Real Numbers

Imagine you have two ribbons, one 420 cm long and the other 130 cm long, and you must cut both into equal pieces of the greatest possible length with nothing wasted. That "greatest common length" is exactly the HCF of 420 and 130 — and around 300 BCE the Greek mathematician Euclid gave a remarkably clean, repeat-until-done recipe to find it that we still use today.

Definition: Euclid's Division Lemma states that for any two positive integers a and b, there exist unique whole numbers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b.

Definition: Euclid's Division Algorithm is a repeated application of this lemma to compute the HCF (Highest Common Factor) of two positive integers: keep dividing the previous divisor by the previous remainder until the remainder becomes 0; the last non-zero remainder (the last divisor) is the HCF.

The lemma — division written as an equation

Euclid's Division Lemma is nothing but the ordinary division you have done since primary school, written in one tidy line. When you divide a by b, you get a quotient q and a remainder r, and the relation dividend = divisor × quotient + remainder becomes a = bq + r.

The single most important part is the condition 0 ≤ r < b. It says two things at once:

• The remainder is never negative (r ≥ 0).
• The remainder is always strictly less than the divisor (r < b). If your remainder ever came out equal to or bigger than b, you did not divide far enough.

The word "lemma" simply means a small proven statement used as a stepping-stone to a bigger result. The word "unique" is doing real work here: for a given a and b, there is exactly one pair (q, r) satisfying both a = bq + r and 0 ≤ r < b. You cannot find a second valid pair. For example, take a = 17, b = 5. Then 17 = 5 × 3 + 2, so q = 3 and r = 2, and indeed 0 ≤ 2 < 5. No other whole-number pair works.

Note the lemma is stated for positive integers a and b in the NCERT Class 10 syllabus, and q, r are whole numbers (so they can be 0).

The algorithm — using the lemma again and again to get HCF

To find the HCF of two positive integers using Euclid's Division Algorithm, follow these steps. Take the larger number as a and the smaller as b (so a > b).

1. Apply the lemma to a and b: write a = bq + r with 0 ≤ r < b.
2. If r = 0, then b is the HCF and you stop.
3. If r ≠ 0, apply the lemma again, this time to b and r (the old divisor becomes the new dividend, the old remainder becomes the new divisor).
4. Repeat step 3 — always divide the previous divisor by the previous remainder — until the remainder is exactly 0.
5. When the remainder finally becomes 0, the divisor at that last stage is HCF(a, b).

Why it works (intuition): at every step, HCF(a, b) = HCF(b, r). This is because any number that divides both a and b also divides r = a − bq, and any number that divides both b and r also divides a = bq + r. So the common divisors never change as we go down the ladder. The numbers keep shrinking (the remainder is strictly smaller each time), so the process must end with remainder 0 after finitely many steps — and the last divisor carries the same HCF as the original pair.

Let us find HCF(420, 130) to match the ribbon story:

• 420 = 130 × 3 + 30 (remainder 30, not 0 — continue)
• 130 = 30 × 4 + 10 (remainder 10, not 0 — continue)
• 30 = 10 × 3 + 0 (remainder 0 — stop)

The last divisor is 10, so HCF(420, 130) = 10. The greatest equal piece is 10 cm long: the 420 cm ribbon gives 42 pieces and the 130 cm ribbon gives 13 pieces, with none wasted.

Common forms: writing any integer in terms of a chosen divisor

A favourite exam use of the lemma is to classify integers. If you fix the divisor b, then by the lemma every integer a can be written as a = bq + r with r taking only the values 0, 1, 2, …, (b − 1).

Take b = 2. Then r is 0 or 1, so every positive integer is either 2q (even) or 2q + 1 (odd). This is the formal proof that every integer is even or odd.

Take b = 3. Then r is 0, 1 or 2, so every positive integer is of the form 3q, 3q + 1, or 3q + 2. This is the key to showing results like "the square of any positive integer is of the form 3m or 3m + 1".

Take b = 4. Then r is 0, 1, 2 or 3, so every positive integer is 4q, 4q + 1, 4q + 2, or 4q + 3. Since 4q and 4q + 2 are even, every odd integer must be 4q + 1 or 4q + 3.

The method is always the same: apply the lemma with the chosen b, list the finite possibilities for r, then reason case by case. The picture below shows how b = 3 sorts the number line into three tidy families.

Handling the details: uniqueness and the "last divisor" trap

Two points decide most marks on this topic.

Uniqueness of q and r. Suppose two pairs worked: a = bq₁ + r₁ and a = bq₂ + r₂, both with the remainder between 0 and b − 1. Subtracting gives b(q₁ − q₂) = r₂ − r₁. The right side is a difference of two numbers each in the range 0 to b − 1, so its size is less than b; but the left side is a multiple of b. The only multiple of b whose size is less than b is 0. Hence r₂ = r₁, and then q₁ = q₂. So the pair (q, r) is unique — exactly what the lemma claims.

The last-divisor trap. The HCF is the last divisor (equivalently, the last non-zero remainder), NOT the final remainder 0 and NOT the last quotient. In HCF(420, 130) above, the answer is the divisor 10 at the step "30 = 10 × 3 + 0". Students who write "0" or who stop at a non-zero remainder lose the mark. Always continue until the remainder is exactly 0, then look back one line for the divisor.

