Chapter 5 of 19

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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board SSC Mathematics · 4 marks · Paper: Section A (2-mark) + Section B (2-mark proof)

Real Numbers

RATIONAL numbers (terminating and recurring decimals). IRRATIONAL numbers (proof that √2 is irrational). Euclid's Division Lemma. Fundamental Theorem of Arithmetic. HCF and LCM. AP Board Class 10 Mathematics.

⏱ 10 min readEasy difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply Euclid's Division Lemma: a = bq + r (0 ≤ r < b) to find HCF of any two numbers
2Find HCF and LCM using prime factorisation; apply HCF × LCM = product of two numbers
3Determine whether a fraction is terminating or non-terminating from the prime factors of its denominator
4Prove that √2, √3, √5, and expressions like 3 + 2√5 are irrational using contradiction
5Express any positive integer in specific forms (e.g., 4q+1 for squares of odd numbers)

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Why this chapter matters

Real Numbers is the gateway chapter of AP SSC Mathematics. Euclid's Division Lemma powers the HCF algorithm tested in every AP exam. The irrationality proofs (√2, √3, √5) are a guaranteed 2-mark question. Mastering this chapter in 2 focused hours secures reliable marks with zero surprises.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Prime factorisation (Class 8)

Factorising numbers into primes — foundation for HCF/LCM

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LCM and HCF (Class 6-7)

Basic concepts; Class 10 extends them algorithmically

Real Numbers

Rational Numbers — Numbers as p/q (q≠0)

Terminating decimal: Denominator has ONLY 2 and 5 as prime factors. 3/8 = 0.375. Non-terminating recurring: Denominator has factors OTHER than 2 and 5. 1/3 = 0.333...

Irrational Numbers — Cannot be expressed as p/q

√2, √3, √5, π. Proof that √2 is irrational: Assume √2 = p/q in simplest form → 2q² = p² → p² is even → p is even → 2q² = 4k² → q² = 2k² → q is even → p,q both even → CONTRADICTION.

Fundamental Theorem of Arithmetic

Every composite number = UNIQUE product of primes. 84 = 2² × 3 × 7.

HCF and LCM Using Prime Factorisation

HCF = product of SMALLEST powers of common primes. LCM = product of GREATEST powers of all primes. For any two numbers: HCF × LCM = Product of the two numbers.

Common Mistakes

Using the HCF×LCM formula for THREE numbers — it ONLY works for TWO.

Euclid's Division Lemma and Algorithm

Lemma

For any positive integers a and b, there exist UNIQUE integers q and r such that a = bq + r, where 0 ≤ r < b.

'This is the formal way of saying: divide a by b, get quotient q and remainder r. The remainder is ALWAYS less than b.'

Algorithm — Finding HCF

To find HCF of two numbers:

Apply the lemma to the larger number (a) and smaller number (b): a = bq + r.
If r = 0, HCF = b. If r ≠ 0, apply the lemma to b and r.
REPEAT until the remainder is zero. The divisor at this step is the HCF.

Example 1: Find the HCF of 135 and 225 using Euclid's algorithm. 225 = 135 × 1 + 90 135 = 90 × 1 + 45 90 = 45 × 2 + 0 HCF = 45.

Example 2: Find HCF of 4052 and 12576. 12576 = 4052 × 3 + 420 4052 = 420 × 9 + 272 420 = 272 × 1 + 148 272 = 148 × 1 + 124 148 = 124 × 1 + 24 124 = 24 × 5 + 4 24 = 4 × 6 + 0 HCF = 4.

Example 3: Show that every positive even integer is of the form 2q and every odd integer is of the form 2q+1. Let a be any positive integer, b = 2. Then a = 2q + r, where r = 0 or 1. If r = 0: a = 2q (even). If r = 1: a = 2q + 1 (odd).

Example 4: Show that the square of any odd integer is of the form 4q + 1. Any odd integer = 2m + 1. Square = (2m+1)² = 4m² + 4m + 1 = 4m(m+1) + 1 = 4q + 1 (since m(m+1) is always even).

Fundamental Theorem of Arithmetic

Every composite number can be expressed as a UNIQUE product of primes (ignoring the order).

