2125}{1000} = \frac{2125}{10^3}$ . Since $10 = 2 \times 5$ , any power of 10 only ever has 2s and 5s as prime factors.

So for $\frac{p}{q}$ to sit nicely over a power of 10, the denominator $q$ must also only have 2s and 5s. If $q$ has any other prime factor (like 3, 7, 11…), you can never write it as $\frac{\text{something}}{10^n}$ , no matter how large $n$ you pick.

The non-terminating part is forced: to divide by 6, long division eventually loops because $10^n$ is never divisible by 3.

Alternative Method

Instead of the “multiply to make $10^n$ ” trick, you can use straight long division for $17 \div 8$ :

17.000 ÷ 8
= 2 remainder 1
10 ÷ 8 = 1 remainder 2
20 ÷ 8 = 2 remainder 4
40 ÷ 8 = 5 remainder 0  ← remainder hits 0, so it terminates

Result: 2.125. The moment remainder becomes 0, the decimal terminates. For $7 \div 6$ , the remainder cycles through $\{4, 4, 4, \ldots\}$ and never hits zero — confirming non-termination.

Board exam shortcut: Don’t do long division to decide IF it terminates — just factorize the denominator. Long division only to FIND the actual value. This saves time in a 3-mark CBSE question.

Common Mistake

Not reducing the fraction first. Suppose the question gives you $\frac{14}{12}$ . Students check 12 = 2² × 3 and say “non-terminating.” But $\frac{14}{12} = \frac{7}{6}$ in lowest terms — and you’d get the same answer here. The theorem requires $\frac{p}{q}$ to be in lowest terms (HCF = 1). A different example: $\frac{6}{15} = \frac{2}{5}$ after reducing, which does terminate even though 15 = 3 × 5 seems to fail the test. Always reduce first.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next