---

Work through these examples. They begin with a plain HCF computation and build to the twists CBSE boards enjoy — proofs by cases, word problems, and an assertion-reason trap. Keep the two tools ready: the lemma a = bq + r (0 ≤ r < b) and the algorithm (last divisor = HCF).

Example 1 — HCF by Euclid's algorithm (basic)

Q: Find the HCF of 867 and 255 using Euclid's division algorithm.
Given: a = 867, b = 255 (larger taken as a).
Solve:
867 = 255 × 3 + 102 (remainder 102 ≠ 0 — continue)
255 = 102 × 2 + 51 (remainder 51 ≠ 0 — continue)
102 = 51 × 2 + 0 (remainder 0 — stop)
The last divisor is 51.
Answer: HCF(867, 255) = 51. Check: 867 = 51 × 17, 255 = 51 × 5, and 17, 5 share no common factor except 1 ✓

Example 2 — HCF of a smaller pair

Q: Find the HCF of 96 and 404 by Euclid's algorithm.
Given: a = 404, b = 96.
Solve:
404 = 96 × 4 + 20 (remainder 20)
96 = 20 × 4 + 16 (remainder 16)
20 = 16 × 1 + 4 (remainder 4)
16 = 4 × 4 + 0 (remainder 0 — stop)
Last divisor = 4.
Answer: HCF(96, 404) = 4. Check: 404 = 4 × 101, 96 = 4 × 24 ✓

Example 3 — Apply the lemma once

Q: Using the division lemma, write 4032 in the form 12q + r with 0 ≤ r < 12.
Formula: a = bq + r, 0 ≤ r < b, with b = 12.
Solve: 4032 ÷ 12 = 336 exactly, so q = 336 and r = 0. Thus 4032 = 12 × 336 + 0.
Answer: q = 336, r = 0, so 4032 = 12 × 336. Since r = 0, 12 divides 4032 exactly ✓

Example 4 — Tricky: every positive odd integer is 4q + 1 or 4q + 3

Q: Show that every positive odd integer is of the form 4q + 1 or 4q + 3, where q is a whole number.
Solve: Let a be any positive integer and take b = 4 in the lemma. Then a = 4q + r with 0 ≤ r < 4, so r ∈ {0, 1, 2, 3}. Hence a is 4q, 4q + 1, 4q + 2, or 4q + 3. Now 4q = 2(2q) is even and 4q + 2 = 2(2q + 1) is even, so neither can be odd. The remaining forms 4q + 1 and 4q + 3 are each "even + 1", hence odd.
Answer: A positive odd integer cannot be 4q or 4q + 2, so it must be 4q + 1 or 4q + 3. Proved. Sanity check: 7 = 4×1 + 3 (odd) and 9 = 4×2 + 1 (odd) ✓

Example 5 — Tricky: square of any positive integer is 3m or 3m + 1

Q: Use Euclid's division lemma to show that the square of any positive integer is of the form 3m or 3m + 1 for some integer m.
Solve: Let a be a positive integer. With b = 3, a = 3q + r where r ∈ {0, 1, 2}. Square each case:
• a = 3q ⇒ a² = 9q² = 3(3q²) = 3m, taking m = 3q².
• a = 3q + 1 ⇒ a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, taking m = 3q² + 2q.
• a = 3q + 2 ⇒ a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1, taking m = 3q² + 4q + 1.
Answer: In every case a² is 3m or 3m + 1 — never 3m + 2. Proved. Check: 4² = 16 = 3×5 + 1 (form 3m+1); 6² = 36 = 3×12 (form 3m) ✓

Example 6 — Tricky: word problem (maximum equal columns)

Q: Two groups of 616 army cadets and 32 army members are to march behind their band such that both groups have the same number of columns, and this number is the greatest possible. Find the maximum number of columns.
Given: We need the greatest common measure of 616 and 32, i.e. HCF(616, 32).
Solve:
616 = 32 × 19 + 8 (remainder 8)
32 = 8 × 4 + 0 (remainder 0 — stop)
Last divisor = 8.
Answer: The maximum number of columns is 8. Check: 616 = 8 × 77 and 32 = 8 × 4, so both groups split evenly into 8 columns ✓

Example 7 — Tricky: assertion–reason trap

Q: Assertion (A): HCF(455, 42) = 7. Reason (R): In Euclid's algorithm, the HCF is the last remainder obtained (that is, 0). State whether A and R are true, and whether R correctly explains A.
Solve: Compute the HCF. 455 = 42 × 10 + 35; 42 = 35 × 1 + 7; 35 = 7 × 5 + 0. Last divisor = 7, so A is TRUE. Now examine R: the HCF is the last non-zero remainder / last divisor, which is 7 here — it is not the last remainder 0. So R is FALSE (it states the wrong rule).
Answer: A is true, R is false. The correct reason is that the HCF is the last non-zero remainder (the last divisor), not the final remainder 0. Trap dodged ✓