Example 5: Prime factorisation: 84 = 2² × 3 × 7 500 = 2² × 5³ 1764 = 2² × 3² × 7²

'Write the product in ascending order of primes. This is the STANDARD form expected in exams.'

HCF and LCM Using Prime Factorisation

Steps

Find the prime factorisation of each number.
HCF = product of SMALLEST powers of COMMON primes.
LCM = product of GREATEST powers of ALL primes present.

Example 6: Find HCF and LCM of 12, 15, and 21. 12 = 2² × 3 15 = 3 × 5 21 = 3 × 7 HCF = 3¹ = 3 (only common prime is 3, smallest power = 1). LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420.

Example 7: Find HCF and LCM of 6, 72, and 120. 6 = 2 × 3 72 = 2³ × 3² 120 = 2³ × 3 × 5 HCF = 2¹ × 3¹ = 6 LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Key Identity: For ANY two positive integers a and b: HCF(a,b) × LCM(a,b) = a × b. Check: 6 × 360 = 2160. 6 × 72 × 120 = 51840... wait. This identity works ONLY for TWO numbers. For 6 and 72: HCF=6, LCM=72, a×b=432, HCF×LCM=6×72=432 ✓.

Proofs of Irrationality

Proof that √2 is irrational

Assume √2 = p/q (in lowest terms, q ≠ 0, HCF(p,q)=1). 2 = p²/q² → 2q² = p² → p² is even → p is even → p = 2k. 2q² = (2k)² = 4k² → q² = 2k² → q² is even → q is even. Thus both p and q are even → CONTRADICTING that p/q was in lowest terms. Hence √2 CANNOT be rational. Therefore √2 is irrational.

Proof that √3 is irrational

Assume √3 = p/q (in lowest terms). 3q² = p². p² is divisible by 3 → p is divisible by 3 → p = 3k. 3q² = 9k² → q² = 3k² → q² is divisible by 3 → q is divisible by 3. Both p and q divisible by 3 → CONTRADICTION. Hence √3 is irrational.

Proof that √5 is irrational

Assume √5 = p/q (in lowest terms). 5q² = p². p² is divisible by 5 → p is divisible by 5 → p = 5k. 5q² = 25k² → q² = 5k² → q is divisible by 5. Both divisible by 5 → CONTRADICTION. Hence √5 is irrational.

'This proof pattern works for ANY prime number: assume rational → square both sides → deduce numerator and denominator share the prime factor → contradiction.'

Proof that 3 + 2√5 is irrational

Assume 3 + 2√5 = p/q → 2√5 = p/q − 3 = (p−3q)/q → √5 = (p−3q)/(2q). If p,q are integers, then (p−3q)/(2q) is RATIONAL, implying √5 is rational. CONTRADICTION (√5 is irrational). Hence 3 + 2√5 is irrational.

Rational Numbers and Their Decimal Expansions

Theorem: Let x = p/q be a rational number in LOWEST TERMS.

If q = 2ᵐ × 5ⁿ (only prime factors 2 and 5), then x has a TERMINATING decimal expansion.
If q has a prime factor OTHER than 2 or 5, then x has a NON-TERMINATING RECURRING decimal expansion.

Terminating Decimals

Example 8: 7/8 = 7/2³ = 7 × 5³/2³×5³ = 875/1000 = 0.875. (Terminates because denominator = 2³.) Example 9: 13/125 = 13/5³ = 13×2³/5³×2³ = 104/1000 = 0.104. (Terminates because 125 = 5³.) Example 10: 17/80 = 17/(2⁴×5). Denominator is 2⁴×5 = 80. Only primes 2 and 5 → terminates. 17/80 = 0.2125.

Non-Terminating Recurring Decimals

Example 11: 1/3 = 0.333... = 0.\overline{3} (denominator 3 has prime factor 3). Example 12: 1/7 = 0.142857142857... = 0.\overline{142857} (denominator 7 has prime factor 7).

Identifying Without Division

'To check if p/q terminates: factorise q into primes. If only 2 and 5 appear, it terminates. If any other prime appears, it recurs.'

Example 13: Without actual division, classify 29/343. 343 = 7³. 7 is a prime factor other than 2 or 5 → NON-TERMINATING RECURRING.

Example 14: Without actual division, classify 77/210. 210 = 2 × 3 × 5 × 7. 3 and 7 are primes other than 2 or 5 → NON-TERMINATING RECURRING.

