Decimal Expansions Of Rationals

Look at the price sticker 0.25 on a chocolate — that decimal simply stops. Now punch 1 ÷ 3 into a calculator and you get 0.3333333… — it never stops, yet the same digit keeps repeating forever. These two behaviours are not accidents: a single rule, resting only on the prime factors of the denominator, tells you which fraction will stop and which will run on — often before you divide even once.

Definition: A rational number is any number that can be written as p/q, where p and q are integers and q ≠ 0.

Definition: A decimal is terminating if it ends after finitely many digits (0.4, 0.875, 2.125). It is non-terminating but repeating (recurring) if it goes on forever with a fixed block of digits repeating (0.333…, 0.142857142857…, 0.1666…).

Key theorem (NCERT): Let x = p/q be a rational number in its lowest terms (HCF of p and q is 1). Then x has a terminating decimal expansion if and only if the denominator q can be written in the form q = 2ⁿ × 5ᵐ, where n and m are non-negative integers. If q has any prime factor other than 2 or 5, then the decimal is non-terminating and repeating.

The big idea — only 2s and 5s decide the fate of a fraction

Every rational number p/q, once reduced to lowest terms, has a decimal expansion that is either terminating or non-terminating repeating. There is genuinely no third possibility for a rational — a rational number can never produce a non-terminating, non-repeating decimal (that behaviour belongs only to irrationals such as √2 and π).

Whether a rational terminates is fixed entirely by which primes sit in the denominator. Reduce first, then look at the bottom:

• Denominator's primes are only 2 and 5 ⇒ terminating.
• Denominator has even one other prime (3, 7, 11, 13, …) ⇒ non-terminating repeating.

This is a two-way (iff) statement. It works in both directions: knowing the form of q tells you the type of decimal, and knowing the decimal type tells you the form of q. That is why CBSE can ask, in a one-mark MCQ, "Is 23/(2³ × 5² × 7) terminating?" and expect you to answer in ten seconds without a single division.

Why only 2 and 5? — the base-10 reason

The secret is buried in how the decimal system is built. Our numbers are written in base 10, and 10 = 2 × 5. Converting a fraction p/q to a decimal is exactly the act of rewriting it with a denominator that is a power of 10, because 0.d₁d₂d₃… simply means d₁d₂d₃…/10ᵏ. Now,

10ᵏ = (2 × 5)ᵏ = 2ᵏ × 5ᵏ.

So a power of 10 contains only the primes 2 and 5, in equal amounts. We can turn q into a power of 10 (by multiplying top and bottom by extra 2s or 5s to balance the powers) only when q itself is built from just 2s and 5s. The moment a stray prime such as 3 appears in q, we are stuck: no power of 10 ever has a factor of 3, so q can never be matched to a 10ᵏ, and the division is forced to run on forever.

Worked reasoning for 7/80: here 80 = 2⁴ × 5¹, which is short of three 5s to become 2⁴ × 5⁴ = 10⁴. Supply them:
7/80 = (7 × 5³)/(2⁴ × 5¹ × 5³) = (7 × 125)/10⁴ = 875/10000 = 0.0875 — it terminates.

Contrast 7/30: here 30 = 2 × 3 × 5. The prime 3 blocks every attempt to reach a power of 10, so 7/30 = 0.2333… is non-terminating repeating.

The three-step method — no long division needed

Step 1 — Reduce p/q to lowest terms. Cancel every common factor first. This step is not optional; skipping it is the single biggest exam mistake (see the 6/15 trap below).

Step 2 — Prime-factorise the denominator q of the reduced fraction.

Step 3 — Inspect the primes. Only 2 and 5 present ⇒ terminating. Any other prime present ⇒ non-terminating repeating.

Question: Without dividing, decide whether 13/3125 terminates.
Solution: 13 is prime and does not divide 3125, so the fraction is already in lowest terms. Now 3125 = 5⁵ = 2⁰ × 5⁵. The only prime is 5, so the form fits and the decimal terminates. (Indeed 13/3125 = (13 × 2⁵)/(5⁵ × 2⁵) = 416/10⁵ = 0.00416.)

Question: Decide for 23/(2³ × 5² × 7).
Solution: 23 is prime and shares no factor with the denominator, so it is in lowest terms. The denominator carries the prime 7, which is neither 2 nor 5, so the decimal is non-terminating repeating — settled purely from the primes, with no division.

Counting the digits, and reading the number line

A useful bonus: when q = 2ⁿ × 5ᵐ in lowest terms, the decimal terminates after exactly max(n, m) places. For 7/80, q = 2⁴ × 5¹, so max(4, 1) = 4 places — matching 0.0875 exactly. This answers "how many decimal places?" questions instantly.