Example 8 — Tricky: three numbers via HCF

Q: Find the largest number that divides 245 and 1029 leaving remainder 5 in each case.
Given: If the number leaves remainder 5, then it divides (245 − 5) = 240 and (1029 − 5) = 1024 exactly. So we need HCF(240, 1024), and the answer must also be greater than 5 (so that a remainder of 5 is possible).
Solve:
1024 = 240 × 4 + 64 (remainder 64)
240 = 64 × 3 + 48 (remainder 48)
64 = 48 × 1 + 16 (remainder 16)
48 = 16 × 3 + 0 (remainder 0 — stop)
Last divisor = 16.
Answer: The largest such number is 16 (and 16 > 5, so a remainder of 5 is valid). Check: 245 = 16 × 15 + 5 and 1029 = 16 × 64 + 5 ✓

:::keypoints

• Euclid's Division Lemma: for positive integers a, b there are unique whole numbers q, r with a = bq + r and 0 ≤ r < b.
• The remainder is never negative and always strictly less than the divisor (0 ≤ r < b).
• Euclid's Algorithm for HCF: divide a by b, then keep dividing the previous divisor by the previous remainder until remainder = 0.
• The HCF is the last non-zero remainder = the last divisor (never the final 0).
• At every step HCF(a, b) = HCF(b, r), which is why the algorithm works.
• Fixing b in the lemma classifies integers: b = 2 gives 2q / 2q+1; b = 3 gives 3q / 3q+1 / 3q+2; b = 4 gives 4q … 4q+3.
• Every positive odd integer is of the form 4q + 1 or 4q + 3.
• The square of any positive integer is of the form 3m or 3m + 1 (never 3m + 2).
  :::

:::memory
"Divide, Down, Divisor" — DIVIDE larger by smaller, step DOWN (old divisor and remainder become the new pair), and when the remainder hits 0 the HCF is the last DIVISOR. Three D's: Divide, Down, Divisor.
:::

:::recap

• Lemma: a = bq + r with 0 ≤ r < b, and q, r are unique.
• Algorithm: repeat "divisor ÷ remainder" until remainder is 0.
• HCF = last non-zero remainder = last divisor.
• Choosing a divisor b splits every integer into the classes bq, bq+1, …, b(q)+(b−1) — the engine behind the odd-form and square-form proofs.
  :::

Two friends are cutting a big block of chocolate: whichever way one of them snaps it into pieces, the smallest square blocks are always identical — and there is always the same number of them. Whole numbers behave exactly like that chocolate. Every composite number can be snapped apart into the same unchanging set of prime "square blocks", and that single fact quietly powers most of the Real Numbers chapter — HCF, LCM, decimal-expansion tests and irrationality proofs are all just this one idea in disguise.

Definition: A prime number is a natural number greater than 1 whose only divisors are 1 and itself (2, 3, 5, 7, 11, 13, …). A composite number is a natural number greater than 1 that has at least one divisor other than 1 and itself (4, 6, 8, 9, 10, 12, …).

Definition: The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors are written. In symbols, for every composite N there is exactly one collection of primes such that N = p₁ × p₂ × … × pₖ (the same primes, each appearing the same number of times).

Section 1 — What the theorem really says

The theorem carries two claims, and a board answer must respect both. The existence claim says: pick any composite number and you can break it into primes. The uniqueness claim says: no matter how you break it, the bag of primes you end up with is exactly the same — same primes, same number of copies of each. The only thing free to change is the order in which you write them.

Take 60. Split it as 60 = 6 × 10, or as 60 = 4 × 15, or as 60 = 2 × 30 — expand each fully and you always land on two 2s, one 3 and one 5. There is no secret alternative factorisation hiding anywhere. That is why we write the answer in exponent form with primes in ascending order:

60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.

This compact form captures the "bag of primes" cleanly and is exactly what an examiner expects to see — write 2² × 3 × 5, never 5 × 3 × 2 × 2.

Why 1 is deliberately not a prime. If 1 were allowed as a prime, then 60 = 2² × 3 × 5 = 1 × 2² × 3 × 5 = 1 × 1 × 2² × 3 × 5, and so on — infinitely many "different" factorisations. Uniqueness would collapse. Mathematicians exclude 1 from the primes precisely to keep the theorem clean; a prime needs two distinct divisors, and 1 has only one.

Section 2 — The factor-tree method, step by step

A factor tree is the standard tool to find the prime factorisation. Follow a fixed routine so you never miss a step.

Step 1: Write the number at the top.
Step 2: Split it into any product of two factors (grab a small prime like 2, 3 or 5 if you can spot one).
Step 3: For each branch ask, "Is this prime?" If yes, stop that branch. If no, split it further.
Step 4: Continue until every branch ends in a prime.
Step 5: Collect all the primes and write them with exponents, in ascending order.

The tree below factorises 540. Notice how every path — whichever split you begin with — funnels down to the same primes: 540 = 2² × 3³ × 5.

Before you put your pen down, run three checks: (1) Are all my factors prime — no 4, 6, 8, 9, 15, 21, 25 left hiding? (2) If I multiply them back, do I recover the original number? (3) Have I written the answer in exponent form with primes in ascending order? A common slip is leaving a composite factor inside the answer: writing "60 = 4 × 15" is a factorisation but not a prime factorisation, because 4 and 15 are still composite. Always push every branch down to a prime.

Section 3 — HCF and LCM from prime factorisation

Once numbers are in prime form, HCF (Highest Common Factor) and LCM (Lowest Common Multiple) become almost mechanical.