Example 15: Without actual division, classify 987/10500. 10500 = 105 × 100 = (3×5×7) × (2²×5²) = 2² × 3 × 5³ × 7. Has 3 and 7 → NON-TERMINATING RECURRING.

AP Exam Focus

Topic	Marks	Frequency
Euclid's division algorithm for HCF	3-4	Common
Prime factorisation (HCF/LCM)	3	Very Common
Proof that √2/√3/√5 is irrational	4-5	Very Common
Decimal expansion (terminating/recurring)	2-3	Common
HCF×LCM identity application	3	Common

Self-Test Questions

Find HCF of 867 and 255 using Euclid's algorithm. (Answer: 867=255×3+102, 255=102×2+51, 102=51×2+0 → HCF=51)
Express 156 as a product of primes. (Answer: 156 = 2² × 3 × 13)
Find LCM and HCF of 336 and 54, and verify HCF×LCM = product. (Answer: 336=2⁴×3×7, 54=2×3³. HCF=6, LCM=3024. 6×3024=18144=336×54 ✓)
Prove that √3 is irrational.
Without actual division, classify 51/150. (Answer: 150=2×3×5². Has factor 3 → Non-terminating recurring)
Find LCM of 12, 18, and 27 using prime factorisation. (Answer: 12=2²×3, 18=2×3², 27=3³. LCM=2²×3³=108)
Show that 5−√3 is irrational. (Answer: Assume 5−√3 = p/q → √3 = 5 − p/q = (5q−p)/q → rational → contradiction since √3 is irrational)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Euclid's Division Lemma

a = bq + r, where 0 ≤ r < b

a = dividend, b = divisor, q = quotient, r = remainder. Unique q and r exist for every pair a, b.

Euclid's HCF Algorithm

Repeatedly apply a = bq + r until r = 0; last non-zero remainder = HCF

Example: HCF(225, 135) → 225=135×1+90 → 135=90×1+45 → 90=45×2+0 → HCF = 45.

HCF × LCM formula

HCF(a, b) × LCM(a, b) = a × b

ONLY valid for EXACTLY TWO numbers. NOT applicable to three or more numbers.

Prime Factorisation — HCF

HCF = product of SMALLEST powers of COMMON prime factors

For 12 = 2²×3 and 18 = 2×3²: HCF = 2¹×3¹ = 6.

Prime Factorisation — LCM

LCM = product of GREATEST powers of ALL prime factors

For 12 and 18: LCM = 2²×3² = 36.

Terminating Decimal Condition

p/q terminates ⟺ q = 2ᵐ × 5ⁿ (m, n ≥ 0)

Fully reduce p/q first. If denominator has ONLY 2s and 5s as factors → terminates.

Proof: √2 is irrational

Assume √2 = p/q (HCF=1) → 2q²=p² → p even → p=2k → q even → contradiction

Pattern works for ANY prime p: assume √p rational → both numerator and denominator share factor p → contradiction.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Applying HCF × LCM = product for three numbers

✓ This formula is ONLY valid for two numbers. For three numbers a, b, c: use prime factorisation to find HCF and LCM separately.

WATCH OUT

✗ Changing subscripts in formulas to balance — or forgetting to fully reduce before checking termination

✓ For terminating decimal check: ALWAYS simplify p/q to its lowest form FIRST. Then check the denominator's prime factors.

WATCH OUT

✗ In irrationality proof, writing 'p is even' without explaining why

✓ Explain: p² is even → p must be even (because if p were odd, p² would be odd). This is the key logical step examiners check.

WATCH OUT

✗ Euclid's algorithm: applying to wrong pair

✓ Always divide the LARGER number by the SMALLER. In each step, the previous divisor becomes the new dividend and the previous remainder becomes the new divisor.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Real Numbers?

