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

  • 1Classify numbers in the hierarchy: N ⊂ W ⊂ Z ⊂ Q ⊂ R
  • 2Identify terminating vs non-terminating recurring decimals from the denominator's prime factors
  • 3Convert a recurring decimal to a fraction using algebraic manipulation
  • 4Prove that √2 (and √3, √5) are irrational using proof by contradiction
  • 5Represent √n on the number line using the Pythagoras-based construction
  • 6Rationalise denominators of the form 1/√a and 1/(√a ± √b)
  • 7Apply laws of exponents including fractional and negative exponents
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Why this chapter matters
Real Numbers is the foundational algebra chapter in AP Class 9, introducing the complete number line and proof-based reasoning. The proof that √2 is irrational (by contradiction) is a standard 4-5 mark question — it appears in nearly every AP Class 9 exam and teaches the 'assume and contradict' proof technique used throughout higher mathematics. Rationalisation of denominators (using conjugate multiplication) is a 3-4 mark question. Laws of exponents for fractional and negative powers appear in 2-3 mark simplification questions. Identifying terminating vs non-terminating decimals and converting recurring decimals to fractions complete the chapter.

Before you start — revise these

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

Real Numbers — Class 9 Mathematics

1. The Number System — A Hierarchy

Natural Numbers (N) : 1, 2, 3, ... (counting numbers). Whole Numbers (W) : 0, 1, 2, 3, ... (N + 0). Integers (Z) : ... −2, −1, 0, 1, 2, ... (W + negatives). Rational Numbers (Q) : Numbers expressible as p/q (p,q ∈ Z, q ≠ 0). Irrational Numbers : Cannot be expressed as p/q. Real Numbers (R) : Q ∪ Irrationals.


2. Rational Numbers (Q)

Every rational number, when expressed as a decimal: Either TERMINATES (ends) — denominator has ONLY 2 and 5 as prime factors. Example: 3/8 = 0.375 (8 = 2³). 7/20 = 0.35 (20 = 2² × 5). Or is NON-TERMINATING RECURRING — denominator has prime factors OTHER than 2 and 5. Example: 1/3 = 0.333... = 0.3̄. 1/7 = 0.142857142857... = 0.142857̅.

Conversion: Recurring Decimal → Fraction

Let x = 0.333... → 10x = 3.333... → 10x − x = 3 → 9x = 3 → x = 3/9 = 1/3.


3. Irrational Numbers

Numbers that CANNOT be expressed as p/q. Non-terminating, NON-RECURRING decimals. Examples: √2 = 1.414213562... π = 3.141592653... e = 2.718281828...

Proof that √2 is Irrational (by Contradiction)

Assume √2 = p/q where p,q are coprime (no common factor except 1). Then 2 = p²/q² → p² = 2q² → p² is EVEN → p is EVEN → p = 2k. Substitute: (2k)² = 2q² → 4k² = 2q² → q² = 2k² → q² is EVEN → q is EVEN. CONTRADICTION: p and q are both even — they have a common factor 2. But we assumed they are coprime. Therefore, √2 CANNOT be expressed as p/q. Hence, √2 is IRRATIONAL.


4. Representing √n on the Number Line

Use Pythagoras Theorem. √2: Construct a right triangle with legs of 1 unit each → hypotenuse = √2. Transfer this length to the number line using a compass. √3: Construct a right triangle with legs √2 and 1 → hypotenuse = √3.


5. Operations on Real Numbers

  • Sum/difference of rational and irrational = IRRATIONAL. 2 + √3 (irrational).
  • Product of rational (≠0) and irrational = IRRATIONAL. 2√3 (irrational).
  • Product of two irrationals: MAY be rational or irrational. √2 × √2 = 2 (rational). √2 × √3 = √6 (irrational).

6. Rationalisation of Denominators

To rationalise 1/(√a + √b): multiply numerator AND denominator by the CONJUGATE (√a − √b).

Example: Rationalise 1/(√5 + √3). Multiply by (√5 − √3)/(√5 − √3) = (√5 − √3)/(5 − 3) = (√5 − √3)/2.

Example: Rationalise 5/(3 − √2). Multiply by (3 + √2)/(3 + √2) = 5(3+√2)/(9−2) = 5(3+√2)/7.


