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Real Numbers Class 10: Formulas, Solved Examples & Common Mistakes (2026 Guide)

📅 Updated Sep 2, 2026 ⏱ 13 min read 📘 CBSE / NCERT Class 10, Chapter 1 ✍️ MyTestSeries Expert Team
Real Numbers Class 10 formulas and Euclid's Division Lemma explained by MyTestSeries

Ask any Class 10 student which chapter feels the trickiest to explain out loud — even when they can solve the questions — and "Real Numbers" comes up more often than you'd expect. It isn't hard because the ideas are complicated. It's hard because the chapter is proof-heavy and definition-heavy in a way earlier chapters weren't: Euclid's Division Lemma, the Fundamental Theorem of Arithmetic, irrationality proofs, and the terminating-decimal rule all ask you to reason precisely, not just plug into a formula.

This guide walks through every formula in the chapter with a solved example next to it, then lists the exact mistakes examiners see year after year — so you know precisely where marks are usually lost, and how to avoid losing yours.

5Core Concepts
6Solved Examples
7Common Mistakes
1Quick Formula Table
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1. What Are Real Numbers?

Real numbers are every number that can sit on a number line — every rational number (numbers expressible as p/q where q ≠ 0, including whole numbers, integers, fractions and terminating/repeating decimals) and every irrational number (non-terminating, non-repeating decimals like √2, π, and e). Class 10's Real Numbers chapter doesn't introduce this definition for the first time — it builds three specific tools on top of it: a method to find HCF cleanly (Euclid's algorithm), a theorem that guarantees every composite number has one unique prime "fingerprint" (the Fundamental Theorem of Arithmetic), and the logical machinery to prove certain numbers are irrational.

2. Euclid's Division Lemma

a = bq + r,   where 0 ≤ r < b a, b are positive integers · q = quotient · r = remainder · a is the "dividend", b is the "divisor"

In plain language: divide any positive integer a by any positive integer b, and you'll always get a unique quotient and a remainder that is strictly smaller than the divisor. This looks like ordinary long division — because it is — but stating it formally as a lemma lets it be used to prove other results, which is exactly what the next section does.

💡 Quick check: If a = 71 and b = 8, then 71 = 8 × 8 + 7. Here q = 8, r = 7, and indeed 0 ≤ 7 < 8. That's a valid application of the lemma.

3. Euclid's Division Algorithm — Finding HCF

Euclid's Division Algorithm is simply the lemma applied repeatedly. To find the HCF of two positive integers c and d, where c > d:

Step 1: c = dq + r  →  Step 2: if r = 0, HCF = d. If r ≠ 0, apply the lemma to d and r, and continue until the remainder is 0. The divisor at the final step (where remainder = 0) is the HCF.

This method is faster than listing all factors, especially for large numbers, and it's the technique CBSE examiners specifically expect to see written out step-by-step — not just the final answer.

4. Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order of the factors.

For example, 3960 can only ever break down into 2³ × 3² × 5 × 11 — no other combination of primes multiplies to give 3960. This uniqueness is what makes prime factorisation such a reliable tool: it's the reason the HCF-LCM shortcuts below always work, and it's also used to prove irrationality (Section 6).

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5. HCF & LCM by Prime Factorisation

HCF = product of the smallest power of each common prime factor
LCM = product of the greatest power of each prime factor present
HCF(a, b) × LCM(a, b) = a × b  (valid only for exactly two numbers)
NumberPrime Factorisation
62¹ × 3¹
202² × 5¹
HCF2¹ = 2 (smallest common power)
LCM2² × 3¹ × 5¹ = 60 (greatest powers)
💡 Sanity check: HCF × LCM = 2 × 60 = 120, and 6 × 20 = 120. They match — this is a fast way to double-check your working in an exam.

6. Proving Numbers Irrational (√2, √3, √5)

An irrational number cannot be written as p/q for any integers p, q (q ≠ 0). CBSE's standard proof method is contradiction — assume the opposite is true, then show that assumption breaks down.