7 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

7 questions~5 min worth ~4 marks in Andhra Pradesh (BIEAP) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Euclid's Division Lemma: a = bq + r (0 ≤ r < b). Apply repeatedly until r = 0; last non-zero remainder = HCF.
•Fundamental Theorem of Arithmetic: every composite number has a UNIQUE prime factorisation.
•HCF × LCM = product of two numbers (ONLY for two numbers, not three).
•HCF from prime factorisation = SMALLEST powers of COMMON primes.
•LCM from prime factorisation = GREATEST powers of ALL primes.
•Terminating decimal: denominator (in lowest form) has ONLY 2s and 5s as prime factors.
•Non-terminating recurring: denominator has a prime factor OTHER than 2 or 5.
•Irrationality proof template: assume rational p/q in lowest terms → square → both p and q share a prime factor → contradiction with HCF = 1.
•For composite irrationality (e.g., 3+2√5): assume rational → isolate √5 → show it equals a rational → contradiction.
•Euclid's Lemma: if prime p divides a², then p divides a. This is the KEY step in irrationality proofs.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 4 marks

Question type	Marks each	Typical count	What it tests
Very Short Answer (VSA)	1	2	Definition, formula recall
Short Answer (SA)	2	2	Direct application
Long Answer (LA)	4	1	Multi-step problem with full working

Prep strategy

AP SSC board rewards clearly shown working — never skip steps
Memorise all formulas: AP board MCQs test formula recall directly
Practise 4-mark questions with full working under timed conditions
For word problems: identify variables, write equation, solve, state answer with units

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Gear ratios in machines

HCF determines the synchronisation of gears with different numbers of teeth — used in watches, bicycles, and engines.

Tile floor planning

The largest square tile that fits a rectangular room without cutting is the HCF of the room's dimensions.

Music rhythm cycles

LCM tells musicians when two rhythmic patterns will align — crucial in polyrhythmic music.

Cryptography (RSA algorithm)

The irrational-looking property of large prime numbers underpins internet encryption via the RSA algorithm.

Calendar systems

LCM of lunar (29.5 days) and solar (365.25 days) cycles determines when calendars realign — the basis of leap year calculations.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Euclid's algorithm: write each step fully — AP examiners award step marks. Show the final 'divisor when remainder = 0' as the HCF explicitly.
HCF/LCM: factor tree or ladder method — circle the prime, write powers systematically. HCF takes MIN powers; LCM takes MAX powers.
Irrationality proof: 5-step structure — (1) Assume rational, (2) Write p/q in lowest terms, (3) Square and deduce numerator divisible by prime, (4) Substitute to get denominator also divisible, (5) State contradiction. AP board awards 1 mark per correct step.
Terminating decimal: first simplify the fraction. Then factorise ONLY the denominator. Write a 1-line conclusion.
Time saver: HCF(a, b) × LCM(a, b) = a × b. Use this to find LCM when you only need it for two numbers.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

The Euclidean algorithm runs in O(log min(a,b)) steps — this is why it's efficient for large numbers in computer science.
Bézout's Identity: for any two integers a, b, there exist integers x, y such that ax + by = HCF(a, b). Useful in number theory olympiads.
A rational number p/q (in lowest terms) has a terminating decimal iff q = 2ᵐ × 5ⁿ. The number of decimal places = max(m, n).
The set of all irrational numbers is uncountably infinite — far 'larger' than the set of rationals (which is countably infinite). This is Cantor's diagonal argument.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSCGuaranteed 2-4 marks — Euclid's algorithm and irrationality proof appear in nearly every paper

EAMCET FoundationNumber theory basics; HCF/LCM used in simplification problems

IMO (Junior)Modular arithmetic and divisibility extend directly from Euclid's lemma

NTSEHCF/LCM problems in both MAT and SAT sections

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Questions students ask

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When the remainder is 0, the DIVISOR at that step divides the previous remainder exactly — meaning it is the LARGEST number that divides both original numbers. That is, by definition, the HCF.

The proof uses the fact that for two numbers: every prime factor of one divides the other at most to one power in HCF and to the maximum power in LCM. For three numbers, this pairing breaks down.

No! 22/7 is a non-terminating recurring decimal (≈ 3.142857142857...). 7 is not of the form 2ᵐ × 5ⁿ. Although 22/7 is often used as an approximation for π, they are not equal.

Yes! √2 × √2 = 2 (rational). The product or sum of two irrationals can be rational or irrational depending on the specific numbers.

Euclid's method is ALGORITHMIC and does not require full factorisation — faster for large numbers. Prime factorisation is more transparent and simultaneously gives both HCF and LCM.

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