7. Laws of Exponents for Real Numbers

aᵐ × aⁿ = aᵐ⁺ⁿ. aᵐ ÷ aⁿ = aᵐ⁻ⁿ. (aᵐ)ⁿ = aᵐⁿ. aᵐ × bᵐ = (ab)ᵐ. a⁰ = 1 (a ≠ 0). a⁻ⁿ = 1/aⁿ.

Example: (2³ × 2⁴) / 2² = 2⁷ / 2² = 2⁵ = 32.


8. Common Mistakes

  1. 'π = 22/7 exactly' — 22/7 is an APPROXIMATION (3.142857...). π = 3.141592... They DIFFER.
  2. Forgetting to multiply BOTH numerator and denominator by the conjugate in rationalisation.
  3. 'All square roots are irrational' — √4 = 2 (rational). √9 = 3 (rational). Only square roots of NON-PERFECT SQUARES are irrational.

9. AP Exam Focus

TopicMarks
Rational vs Irrational identification2-3
Proof that √2 is irrational4-5
Rationalisation3-4
Laws of exponents2-3

Deep Dive — Proof that √3 is Irrational

The proof follows the EXACT same structure as √2. Assume √3 = p/q (p,q coprime). Square: 3 = p²/q² → p² = 3q² → p² is divisible by 3 → p is divisible by 3 → p = 3k. Substitute: 9k² = 3q² → q² = 3k² → q² is divisible by 3 → q is divisible by 3. CONTRADICTION: Both p and q are divisible by 3. But we assumed they are coprime. Hence, √3 is irrational. 'The AP exam often asks you to prove √3 or √5 is irrational. The structure is IDENTICAL to the √2 proof — just change the prime number. Practice writing it yourself.'

General method: To prove √n is irrational (where n is NOT a perfect square), replace "even" with "divisible by some prime factor of n."

More Worked Examples — Rationalisation

Example — Rationalise 1/(√7 − √6): Multiply numerator and denominator by (√7 + √6): = (√7+√6)/(7−6) = √7 + √6. 'When the denominator is √a − √b, multiplying by √a + √b gives a−b, which is rational.'

Example — Rationalise 3/(2√5 + 3√2): Multiply by (2√5 − 3√2): = 3(2√5−3√2)/((2√5)²−(3√2)²) = 3(2√5−3√2)/(20−18) = 3(2√5−3√2)/2.

Example — Rationalise (√2+1)/(√2−1): Multiply by (√2+1): = (√2+1)²/(2−1) = (2+2√2+1)/1 = 3+2√2.

Exponent Problems for Practice

Simplify: (a) 2^(1/2) × 2^(3/2) = 2^(4/2) = 2² = 4. (b) (3^(1/3))⁶ = 3² = 9. (c) 16^(3/4) = (16^(1/4))³ = 2³ = 8. (d) [5(8^(1/3)+27^(1/3))³]^(1/4) — work from inside out: 8^(1/3)=2, 27^(1/3)=3 → 5(5)³ = 5×125 = 625 → 625^(1/4) = 5.

Finding Rational Numbers Between Two Numbers

Method 1 — Average method: Between a and b, (a+b)/2 is always a rational number if a and b are rational. Method 2 — Decimal method: Write numbers as decimals and insert terminating decimals between them.

Example: Find 5 rational numbers between 2/5 and 3/5. 2/5=0.4, 3/5=0.6. Five numbers: 0.41, 0.42, 0.45, 0.5, 0.55 → 41/100, 42/100, 45/100, 50/100, 55/100.

Example: Find 3 irrational numbers between 2 and 3. √4.5, √5, √7, π, 2+√0.1... Any non-recurring decimal between them works. 'Irrational numbers are DENSE on the number line — there are infinitely many between any two distinct real numbers.'

Number Line Construction — Extending √n Spirals

The beautiful 'Square Root Spiral' (or 'Wheel of Theodorus'): Start with a right triangle of legs 1 and 1 → hypotenuse = √2. Build a new right triangle with legs √2 and 1 → hypotenuse = √3. Continue: √4=2, √5, √6, ... The spiral creates ALL square roots geometrically. 'This is a 4-mark practical geometry question in the AP exam. Bring a compass and a sharp pencil.'

