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Chapter 1 — Real Numbers

Class 10 · Mathematics

Overview

Chapter 'Real Numbers' builds the foundation for understanding the classification and properties of numbers beyond integers. Starting with Euclid's Division Lemma and the Fundamental Theorem of Arithmetic (unique prime factorisation), the chapter develops methods to find HCF and LCM, characterise decimal expansions of rational numbers, and distinguish rational from irrational numbers with rigorous proofs (for example, √2 is irrational). This chapter is important because it introduces proof techniques, number-theoretic tools used throughout mathematics, and the precise meaning of 'real numbers' as the union of rationals and irrationals.

Learning Objectives

  • Define Euclid's division lemma and apply it to find quotient and remainder in given problems
  • State and prove the Fundamental Theorem of Arithmetic (unique prime factorisation) for integers greater than 1
  • Apply prime factorisation to compute HCF and LCM of two or more numbers
  • Use Euclid's algorithm to determine the HCF of large integers efficiently
  • Determine whether a given real number is rational or irrational and justify the answer
  • Prove the irrationality of √2 and extend the proof to √p for any prime p
  • Explain the decimal representation of rational numbers (terminating or recurring) and of irrational numbers (non‑terminating, non‑repeating)
  • Illustrate how to convert a recurring decimal into a rational number

Topics in this chapter

10 topics · tap a topic title to jump straight to it.

1

Euclid's Division Lemma

📐 MATHEMATICAL FORMULA / THEOREM

Euclid's Division Lemma

Key Point: Euclid's Division Lemma: a = bq + r, where b > 0 and 0 ≤ r < b

Statement: For any integers a and b with b > 0, there exist unique integers q (quotient) and r (remainder) such that

a = bq + r,   where   0 ≤ r < b.

Explanation / Existence: Think of dividing a objects into groups of size b. The number of complete groups is q and the leftover objects are r. Formally, take q = floor(a/b) (the greatest integer ≤ a/b). Then r = a − bq satisfies 0 ≤ r < b, so such q and r exist.

Uniqueness: If a = bq1 + r1 = bq2 + r2 with 0 ≤ r1, r2 < b, then b(q1 − q2) = r2 − r1. The left side is a multiple of b, but the right side has absolute value < b unless r1 = r2. Hence r1 = r2 and therefore q1 = q2; q and r are unique.

Key consequence (Euclidean algorithm): If a = bq + r, then gcd(a,b) = gcd(b,r). Repeating this step (replacing a by b and b by r) reduces the numbers until the remainder becomes 0; the last nonzero remainder is gcd(a,b). This yields an efficient algorithm to compute GCDs.

Notes: CBSE usually treats a as a nonnegative integer and b > 0. The lemma is the foundation for division, gcd computations, and the Fundamental Theorem of Arithmetic (unique prime factorization).

📌 Examples
  • Simple numeric: 25 = 7*3 + 4 (here a=25, b=7, q=3, r=4; 0 ≤ 4 < 7).
  • Negative dividend (extended view): -20 = 7*(-3) + 1 (q = -3, r = 1; 0 ≤ r < 7).
  • Euclidean algorithm (gcd): To find gcd(252,105): 252 = 105*2 + 42; 105 = 42*2 + 21; 42 = 21*2 + 0 ⇒ gcd = 21.
  • Real-life: Distributing 53 apples into baskets of 6 apples each gives q = 8 full baskets and r = 5 leftover apples (53 = 6*8 + 5).
  • Another real-life: If a bus has 47 passengers and each row seats 4, there are 11 full rows and 3 passengers in an incomplete row (47 = 4*11 + 3).
🧮 Formulas
  1. \[Euclid's Division Lemma: a = bq + r\]
    \[where b > 0 and 0 ≤ r < b\]
  2. \[Divisibility special case: if r = 0 then b | a (b divides a)\]
  3. \[GCD relation: gcd(a\]
    \[b) = gcd(b\]
    \[r)\]
  4. \[Quotient as floor: q = ⌊a / b⌋ and r = a − b⌊a / b⌋\]
2

Euclidean Algorithm (GCD by repeated division)

📐 MATHEMATICAL FORMULA / THEOREM

Euclidean Algorithm (GCD by repeated division)

Key Point: Division identity: a = bq + r, where 0 ≤ r < b (used each step).

What is GCD? The greatest common divisor (GCD or HCF) of two integers a and b (not both zero) is the largest positive integer that divides both a and b exactly.

Euclidean algorithm (GCD by repeated division) — idea in one line: If a = bq + r (0 ≤ r < b), then gcd(a, b) = gcd(b, r). Replacing the pair (a, b) by (b, r) repeatedly reduces the numbers until the remainder becomes 0; the last nonzero remainder is the gcd.