• HCF = product of the smallest power of each prime that is common to all the numbers.
• LCM = product of the greatest power of every prime that appears in any of the numbers.

The strip diagram below shows the shared and extra primes of 60 = 2² × 3 × 5 and 72 = 2³ × 3². The overlap (the smaller powers 2² and 3) is the HCF; the full union (largest of every prime) is the LCM.

For two positive integers a and b there is a golden identity:

HCF(a, b) × LCM(a, b) = a × b.

This gives instant cross-checks and shortcuts: LCM = (a × b) ÷ HCF, and HCF = (a × b) ÷ LCM. Important caution — this product rule holds only for two numbers. For three or more it fails in general; for 6, 72, 120 we get HCF = 6 and LCM = 360, but 6 × 360 = 2160 while 6 × 72 × 120 = 51840, nowhere near equal.

Section 4 — Why uniqueness is the bedrock (with the irrationality lemma)

Why is the uniqueness clause such a big deal? Because nearly every result in Real Numbers quietly leans on it. When you declare "HCF(60, 72) is definitely 12", you are relying on each number having one and only one prime skeleton; a second hidden factorisation could shift the answer.

The clearest payoff is the proof that √2 is irrational, which NCERT builds on a single lemma:

Lemma: If a prime p divides a², then p divides a. (This is a direct consequence of the Fundamental Theorem: the prime p must appear in the unique factorisation of a².)

Proof that √2 is irrational (by contradiction).
Assume, to the contrary, that √2 is rational. Then √2 = p/q for some integers p, q with q ≠ 0 and HCF(p, q) = 1 (lowest terms).
Squaring: 2 = p²/q², so p² = 2q². Hence 2 divides p². By the lemma (with p = 2), 2 divides p. Write p = 2m.
Substitute: (2m)² = 2q² ⇒ 4m² = 2q² ⇒ q² = 2m². So 2 divides q², and again by the lemma 2 divides q.
But now 2 divides both p and q, contradicting HCF(p, q) = 1. The assumption is false, so √2 is irrational. ∎

The identical argument proves √3, √5, √7, … irrational (replace the prime 2 by 3, 5, 7). Two corollaries you must remember:

• Rational ± Irrational = Irrational (e.g., 5 + √2 is irrational).
• Non-zero rational × Irrational = Irrational (e.g., 3√2 is irrational).

Beware the trap "sum of two irrationals is always irrational" — false, since √2 + (−√2) = 0 is rational.

Section 5 — Everyday power of the theorem

The theorem is not just exam theory; it is the reason repeating events sync up. Imagine two LED signs outside a shop: one blinks every 60 seconds, the other every 72 seconds, both starting together at 12:00:00. They next blink together after LCM(60, 72) = 360 seconds = 6 minutes, i.e. at 12:06:00. RFID tags, traffic-signal timers and railway announcement cycles all ride on this LCM logic — and it returns a single definite answer only because the prime skeleton is unique.

The same skeleton explains a neat divisibility fact: a number ending in 0 must have both 2 and 5 in its factorisation (since 10 = 2 × 5). So 6ⁿ can never end in 0, because 6ⁿ = 2ⁿ × 3ⁿ contains no factor 5 — no amount of raising to powers can conjure a prime that was not there to begin with. This "you cannot invent a missing prime" idea is the engine behind a whole family of board HOTS questions.

---

Work through these examples. They begin with plain factorisation and HCF/LCM, then build to the twists NCERT boards enjoy — verification, three-number cases, irrationality proofs, and "invent-a-missing-prime" reasoning. Keep handy: HCF = smallest powers of common primes, LCM = greatest powers of all primes, and HCF × LCM = a × b (two numbers only).

Example 1 — Factorise using a factor tree (basic)

Q: Express 96 as a product of its prime factors.
Solve: 96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3.
Collect the primes: five 2s and one 3.
Answer: 96 = 2⁵ × 3. Check: 2⁵ × 3 = 32 × 3 = 96 ✓

Example 2 — HCF and LCM by prime factorisation

Q: Find the HCF and LCM of 96 and 404, and verify HCF × LCM = product.
Given: 96 = 2⁵ × 3 and 404 = 2² × 101 (101 is prime).
Solve:
HCF = smallest power of each common prime. Only common prime is 2, smallest power 2² ⇒ HCF = 2² = 4.
LCM = greatest power of every prime = 2⁵ × 3 × 101 = 32 × 3 × 101 = 9696.
Verify: HCF × LCM = 4 × 9696 = 38784, and 96 × 404 = 38784.
Answer: HCF = 4, LCM = 9696, and both products equal 38784 ✓

Example 3 — Find the missing number from HCF and LCM

Q: The HCF of two numbers is 9 and their LCM is 360. If one number is 45, find the other.
Given: HCF = 9, LCM = 360, one number a = 45.
Formula: HCF × LCM = a × b (valid for two numbers).
Solve: 9 × 360 = 45 × b ⇒ 3240 = 45b ⇒ b = 3240 ÷ 45 = 72.
Answer: The other number is 72. Check: 45 = 3² × 5 and 72 = 2³ × 3² ⇒ HCF = 3² = 9 and LCM = 2³ × 3² × 5 = 360 ✓