Common Errors in Exponential Simplification

  1. Writing 2¹ᐟ² + 2¹ᐟ² = 2 → 2¹ᐟ² + 2¹ᐟ² = 2×2¹ᐟ² = 2^(3/2), NOT 2.
  2. Confusing (a+b)^(1/2) with a^(1/2) + b^(1/2) — NO, √(a+b) ≠ √a + √b.
  3. Writing (−8)^(1/3) as invalid — NO, (−8)^(1/3) = −2 because (−2)³ = −8. Odd roots of negative numbers are REAL.

Quick Self-Test

  1. Is 0.101001000100001... rational or irrational? (Answer: Irrational — non-terminating, non-recurring.)
  2. Rationalise: 1/(2−√3). (Answer: Multiply by (2+√3): = (2+√3)/(4−3) = 2+√3.)
  3. Simplify: (8^(1/3) × 27^(1/3)). (Answer: 2 × 3 = 6.)
  4. Prove √5 is irrational. (Answer: Follow the √2 proof pattern with divisibility by 5.)
  5. Express 0.6̄ as p/q. (Answer: x=0.666..., 10x=6.666..., 9x=6, x=2/3.)
  6. Between any two rational numbers, how many irrational numbers exist? (Answer: Infinitely many.)

Key formulas & results

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

Real Numbers, Irrationality and Exponents
NUMBER HIERARCHY: N (natural) ⊂ W (whole) ⊂ Z (integers) ⊂ Q (rational) ⊂ R (real). RATIONAL NUMBERS (Q): p/q where q≠0. Decimal forms: TERMINATING ↔ denominator has ONLY 2s and 5s as prime factors (3/8 = 0.375; 8 = 2³). NON-TERMINATING RECURRING ↔ denominator has prime factors other than 2 and 5 (1/3 = 0.333...; 1/7 = 0.142857142857...). CONVERTING RECURRING DECIMAL TO FRACTION: Let x = 0.666... → 10x = 6.666... → 9x = 6 → x = 2/3. For 0.12̄: x = 0.1222..., 10x = 1.222..., 100x = 12.222..., 90x = 11, x = 11/90. IRRATIONAL NUMBERS: Non-terminating, NON-RECURRING. Examples: √2, √3, √5, π, e. PROOF √2 IS IRRATIONAL (BY CONTRADICTION): Assume √2 = p/q (p, q coprime, q≠0). Square: 2q² = p² → p² is even → p is even → p = 2k. Substitute: 2q² = 4k² → q² = 2k² → q² is even → q is even. CONTRADICTION: p and q are both even, contradicting coprimeness. ∴ √2 is irrational. SAME PROOF WORKS FOR √3, √5 (substitute divisibility by 3 or 5). OPERATIONS: rational + irrational = irrational. rational × irrational (≠0) = irrational. irrational × irrational: can be rational (√2 × √2 = 2) or irrational (√2 × √3 = √6). RATIONALISATION: 1/√a = √a/a (multiply by √a/√a). 1/(√a+√b): multiply by (√a−√b)/(√a−√b) → result/(a−b). 1/(√a−√b): multiply by (√a+√b)/(√a+√b). EXAMPLES: 1/(√5+√3) × (√5−√3)/(√5−√3) = (√5−√3)/2. 5/(3−√2) × (3+√2)/(3+√2) = 5(3+√2)/7. LAWS OF EXPONENTS: aᵐ × aⁿ = aᵐ⁺ⁿ. aᵐ/aⁿ = aᵐ⁻ⁿ. (aᵐ)ⁿ = aᵐⁿ. (ab)ᵐ = aᵐbᵐ. a⁰ = 1. a⁻ⁿ = 1/aⁿ. a^(1/n) = ⁿ√a. a^(m/n) = (ⁿ√a)ᵐ. EXAMPLE: 16^(3/4) = (16^(1/4))³ = 2³ = 8.
AP EXAM KEY TRAPS: (1) π = 22/7 is an APPROXIMATION — they differ at the 4th decimal place. π is irrational; 22/7 is rational. (2) √4 = 2 (RATIONAL), √9 = 3 (RATIONAL) — not all square roots are irrational. Only √n where n is a NON-PERFECT SQUARE is irrational. (3) In rationalisation: multiply BOTH numerator AND denominator by conjugate. (4) PROOF STRUCTURE for irrationality: Assume rational (p/q coprime) → square both sides → show p is divisible by prime → substitute → show q is divisible by same prime → CONTRADICTION (coprime). (5) (−8)^(1/3) = −2 (valid, odd root of negative number). (−8)^(1/2) is NOT a real number (even root of negative number).
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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
Saying π = 22/7 exactly, or that all square roots are irrational
TWO CORRECTIONS: (1) π ≠ 22/7. Pi is an IRRATIONAL number: π = 3.14159265358979... It cannot be expressed as p/q. 22/7 = 3.142857142857... It is a RATIONAL approximation, accurate to 2 decimal places but NOT exact. They differ: π ≈ 3.14159..., 22/7 ≈ 3.14285... They differ from the 3rd decimal place onward. Always say '22/7 is an approximation of π.' (2) NOT ALL SQUARE ROOTS ARE IRRATIONAL. Square roots of PERFECT SQUARES are rational: √1 = 1, √4 = 2, √9 = 3, √16 = 4, √25 = 5, √100 = 10, etc. Only square roots of NON-PERFECT SQUARES are irrational: √2, √3, √5, √6, √7, √8, √10, etc. Rule: √n is irrational if and only if n is NOT a perfect square.