Standard structure of the proof (using √2 as the model)

1. Assume √2 is rational: √2 = p/q, where p and q are coprime integers (no common factor except 1).
2. Squaring both sides: 2q² = p² — so p² is even, which means p itself must be even.
3. Write p = 2m. Substituting back: 2q² = 4m² → q² = 2m² — so q² is even, meaning q is also even.
4. But if both p and q are even, they share a common factor of 2 — contradicting the assumption that they were coprime.
5. Since the assumption leads to a contradiction, √2 cannot be rational. Hence, √2 is irrational.

The exact same five-step structure proves √3, √5, √7, and any √p where p is prime. It's also used for expressions like "3 + 2√5 is irrational" — you assume it equals a rational number, isolate the √5 term, and show that would force √5 itself to be rational (a contradiction).

⚠️ Watch out: Never skip stating "p and q are coprime" — this exact phrase is what the contradiction depends on, and examiners specifically check for it in the answer.
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7. Decimal Expansion: Terminating vs Non-Terminating

A fraction p/q (in lowest terms) has a terminating decimal expansion if and only if q = 2m × 5n, for non-negative integers m, n. If q has any prime factor other than 2 or 5, the decimal is non-terminating and repeating.
FractionDenominator's Prime FactorsType of Decimal
7/8Terminating (0.875)
13/31255⁵Terminating
1/3Non-terminating, repeating (0.333…)
17/62¹ × 3¹Non-terminating, repeating
⚠️ Watch out: Always simplify the fraction to lowest terms first. A denominator like 30 looks "bad", but 15/30 simplifies to 1/2 — which terminates. Checking the un-simplified denominator is one of the most common errors here.

8. Six Fully Solved Examples

Example 1 — HCF using Euclid's Division Algorithm

Find the HCF of 135 and 225.

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
HCF = 45 (the divisor when remainder becomes 0)

Example 2 — HCF & LCM by prime factorisation

Find the HCF and LCM of 6 and 20, and verify the product rule.

6 = 2 × 3,   20 = 2² × 5
HCF = 2  |  LCM = 2² × 3 × 5 = 60
Check: HCF × LCM = 2 × 60 = 120 = 6 × 20 ✅

Example 3 — Proving irrationality

Prove that √5 is irrational.

Assume √5 = p/q (p, q coprime). Then 5q² = p², so p² is divisible by 5, hence p is divisible by 5. Let p = 5m. Substituting: 5q² = 25m² → q² = 5m², so q is also divisible by 5. This contradicts p, q being coprime. Hence √5 is irrational.

Example 4 — Terminating decimal check

Without actual division, state whether 13/3125 has a terminating or non-terminating decimal expansion.

3125 = 5⁵. The denominator's only prime factor is 5 (i.e., 2⁰ × 5⁵), so by the rule in Section 7, 13/3125 has a terminating decimal expansion.

Example 5 — Applied HCF problem

Find the largest number that divides 615 and 963, leaving remainder 6 in each case.

Subtract the remainder first: 615 − 6 = 609, and 963 − 6 = 957. Now find HCF(609, 957) using Euclid's algorithm:
957 = 609 × 1 + 348
609 = 348 × 1 + 261
348 = 261 × 1 + 87
261 = 87 × 3 + 0
HCF = 87 — the required largest number.

Example 6 — Applying Euclid's Lemma with b = 6

Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, for some integer q.

By Euclid's Division Lemma, for any positive integer a and b = 6: a = 6q + r, where 0 ≤ r < 6, so r can be 0, 1, 2, 3, 4, or 5. This gives a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. Since a is odd, the even-remainder cases (6q, 6q+2, 6q+4) are eliminated, leaving only 6q+1, 6q+3, and 6q+5.