Example 4 — Tricky: three numbers, and why the product rule breaks

Q: Find HCF and LCM of 6, 72 and 120 by prime factorisation. Does HCF × LCM = product of the three numbers?
Given: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5.
Solve:
HCF = smallest powers of primes common to all three: 2¹ and 3¹ ⇒ HCF = 2 × 3 = 6.
LCM = greatest powers of every prime: 2³ × 3² × 5 = 8 × 9 × 5 = 360.
Test the product rule: HCF × LCM = 6 × 360 = 2160, but 6 × 72 × 120 = 51840. Not equal.
Answer: HCF = 6, LCM = 360. The identity HCF × LCM = product holds only for two numbers, so it correctly fails here. Trap: never apply the two-number shortcut to three or more numbers.

Example 5 — Tricky: word problem (LCM of blinking / bells)

Q: Three temple bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 6:00 a.m., at what time will they next ring together?
Given: Intervals 9 = 3², 12 = 2² × 3, 15 = 3 × 5.
Formula: Next common ring after LCM(9, 12, 15) minutes.
Solve: LCM = greatest powers = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes = 3 hours.
6:00 a.m. + 3 hours = 9:00 a.m.
Answer: They next ring together at 9:00 a.m. Sanity check: 180 ÷ 9 = 20, 180 ÷ 12 = 15, 180 ÷ 15 = 12 — all whole numbers, so 180 is indeed a common multiple ✓

Example 6 — Tricky: can a power end in zero? (HOTS)

Q: Check whether 6ⁿ can end with the digit 0 for any natural number n. Justify using the Fundamental Theorem of Arithmetic.
Solve: A number ends in 0 only if it is divisible by 10 = 2 × 5, i.e. its prime factorisation contains both 2 and 5. Now 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. By the uniqueness of prime factorisation, the only primes in 6ⁿ are 2 and 3 — the prime 5 never appears. Since 5 is absent, 6ⁿ is not divisible by 10.
Answer: No, 6ⁿ can never end with 0, because its unique prime factorisation contains no factor 5. Key idea: you cannot invent a prime that was never in the number.

Example 7 — Tricky: prove √5 is irrational

Q: Prove that √5 is irrational.
Solve (contradiction): Suppose √5 is rational, so √5 = p/q with integers p, q, q ≠ 0 and HCF(p, q) = 1.
Squaring: 5 = p²/q² ⇒ p² = 5q². So 5 divides p². By the lemma "prime p ∣ a² ⇒ p ∣ a", 5 divides p. Write p = 5m.
Then (5m)² = 5q² ⇒ 25m² = 5q² ⇒ q² = 5m². So 5 divides q², hence 5 divides q.
Now 5 divides both p and q, contradicting HCF(p, q) = 1.
Answer: The assumption is false, so √5 is irrational. ∎ (The same steps work for √2, √3, √7 — swap in the relevant prime.)

Example 8 — Tricky: assertion–reason on primes and 1

Q: Assertion (A): The number 1 is neither prime nor composite.
Reason (R): A prime has exactly two distinct factors and a composite has more than two; 1 has only one factor. Choose: (a) both true and R explains A; (b) both true but R does not explain A; (c) A true, R false; (d) A false.
Solve: 1 has exactly one factor (itself), so it fails the "two distinct factors" test for a prime and the "more than two factors" test for a composite — hence 1 is neither. That is precisely what R states, so R is the correct reason for A. Both statements are true and R explains A.
Answer: (a) Both A and R are true, and R is the correct explanation of A. (This is also why 1 is excluded from primes — to keep prime factorisation unique.)

:::keypoints

• Fundamental Theorem of Arithmetic: every composite number is a unique product of primes, apart from order.
• 1 is neither prime nor composite; excluding it from primes is what keeps factorisation unique.
• HCF = product of smallest powers of common primes; LCM = product of greatest powers of all primes.
• For two numbers only: HCF × LCM = a × b (fails for three or more).
• Key lemma: if a prime p divides a², then p divides a — the engine of every irrationality proof.
• √2, √3, √5, √7 are irrational, proved by contradiction using that lemma.
• Rational ± Irrational is irrational; non-zero rational × irrational is irrational.
• A number ends in 0 only if both primes 2 and 5 divide it, so e.g. 6ⁿ never ends in 0.
  :::

:::memory
"Prime Skeleton, Same Skeleton." Every composite has one and only one prime skeleton — reorder the bones if you like, but you can never change which bones are there, and you can never add a bone (prime) that was not in the number to start with.
:::

:::recap

• Composite → product of primes, always uniquely (existence + uniqueness).
• Prime factorisation is the single engine behind HCF, LCM and divisibility.
• HCF × LCM = product works for two numbers only.
• The lemma "p ∣ a² ⇒ p ∣ a" drives every proof that a surd is irrational.
• Stop only when every factor is prime, then write it in exponent form, ascending order.
  :::

Build a perfectly square room exactly 1 metre on each side and stretch a tape across the diagonal — no matter how finely your tape is marked, you can never read the diagonal exactly. Its length is √2 metres, a number that refuses to be written as any ratio of whole numbers. Numbers like this are not mistakes or approximations; they are a whole different species called irrational numbers, and Class 10 asks you to prove that they exist.