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?

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

1 questions~2 min

5-minute revision

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

  • NUMBER HIERARCHY: Natural (N: 1,2,3,...) ⊂ Whole (W: 0,1,2,...) ⊂ Integers (Z: ...,−2,−1,0,1,2,...) ⊂ Rational (Q: p/q form) ⊂ Real (R: rational + irrational). Real numbers fill the entire number line completely.
  • RATIONAL NUMBERS (Q): Can be written as p/q where p and q are integers and q ≠ 0. Decimal forms: either TERMINATING (ends, like 0.375) or NON-TERMINATING RECURRING (repeats, like 0.333...).
  • TERMINATING vs NON-TERMINATING TEST: A fraction p/q (in lowest form) is TERMINATING if and only if q has ONLY 2s and 5s as prime factors. Examples: 3/8 (8=2³) terminating; 7/20 (20=2²×5) terminating. 1/3 (3) non-terminating recurring; 5/6 (6=2×3) non-terminating recurring.
  • CONVERT RECURRING DECIMAL TO FRACTION: Let x = decimal. Multiply by 10ⁿ (n = length of recurring block) and subtract original. Example: x = 0.6̄ → 10x = 6.6̄ → 9x = 6 → x = 2/3. For 0.27̄: x = 0.272727..., 100x = 27.272727..., 99x = 27, x = 27/99 = 3/11.
  • IRRATIONAL NUMBERS: Cannot be expressed as p/q. Decimal form is NON-TERMINATING and NON-RECURRING. Examples: √2, √3, √5, π, e, the golden ratio φ. Important: √n is irrational only when n is NOT a perfect square. √4=2, √9=3 are rational.
  • PROOF √2 IS IRRATIONAL (by contradiction): Assume √2 = p/q (p,q coprime, q≠0). Square: 2q² = p² → p² is even → p is even → write p = 2k. Substitute: 2q² = 4k² → q² = 2k² → q² is even → q is even. CONTRADICTION: p and q both even contradicts coprimeness. Therefore √2 is irrational. Same proof for √3 (substitute 'divisible by 3') and √5.
  • OPERATIONS WITH IRRATIONALS: rational + irrational = irrational. rational × irrational (≠0) = irrational. irrational × irrational = sometimes rational (√2 × √2 = 2), sometimes irrational (√2 × √3 = √6).
  • REPRESENTING √n ON NUMBER LINE: Use Pythagoras construction. For √2: draw unit square with one corner at origin. Diagonal = √2. Use compass to mark √2 on the number line. For √3: hypotenuse of right triangle with sides 1 and √2. For √5: sides 1 and 2.
  • RATIONALISATION: 1/√a = √a/a (multiply by √a/√a). 1/(√a+√b): multiply by conjugate (√a−√b)/(√a−√b), denominator becomes (a−b). Identity: (√a+√b)(√a−√b) = a−b. Example: 1/(√5+√3) × (√5−√3)/(√5−√3) = (√5−√3)/(5−3) = (√5−√3)/2.
  • LAWS OF EXPONENTS: aᵐ × aⁿ = aᵐ⁺ⁿ. aᵐ ÷ aⁿ = aᵐ⁻ⁿ. (aᵐ)ⁿ = aᵐⁿ. (ab)ᵐ = aᵐbᵐ. a⁰ = 1 (a≠0). a⁻ⁿ = 1/aⁿ. a^(1/n) = ⁿ√a (nth root). a^(m/n) = (ⁿ√a)ᵐ. Example: 16^(3/4) = (16^(1/4))³ = 2³ = 8.