9. Common Mistakes Students Make

⚠️ Mistake 1 — Wrong remainder range. Writing r without checking 0 ≤ r < b. A "remainder" of 15 when dividing by 8 is invalid — keep dividing until it's smaller than the divisor.
⚠️ Mistake 2 — Confusing the lemma with the algorithm. The Lemma is one division step (a = bq + r). The Algorithm is the repeated process used specifically to find HCF. Examiners deduct marks when a question asking for HCF via the "algorithm" is answered with only one division step.
⚠️ Mistake 3 — Misusing HCF × LCM = product. This shortcut works for exactly two numbers only. It does not extend to three or more numbers — a very common trap in MCQs.
⚠️ Mistake 4 — Incomplete prime factorisation. Missing a repeated prime factor (e.g., writing 72 = 2³ × 3² instead of 2³ × 3² — check carefully: 72 = 8 × 9 = 2³ × 3²) changes both the HCF and LCM answers.
⚠️ Mistake 5 — Skipping the "coprime" assumption. In irrationality proofs, forgetting to state that p and q share no common factor removes the very contradiction the proof depends on.
⚠️ Mistake 6 — Checking the wrong denominator. Testing whether a decimal terminates using the original (un-simplified) fraction instead of reducing it to lowest terms first.
⚠️ Mistake 7 — Assuming irrational × irrational is always irrational. It's not always true — √2 × √2 = 2, which is rational. Don't generalise without checking.
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10. Quick Formula Recap Table

ConceptFormula
Euclid's Division Lemmaa = bq + r, 0 ≤ r < b
Fundamental Theorem of ArithmeticUnique prime factorisation of every composite number
HCF (prime factorisation method)Product of smallest powers of common primes
LCM (prime factorisation method)Product of greatest powers of all primes present
HCF–LCM relation (two numbers only)HCF(a,b) × LCM(a,b) = a × b
Terminating decimal conditionq = 2m × 5n (lowest terms)
Irrationality of √p√p is irrational for every prime p
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11. How to Revise This Chapter Effectively

📖 Read the "why" before the formula. Understand why 0 ≤ r < b matters before memorising the lemma — it's what makes every proof in this chapter work.

✍️ Write out full proofs by hand at least twice. Irrationality proofs are marked step-by-step; muscle memory for the five-step structure prevents skipped statements under exam pressure.

🧮 Practise the HCF-remainder pattern. "Find the largest number that divides X and Y leaving remainder r" is one of the most repeated question types — Example 5 above is the template.

🎯 Time yourself on decimal-expansion MCQs. These should take under 20 seconds each once the 2ᵐ×5ⁿ rule is automatic — speed here frees up time for longer geometry and trigonometry answers later in the paper.

🔁 Revisit this page before your unit test. Bookmark it, and use the Quick Formula Recap Table (Section 10) as your final five-minute revision before the exam.

12. Frequently Asked Questions

What is Euclid's Division Lemma in Class 10 Real Numbers?
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. It's the foundation used to build Euclid's Division Algorithm for finding the HCF of two positive integers.
What is the Fundamental Theorem of Arithmetic?
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique except for the order in which the prime factors occur. It underpins the HCF-LCM shortcut methods in this chapter.
What is the relation between HCF and LCM of two numbers?
HCF(a, b) × LCM(a, b) = a × b — but this shortcut only holds for exactly two numbers, not three or more.
How do you prove that root 2 is irrational?
By contradiction: assume √2 = p/q where p and q are coprime. Squaring gives 2q² = p², making p even. Writing p = 2m leads to q² = 2m², making q even too — contradicting the coprime assumption. So √2 cannot be written as p/q, meaning it is irrational.
How do you know if a decimal expansion will terminate?
A fraction p/q, reduced to lowest terms, terminates only if q = 2^m × 5^n for non-negative integers m and n. Any other prime factor in the denominator means the decimal is non-terminating and repeating.
What are the most common mistakes in Class 10 Real Numbers?
Forgetting 0 ≤ r < b in Euclid's Lemma, applying HCF × LCM = product to three or more numbers, skipping the "p and q coprime" step in irrationality proofs, and checking the terminating-decimal condition before reducing the fraction to lowest terms.

Conclusion: Precision Beats Memorisation

Real Numbers rewards students who understand why each rule holds, not just those who've memorised it. Once the logic behind Euclid's Lemma, the Fundamental Theorem of Arithmetic, and the irrationality proofs clicks, the rest of the chapter — HCF-LCM problems, decimal-expansion questions, applied word problems — becomes fast and almost mechanical.

Read through each proof once more, rework the six solved examples without looking at the answers, and then test yourself under timed conditions. That sequence — understand, rework, test — is what actually converts a formula sheet into exam marks.

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Tags: Real Numbers Class 10, Euclid's Division Lemma, Fundamental Theorem of Arithmetic, HCF LCM Class 10, Irrational Numbers Proof, CBSE 2026, NCERT Class 10 Maths Chapter 1

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