Definition: A rational number is any number that can be written as p/q, where p and q are integers and q ≠ 0. Its decimal expansion is either terminating (like 0.75) or non-terminating but repeating (like 0.333…). Examples: 3, −7, 1/2, 11/13, 22/7.

Definition: An irrational number is a real number that cannot be written as p/q with integer p and q (q ≠ 0). Its decimal expansion is non-terminating AND non-repeating. Classic examples: √2, √3, √5, √7, π, e.

Definition: Proof by contradiction (reductio ad absurdum) is a method where we assume the statement we want to prove is false, follow the logic honestly, and reach an impossibility. Since the impossibility grew out of our assumption, the assumption itself must be wrong — so the original statement is true.

The lemma that powers every irrationality proof

Every irrationality proof in NCERT Class 10 rests on one small but mighty result that comes straight from the Fundamental Theorem of Arithmetic (every integer greater than 1 factorises into primes uniquely):

Theorem 1.3 (NCERT): If a prime number p divides a², then p divides a (where a is a positive integer).

Why is this true? If p divides a², then p must appear in the prime factorisation of a². But the primes of a² are exactly the primes of a, each appearing twice (paired up). So a prime sitting inside a² had to be sitting inside a in the first place. There is nowhere else for it to have come from.

You will use this lemma twice in a single proof — once to catch a factor in the numerator, and once more to catch the same factor in the denominator. Keep it at your fingertips: p prime and p | a² ⇒ p | a.

The strategy in plain words

We want to show √2 is irrational. Proving this directly is painful, because "irrational" means "cannot be written as p/q" — we would have to rule out every possible pair p, q. So we flip the script. We assume √2 is rational, write it as p/q in lowest terms, and then squeeze the equation until it breaks the very "lowest terms" condition we started with. That breakage is the contradiction. The only way out is to admit the first assumption was wrong.

Every such proof follows the same four moves:

1. Assume the number is rational: it equals p/q with integers p, q, q ≠ 0, in lowest terms (gcd(p, q) = 1).
2. Square or rearrange to link p, q and the prime you care about.
3. Use the lemma to show the prime divides both p and q.
4. Spot the contradiction — p and q were supposed to share no factor, yet they share one. Hence the assumption was false.

That last line is the punchline. The "lowest terms" opening is the cliff over which the contradiction throws the assumption. If you forget to state "in lowest terms" at the very start, the proof has nothing to contradict.

The proof that √2 is irrational, step by step

We use proof by contradiction.

Step 1 — Assume the opposite. Suppose √2 is rational. Then √2 = p/q, where p and q are integers, q ≠ 0, and the fraction is in its lowest terms, i.e. gcd(p, q) = 1.

Step 2 — Rearrange and square.
√2 = p/q ⇒ q√2 = p ⇒ squaring both sides, 2q² = p². …(i)

Step 3 — Deduce a factor of p. Equation (i) shows 2 divides p². By the lemma (2 is prime), 2 divides p. So write p = 2m for some integer m.

Step 4 — Substitute back. Put p = 2m into (i):
2q² = (2m)² = 4m² ⇒ q² = 2m². …(ii)

Step 5 — Deduce a factor of q. Equation (ii) shows 2 divides q², so by the same lemma, 2 divides q.

Step 6 — Reach the contradiction. From Steps 3 and 5, both p and q are divisible by 2, so 2 is a common factor of p and q. This contradicts our assumption that gcd(p, q) = 1 (lowest terms).

Step 7 — Conclusion. The assumption that √2 is rational must be false. Therefore √2 is irrational. Proved. ∎

Notice what we did not do. We did not compute √2 to many decimals and note it never repeats — that is only an observation, not a proof. We used pure logic: assume, derive, contradict. The very same template proves √3, √5, √7, and √p for any prime p.

The picture below places √2 exactly on the number line as the diagonal of a unit square — a length you can draw precisely but never write as a fraction.

Extending the idea: sums and products with irrationals

Once you know √2, √3 or √5 is irrational, you rarely need a fresh proof for expressions built from them. Two facts, both provable by the same contradiction trick, do the heavy lifting:

Fact A — rational + irrational = irrational. If their sum were rational, then irrational = rational − rational = rational, a contradiction.

Fact B — (non-zero rational) × irrational = irrational. If the product were rational, dividing by the non-zero rational would make the irrational rational — impossible.

These instantly tell you that 5 + √3, 2√7, and (3 − √5)/2 are all irrational. Do note the caution in Fact B: the rational multiplier must be non-zero, because 0 × √2 = 0 is rational. And the "closure" fails in general: √2 + (−√2) = 0 is rational, and √2 × √2 = 2 is rational, so irrationals are not closed under addition or multiplication. That is a favourite trap in assertion-reason questions.

Why irrational numbers matter

Irrational numbers are the rule, not the exception. Between any two rationals on the number line lie infinitely many irrationals. Without them you could not describe the diagonal of a unit square (√2), the circumference-to-diameter ratio of a circle (π), or the natural growth constant in compound interest (e). When the Pythagoreans first proved √2 irrational, it shattered their belief that "all numbers are ratios of whole numbers" — legend says the discoverer, Hippasus, was drowned for revealing it.