Andhra Pradesh (BIEAP) marks blueprint

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

Where this shows up in the real world

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

Computer floating-point arithmetic and decimal approximations

Computers cannot store irrational numbers exactly — they must use approximations. When you compute √2 on a calculator, it returns 1.4142135... truncated to some number of digits. Floating-point arithmetic (IEEE 754) handles these approximations using fractional representations. Every time you use π or √2 in a calculation, you are using a rational approximation. Understanding why these are approximations (not exact) is crucial for software engineering. The growing IT industry in AP requires programmers who understand numerical precision.

Cryptography and prime numbers

Modern cryptography (RSA encryption, used by banks, secure websites, government systems) relies on the difficulty of factoring large numbers into primes. The number theory developed in Real Numbers chapter — primes, divisibility, coprimes (used in the √2 proof) — is the foundation of cryptography. Every secure online transaction in AP (UPI, banking, online shopping) uses cryptographic algorithms built on the number theory you learn here.

Engineering and √n calculations in construction

Civil engineers in AP calculating diagonal beam lengths, slope angles, or distances frequently use square roots. A 3-4-5 right triangle (a Pythagorean triple) creates a 90° angle exactly — used by carpenters and bricklayers for centuries. When the triangle sides are NOT a Pythagorean triple (e.g., 1-1-√2 square diagonal), the answer is irrational and must be approximated to suitable precision. Construction practice teaches that being able to estimate √2 ≈ 1.414, √3 ≈ 1.732 to 3-4 decimals is sufficient for most building work.

Exam strategy

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

  1. √n irrationality proof (4-5 marks): structure as FIVE STEPS: (1) Assume √n is rational, so √n = p/q with p, q coprime, q ≠ 0. (2) Square: nq² = p². (3) Conclude p is divisible by n (using primality argument), write p = nk. (4) Substitute, conclude q is also divisible by n. (5) CONTRADICTION: p, q both divisible by n contradicts coprimeness. Therefore √n is irrational. Five steps = 5 marks.
  2. Rationalisation (3-4 marks): identify the conjugate of the denominator. Multiply numerator AND denominator by the conjugate. Use (a+b)(a−b) = a²−b² for the denominator. Simplify. Always show the conjugate explicitly: 'Multiplying by (√a − √b)/(√a − √b).'
  3. Recurring decimal to fraction (2-3 marks): write x = decimal. Multiply by 10ⁿ where n = length of repeating block. Subtract original. Solve for x. Show every step.
  4. Terminating vs non-terminating (1-2 marks): factor the denominator into primes. If only 2s and 5s → terminating. Otherwise non-terminating. State the rule explicitly before applying.
  5. Exponent simplification (2-3 marks): apply laws of exponents step-by-step. For a^(m/n), rewrite as (ⁿ√a)ᵐ. For negative exponents, write as reciprocals. Don't try to do multiple steps at once — show each application of a law.

Going beyond the textbook

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

  • Research the proof that π is irrational (Lambert, 1761) and that π is transcendental (Lindemann, 1882). Transcendental means NOT the root of any polynomial with integer coefficients — a stronger property than just irrational. The proof of transcendentality settled the ancient problem of 'squaring the circle' (constructing a square with the same area as a circle using only compass and straightedge) — proved IMPOSSIBLE because π is transcendental.
  • Investigate continued fractions — every irrational number has a unique infinite continued fraction expansion. For √2 = [1; 2, 2, 2, 2, ...] (period 1). For the golden ratio φ = [1; 1, 1, 1, 1, ...] (the 'most irrational' number). For π ≈ [3; 7, 15, 1, 292, ...] (no pattern). Continued fractions give the best rational approximations to irrationals. Research how 22/7 and 355/113 arise as continued fraction convergents to π.
  • Explore Cantor's diagonal argument — Georg Cantor (1891) proved that the irrational numbers between 0 and 1 are UNCOUNTABLY INFINITE (more numerous than the natural numbers). His proof: assume all real numbers in (0,1) can be listed; construct a new number by going down the diagonal and changing each digit. This new number cannot be in the list. Therefore the assumption is wrong. This proof opened the entire field of set theory and the study of different infinities.
  • Research the construction of real numbers from rationals — Dedekind cuts (1872) and Cauchy sequences (1880s). Dedekind defined each real number as a 'cut' that separates rational numbers into two sets. Cauchy sequences define reals as equivalence classes of sequences of rationals that 'should converge.' These rigorous constructions made the real number system mathematically respectable. Research the foundations of analysis.