The number line below shows why a terminating decimal is just a fraction with a power-of-10 denominator. The point 7/8 = 0.875 sits neatly at a tenths-and-hundredths mark, while 1/3 = 0.333… can only be approached, never landed on exactly, by a finite decimal.

Why it matters — and the traps to dodge

This tiny theorem pays back at every level. In boards it is a fast MCQ. Later, in competitive maths, the same idea drives converting recurring decimals back into fractions. In computer science it explains why 0.1 terminates in base 10 but not in base 2 (since 10 = 2 × 5 but binary's base is only 2), which is precisely why 0.1 + 0.2 is not exactly 0.3 on a computer.

Real-world example: a shop price of ₹0.875 per gram comes from 7/8 of a rupee, and 8 = 2³, so the price terminates cleanly. But 1/3 kg on a digital scale shows 0.333 because the true value never ends and the display simply rounds.

Common misconception 1 — forgetting to reduce. Take 6/15. A hasty student sees 15 = 3 × 5, spots the 3, and calls it non-terminating. Wrong: 6/15 = 2/5, denominator 5¹, giving 0.4 — clearly terminating. The theorem applies only in lowest terms. Cancel first, then check.

Common misconception 2 — thinking any long, messy decimal must be irrational. False. 0.142857142857… looks wild but is exactly 1/7, a rational with a six-digit repeating block. What makes a decimal rational is the repetition, however long the block; what makes it irrational is a pattern that never repeats at all.

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Work through these examples. They begin with a straight application of the theorem and build to the twists CBSE loves — reducing before testing, counting decimal places, converting a recurring decimal back to a fraction, and assertion–reason traps. Keep the drill handy: reduce → factorise the denominator → check for primes other than 2 and 5.

Example 1 — Classify and convert 13/3125 (basic)

Q: Show that 13/3125 has a terminating decimal expansion and write the decimal.
Given: Fraction 13/3125.
Solve: 13 is prime and does not divide 3125, so the fraction is in lowest terms. Factorise: 3125 = 5⁵ = 2⁰ × 5⁵. Only the prime 5 appears, so the decimal terminates. To find it, make the denominator a power of 10 by supplying 2⁵: 13/3125 = (13 × 2⁵)/(5⁵ × 2⁵) = (13 × 32)/10⁵ = 416/100000 = 0.00416.
Answer: 0.00416 (terminating). Check: max(n, m) = max(0, 5) = 5 places, and 0.00416 has exactly 5 decimal places ✓

Example 2 — Classify and convert 17/8

Q: Show that 17/8 terminates and write its decimal.
Given: Fraction 17/8.
Solve: 17 is prime and does not divide 8, so 17/8 is in lowest terms. Factorise: 8 = 2³ = 2³ × 5⁰. Only the prime 2 appears, so it terminates. Supply 5³: 17/8 = (17 × 5³)/(2³ × 5³) = (17 × 125)/10³ = 2125/1000 = 2.125.
Answer: 2.125 (terminating). Check: max(3, 0) = 3 places ✓

Example 3 — Tricky: prove non-terminating without dividing (64/455)

Q: Show, without long division, that 64/455 is non-terminating repeating.
Given: Fraction 64/455.
Solve: Factorise both: 64 = 2⁶ and 455 = 5 × 91 = 5 × 7 × 13. The numerator has only the prime 2, the denominator has 5, 7, 13 — no common factor, so 64/455 is already in lowest terms. The denominator 5 × 7 × 13 contains the primes 7 and 13, neither of which is 2 or 5, so it is not of the form 2ⁿ × 5ᵐ.
Answer: By the theorem, 64/455 is non-terminating repeating — established purely from the prime factors, with no division performed.

Example 4 — Tricky: reduce first, or fall into the trap (6/15)

Q: Assertion (A): 6/15 is non-terminating because 15 = 3 × 5 contains the prime 3. Reason (R): a fraction terminates only if its denominator has no prime other than 2 or 5. Are A and R correct?
Given: Fraction 6/15; an assertion–reason pair.
Solve: Reason R is the correct theorem. But Assertion A applies R to an unreduced fraction, which is illegal. Reduce first: 6/15 = 2/5, denominator 5¹ = 2⁰ × 5¹, so 6/15 = 0.4, which terminates. Thus A is false while R is true.
Answer: R is true, A is false. The theorem must be applied only in lowest terms — the classic reduce-first trap. Verify: 6 ÷ 15 = 0.4 ✓

Example 5 — Tricky: how many decimal places? (without full division)