For your board exam, expect a direct 3-mark question ("Prove √2 is irrational" or "Prove √5 is irrational") and objective questions that lean on Facts A and B to settle expressions like 7 − 2√3 in a single line. Master the structure once and you unlock several marks across the Real Numbers chapter.

Common misconception: "My calculator shows √2 = 1.41421356, a finite decimal, so it is rational." The calculator merely truncates at its memory limit. The true expansion of √2 never terminates and never repeats — what you see is an approximation, not the number. Also beware: 22/7 is a rational approximation of π, but 22/7 ≠ π. And not all square roots are irrational — √4 = 2, √9 = 3 are perfectly rational. Only the square root of a non-perfect-square positive integer is irrational.

:::compare

Feature	Rational numbers	Irrational numbers
Form	p/q with q ≠ 0	Cannot be written as p/q
Decimal expansion	Terminating OR non-terminating repeating	Non-terminating AND non-repeating
Examples	1/2, 0.75, 0.333…, 22/7	√2, √3, π, e, 0.1010010001…
Closed under +, ×, ÷?	Yes (except ÷ by 0)	No (√2 + (−√2) = 0 is rational)
On the number line	Dense, but countable	Dense, and uncountable
:::

---

Work through these examples. They begin with the core board proofs (√2, √3, √5) and build up to the multi-step twists NCERT loves — sums, products, assertion-reason, reading the figure, and the classic sign/zero traps. Keep the lemma handy: p prime and p | a² ⇒ p | a, and the two shortcuts: rational + irrational = irrational, non-zero rational × irrational = irrational.

Example 1 — Prove √2 is irrational (basic, board 3-marker)

Q: Prove that √2 is irrational.
Solve:
Assume √2 = p/q in lowest terms (gcd(p, q) = 1, q ≠ 0).
Then q√2 = p ⇒ 2q² = p², so 2 | p² ⇒ 2 | p. Write p = 2m.
Substitute: 2q² = 4m² ⇒ q² = 2m², so 2 | q² ⇒ 2 | q.
Now 2 divides both p and q, contradicting gcd(p, q) = 1.
Answer: The assumption fails, so √2 is irrational. ∎ (Sanity check: the contradiction lands precisely on the "lowest terms" line — exactly where it should.)

Example 2 — Prove √3 is irrational

Q: Prove that √3 is irrational.
Solve:
Assume √3 = p/q in lowest terms.
Squaring: 3 = p²/q² ⇒ 3q² = p², so 3 | p² ⇒ 3 | p (3 is prime). Write p = 3m.
Substitute: 3q² = 9m² ⇒ q² = 3m², so 3 | q² ⇒ 3 | q.
Then 3 divides both p and q — contradiction with gcd(p, q) = 1.
Answer: √3 is irrational. ∎ Note how only the prime changed from 2 to 3; the skeleton is identical.

Example 3 — Prove 5 − √3 is irrational (given √3 is irrational)

Q: Given that √3 is irrational, prove that 5 − √3 is irrational.
Solve:
Assume 5 − √3 is rational, say 5 − √3 = a/b (integers a, b; b ≠ 0).
Isolate the surd: √3 = 5 − a/b = (5b − a)/b.
Since a, b are integers, (5b − a)/b is a ratio of integers, i.e. a rational number.
So √3 would be rational — but it is given irrational. Contradiction.
Answer: 5 − √3 is irrational. ∎ (This is just Fact A: rational − irrational = irrational.)

Example 4 — Tricky: prove 3 + 2√5 is irrational

Q: Prove that 3 + 2√5 is irrational, given that √5 is irrational.
Solve:
Assume 3 + 2√5 is rational, say 3 + 2√5 = a/b (integers a, b; b ≠ 0).
Then 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b).
The right side is a ratio of integers (denominator 2b ≠ 0), hence rational.
So √5 would be rational — contradicting the given fact. Contradiction.
Answer: 3 + 2√5 is irrational. ∎ Trap: you must divide by 2 as well as b; the multiplier 2 does not disturb rationality since it is a non-zero rational (Fact B in reverse).

Example 5 — Tricky: prove √5 is irrational, then use it for 1/√5

Q: Prove √5 is irrational, and hence show that 1/√5 is irrational.
Solve:
Part 1: Assume √5 = p/q in lowest terms. Then 5q² = p² ⇒ 5 | p² ⇒ 5 | p. Write p = 5m.
So 5q² = 25m² ⇒ q² = 5m² ⇒ 5 | q² ⇒ 5 | q. Then 5 | p and 5 | q — contradiction. Hence √5 is irrational.
Part 2: Rationalise: 1/√5 = (1/√5) × (√5/√5) = √5/5. Now 1/5 is a non-zero rational and √5 is irrational, so by Fact B, (1/5) × √5 = √5/5 is irrational.
Answer: 1/√5 = √5/5 is irrational. ∎ (Check: if 1/√5 were rational, its reciprocal √5 would be rational too — impossible.)

Example 6 — Tricky (assertion–reason): are irrationals closed under addition?