Where else this chapter is tested

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

AP Board SSC (Class 10) — Real NumbersVery High — Class 9 Real Numbers directly extends to Class 10 Real Numbers (Euclid's algorithm, LCM/HCF, irrationality proofs)
JEE Main and AdvancedVery High — number theory, irrationality, and rationalisation are JEE topics; the proof technique by contradiction is widely used
NTSE (Mathematics)High — real numbers and proof techniques appear in NTSE Stage I and II
Mathematics Olympiad (RMO, INMO)High — number theory (primes, divisibility, irrationality) is one of the four main olympiad areas

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

π is defined as the ratio of a circle's circumference to its diameter. It was PROVED to be irrational in 1761 by Johann Lambert. π = 3.14159265358979323846... — non-terminating and non-recurring. It CANNOT be written as p/q. 22/7 = 3.142857142857... is a RATIONAL approximation. They agree at the first 2 decimal places but differ from the 3rd onward: π ≈ 3.14159..., 22/7 ≈ 3.14285... The difference is about 0.00126. For most school-level problems, using π = 22/7 gives answers accurate to about 2 decimal places — usually sufficient. But 22/7 is NOT π. Other approximations: 3.14, 3.14159, 355/113 (very accurate).

Because square roots do NOT distribute over addition. The rule √(a×b) = √a × √b holds for MULTIPLICATION but NOT for ADDITION. Concretely: √2 ≈ 1.414, √3 ≈ 1.732 → √2 + √3 ≈ 3.146. But √5 ≈ 2.236 ≠ 3.146. To verify algebraically: if √2 + √3 = √5, then squaring: 2 + 2√6 + 3 = 5 → 5 + 2√6 = 5 → 2√6 = 0, which is false. The correct simplifications: √8 = √(4×2) = 2√2 (multiplication splits). But √(4+9) = √13 ≠ √4 + √9 = 2+3 = 5 (addition does NOT split). Memory: 'Roots split under multiplication, not under addition.'

Between any two distinct numbers (real, rational, or irrational), there are INFINITELY MANY irrational numbers. To find two irrational numbers between, say, √2 (≈1.414) and √3 (≈1.732): METHOD: Construct non-terminating, non-recurring decimals between them. Example: 1.5010010001... (the 0s increase by one each time — clearly non-recurring) and 1.6010010001... — both lie between 1.414 and 1.732, both are non-terminating non-recurring → both irrational. ALTERNATIVE: 1.575775777... and 1.6878778777... Many constructions work. The key insight: between any two reals, there is always 'room' for irrationals.

The IDENTITY (a+b)(a−b) = a² − b² is the key. When the denominator is √a + √b (or √a − √b), multiplying by the conjugate gives: (√a + √b)(√a − √b) = (√a)² − (√b)² = a − b. This is a difference of squares — the square roots have been ELIMINATED because squaring √a gives just a. Example: rationalise 1/(2 + √3). Multiply numerator and denominator by 2 − √3: → (2 − √3)/[(2+√3)(2−√3)] = (2 − √3)/(4 − 3) = (2 − √3)/1 = 2 − √3. The denominator is now 1 — completely rational. This works because we converted the irrational √3 into 3 (rational) by squaring.

Around 500 BCE, the Greek Pythagoreans believed all lengths could be expressed as ratios of whole numbers (rational). When they discovered that the DIAGONAL of a unit square (which has length √2 by Pythagoras) CANNOT be expressed as p/q, this caused a profound crisis in Greek mathematics. Legend says Hippasus, who discovered this, was thrown into the sea for revealing it! This discovery proved that NOT ALL lengths are rational — some are 'incommensurable.' This forced the Greeks to develop a new theory of magnitudes (Eudoxus, ~400 BCE) and ultimately led to the modern concept of real numbers. The proof you learn in Class 9 — assume rational, derive contradiction — is the same proof attributed to Hippasus or his contemporaries 2,500 years ago.
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