Q: Without fully dividing, find the number of decimal places in the terminating expansion of 27/1250, and hence write the decimal.
Given: Fraction 27/1250.
Formula: For q = 2ⁿ × 5ᵐ, the number of decimal places = max(n, m).
Solve: 27 = 3³ and 1250 = 2 × 625 = 2¹ × 5⁴; they share no common factor, so the fraction is in lowest terms and has only primes 2 and 5 in the denominator ⇒ terminating. Here n = 1, m = 4, so places = max(1, 4) = 4. To write it, balance to 10⁴ by supplying 2³: 27/1250 = (27 × 2³)/(2⁴ × 5⁴) = (27 × 8)/10⁴ = 216/10000 = 0.0216.
Answer: 4 decimal places, 0.0216. Check: 0.0216 has exactly 4 places ✓

Example 6 — Tricky: convert a recurring decimal to p/q (0.overline-36)

Q: Express the recurring decimal 0.363636… (written 0.overline(36)) as a rational number in lowest terms, and confirm its denominator has a prime other than 2 or 5.
Given: x = 0.363636…, repeating block "36" of length 2.
Solve: Let x = 0.363636…. The block has 2 digits, so multiply by 10² = 100: 100x = 36.363636…. Subtract the first equation: 100x − x = 36.363636… − 0.363636… ⇒ 99x = 36 ⇒ x = 36/99. Reduce by HCF 9: x = 4/11.
Answer: 0.overline(36) = 4/11. Denominator 11 is a prime other than 2 or 5 — consistent with the decimal being non-terminating repeating ✓ (Check: 4 ÷ 11 = 0.3636… ✓)

Example 7 — Tricky: identify which are terminating in a set

Q: Among 3/8, 29/343, 23/(2³ × 5²), 129/(2² × 5⁷ × 7⁵), and 6/15, state which have terminating decimal expansions.
Given: Five fractions.
Solve: Test each in lowest terms.
3/8: 8 = 2³ — only 2 ⇒ terminating.
29/343: 343 = 7³ — prime 7 ⇒ non-terminating.
23/(2³ × 5²): denominator only 2s and 5s, and 23 shares no factor ⇒ terminating.
129/(2² × 5⁷ × 7⁵): the prime 7 is present and 129 = 3 × 43 cannot cancel it ⇒ non-terminating.
6/15 = 2/5 after reducing: denominator 5 ⇒ terminating.
Answer: Terminating: 3/8, 23/(2³ × 5²), and 6/15. Non-terminating: 29/343 and 129/(2² × 5⁷ × 7⁵). Sanity check on the trap: 6/15 counts only because we reduced it first ✓

Example 8 — Determine the missing power (reverse reasoning)

Q: The fraction 7/(2ⁿ × 5) is known to terminate after exactly 4 decimal places. Find n.
Given: 7/(2ⁿ × 5), terminating in 4 places.
Formula: Places = max(n, m); here m = 1.
Solve: 7 is prime and does not divide the denominator, so the fraction is in lowest terms and terminating for every n. We need max(n, 1) = 4. Since n ≥ 1 forces max(n, 1) = n, we require n = 4.
Answer: n = 4. Check: 7/(2⁴ × 5) = 7/80 = 0.0875, which indeed has 4 decimal places ✓

When a fraction gives a terminating decimal and when it repeats.

• A rational number p/q (in lowest terms) has a terminating decimal if and only if q has only 2 and 5 as prime factors — that is, q = 2ᵐ × 5ⁿ.
• If q has any prime factor other than 2 or 5, the decimal is non-terminating but repeating (recurring).
• Every rational number has a decimal that either terminates or repeats.
• Example: 3/8 = 0.375 terminates (8 = 2³); 1/6 = 0.1666... repeats (6 = 2×3).
• To check without dividing, factorise the denominator after simplifying the fraction.
• A non-terminating non-repeating decimal is irrational, not rational.

Cover the answer, recall, then check. 7 cards on decimal expansions.

Q1. When does a rational number p/q (in lowest terms) have a terminating decimal?
A1. When the denominator q has only 2 and 5 as its prime factors (q = 2ᵐ × 5ⁿ).

Q2. When does p/q give a non-terminating repeating decimal?
A2. When q has a prime factor other than 2 or 5.

Q3. Will 3/8 terminate? Why?
A3. Yes — 8 = 2³, so it terminates: 3/8 = 0.375.

Q4. Will 1/6 terminate? Why?
A4. No — 6 = 2 × 3 has a factor 3, so it is non-terminating recurring (0.1666...).

Q5. Is every rational number’s decimal either terminating or repeating?
A5. Yes — a decimal that neither terminates nor repeats is irrational.

Q6. What kind of decimal does 13/125 give?
A6. Terminating, since 125 = 5³ (only 5 as prime factor).

Q7. To decide if 17/8 terminates, what should you do first?
A7. Ensure it is in lowest terms, then factorise the denominator (8 = 2³ → terminates).