Q: Assertion (A): The sum of two irrational numbers is always irrational. Reason (R): √2 is irrational. Choose: (a) both true, R explains A; (b) both true, R does not explain A; (c) A false, R true; (d) A true, R false.
Solve:
Test A with a counter-example: √2 and (−√2) are both irrational, but √2 + (−√2) = 0, which is rational. So A is false. R (√2 is irrational) is a true statement.
Answer: (c) — A is false, R is true. The trap: "irrational + irrational" is NOT always irrational; only "rational + irrational" is guaranteed irrational.

Example 7 — Tricky (figure reading): is the labelled diagonal rational?

Q: In the number-line figure above, a unit square (side 1) sits on the line and its diagonal is swung down with a compass to mark a point P. Is the length OP rational or irrational? Justify using the figure.
Solve:
By Pythagoras on the unit square, diagonal² = 1² + 1² = 2, so the diagonal (and hence OP) = √2.
Since 2 is not a perfect square, √2 is the square root of a non-perfect-square integer, which is irrational (proved in Example 1). The compass only transfers this exact length onto the line; it does not make it a fraction.
Answer: OP = √2, which is irrational. The figure shows a length you can construct precisely yet never express as p/q. (Sanity check: OP ≈ 1.414 lies between the ticks 1 and 2, as the dashed arc lands.)

Example 8 — Tricky: spot the flaw / classify the number

Q: A student claims "0.101001000100001… is rational because it has a clear pattern, while 22/7 is irrational because it is π." Identify and correct both errors.
Solve:
Error 1: 0.101001000100001… has a pattern, but the pattern is non-repeating (the block of 0s keeps growing), and the decimal is non-terminating. Non-terminating AND non-repeating ⇒ it is irrational, not rational. "Has a pattern" is not the test; "repeating block" is.
Error 2: 22/7 is a ratio of two integers (7 ≠ 0), so 22/7 is rational by definition. It is only an approximation of π; 22/7 ≠ π, and π itself is irrational.
Answer: 0.101001000100001… is irrational; 22/7 is rational. Both of the student's claims are wrong. The rigorous criterion is "non-terminating AND non-repeating", never "looks messy" or "has a pattern".

:::keypoints

• Rational = p/q with integers p, q and q ≠ 0; decimals terminate or repeat.
• Irrational decimals are non-terminating AND non-repeating; e.g. √2, √3, π, e.
• Lemma (from Fundamental Theorem of Arithmetic): if prime p | a², then p | a.
• Proof by contradiction: assume rational in lowest terms, then derive a shared prime factor.
• The contradiction always strikes the "lowest terms" (gcd = 1) assumption.
• 22/7 is rational and only approximates π; √4, √9 are rational — only √(non-perfect-square) is irrational.
• rational + irrational = irrational; (non-zero rational) × irrational = irrational.
• Irrationals are NOT closed under +, × (√2 + (−√2) = 0; √2 × √2 = 2).
  :::

:::memory
"ALICE" — Assume rational in Lowest terms, Imply the prime divides p, Chain back to show it divides q, Expose the common factor (contradiction!). Walk Alice through these five steps and the contradiction always lands on gcd(p, q) = 1.
:::

:::recap

• Irrational numbers cannot be written as p/q and have non-terminating, non-repeating decimals.
• Irrationality is proved by contradiction using the prime-divisibility lemma p | a² ⇒ p | a.
• The proofs of √3 and √5 mirror √2 exactly — swap the prime, keep the skeleton.
• rational ± irrational and (non-zero rational) × irrational stay irrational — a shortcut worth many objective marks.
• "Non-terminating AND non-repeating" is the true test of irrationality, not "looks messy" or "has a pattern".
  :::

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.

:::compare

Fraction (lowest terms)	Denominator factorisation	Verdict
7/80	2⁴ × 5	Terminating (0.0875)
13/3125	5⁵	Terminating (0.00416)
17/8	2³	Terminating (2.125)
7/30	2 × 3 × 5	Non-terminating (0.2333…)
1/7	7	Non-terminating (0.142857…)
6/15 → 2/5	5	Terminating (0.4) — reduce first!
:::

---

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 ✓

:::keypoints

• Every rational p/q is either terminating or non-terminating repeating — never non-repeating.
• A rational terminates iff its denominator (in lowest terms) is of the form 2ⁿ × 5ᵐ.
• Always reduce to lowest terms first, then prime-factorise the denominator.
• Any prime other than 2 or 5 in the reduced denominator ⇒ non-terminating repeating.
• Number of decimal places in a terminating expansion = max(n, m).
• The rule works because 10 = 2 × 5, so powers of 10 absorb only 2s and 5s.
• A non-terminating, non-repeating decimal (like √2, π) is irrational, not rational.
• Convert a recurring decimal to p/q by subtracting after multiplying by 10^(block length).
  :::

:::memory
"Only Two and Five let me Survive." If the reduced denominator holds any prime other than 2 or 5, the decimal can never finish. T-F-S: Two, Five, Survive.
:::

:::recap

• Reduce, then prime-factorise the denominator — no division needed to classify.
• Only 2s and 5s ⇒ terminating; any other prime ⇒ non-terminating repeating.
• Terminating length = max(power of 2, power of 5) in the denominator.
• Forgetting to reduce is the top error — 6/15 reduces to 2/5 and does terminate.
• Recurring decimals are still rational (e.g. 0.overline(36) = 4/11).
  :::