Why it works (brief proof):

  • If d divides both a and b then d divides a − bq = r, so every common divisor of a and b is a common divisor of b and r.
  • If d divides both b and r then d divides bq + r = a, so every common divisor of b and r is a common divisor of a and b.
  • Thus the sets of common divisors of (a,b) and (b,r) are identical; hence their greatest element (gcd) is the same.

Algorithm (step-by-step):

  1. Arrange the two positive integers so that a ≥ b (<span style="white-space:nowrap;">if not, swap them).
  2. Divide a by b to get quotient q and remainder r: a = bq + r, where 0 ≤ r < b.
  3. If r = 0, then gcd(a, b) = b and stop.
  4. Replace (a, b) by (b, r) and repeat from step 2.
  5. The algorithm terminates because remainders strictly decrease and are non‑negative; the last nonzero remainder is the gcd.

Notes:

  • For three or more numbers, gcd(a,b,c) = gcd(gcd(a,b), c).
  • Special cases: gcd(a,0) = |a| (for a ≠ 0); gcd(0,0) is undefined or sometimes taken as 0.

When to use: Use this method whenever you need the greatest common divisor of two (or more) integers; it is efficient even for large integers and underlies many number‑theory algorithms.

📌 Examples
  • Example 1 (worked): Find gcd(252, 105) by repeated division. Step 1: 252 = 105 × 2 + 42 (remainder 42) Step 2: 105 = 42 × 2 + 21 (remainder 21) Step 3: 42 = 21 × 2 + 0 (remainder 0) So gcd(252, 105) = 21 (the last nonzero remainder).
  • Example 2 (worked): Find gcd(48, 18). Step 1: 48 = 18 × 2 + 12 Step 2: 18 = 12 × 1 + 6 Step 3: 12 = 6 × 2 + 0 So gcd(48, 18) = 6.
  • Real-life example 1: Tiling a rectangular floor 252 cm by 105 cm with largest possible square tiles. The side of the largest square tile will be gcd(252,105) = 21 cm. You will need (252/21) × (105/21) = 12 × 5 = 60 tiles.
  • Real-life example 2: Distributing 48 apples and 18 oranges into identical fruit baskets so that each basket has the same number of apples and the same number of oranges and nothing left over. Maximum number of baskets = gcd(48,18) = 6 (each basket gets 8 apples and 3 oranges).
  • Practical tip: To find gcd of more than two numbers, compute gcd(a,b) first, then gcd(result, c), and so on.
🧮 Formulas
  1. \[Division identity: a = bq + r\]
    \[where 0 ≤ r &lt\]
    \[b (used each step).\]
  2. \[Key property: If a = bq + r then gcd(a\]
    \[b) = gcd(b\]
    \[r).\]
  3. \[Stopping condition: gcd(a, 0) = |a| (so when remainder becomes 0\]
    \[the other number is the gcd).\]
  4. \[Relation with LCM: gcd(a\]
    \[b) × lcm(a\]
    \[b) = |a × b| (useful to switch between gcd and lcm problems).\]
  5. \[For three numbers: gcd(a\]
    \[b\]
    \[c) = gcd(gcd(a\]
    \[b)\]
    \[c).\]
🔢3

Prime and Composite Numbers

📐 MATHEMATICAL FORMULA / THEOREM

Prime and Composite Numbers

Key Point: Definition: prime ⇔ number n > 1 with exactly two positive divisors: 1 and n.

Definition: A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself (e.g., 2, 3, 5, 7...). A composite number is a natural number greater than 1 that has more than two positive divisors (e.g., 4, 6, 8, 9...). The number 1 is neither prime nor composite.

Basic properties:

  • 2 is the only even prime. All other even numbers are composite.
  • If a number n has a divisor d > 1, then it also has a divisor n/d. To test primality it suffices to check divisors up to √n.
  • Every composite number can be expressed as a product of primes (prime factorization).

Fundamental Theorem of Arithmetic: Every integer greater than 1 can be written uniquely (up to order) as a product of prime powers: n = p1^a1 × p2^a2 × ... × pk^ak, where p1, p2, ..., pk are distinct primes and a1, a2, ..., ak are positive integers.

Prime factorization uses: Finding HCF (GCD) and LCM. For two numbers expressed by prime powers, take minimum exponents for HCF and maximum exponents for LCM. For two numbers a and b, HCF(a,b) × LCM(a,b) = a × b.

Primality checks & simple divisibility tests:

  • Divisible by 2: last digit even.
  • Divisible by 3: sum of digits divisible by 3.
  • Divisible by 5: last digit 0 or 5.
  • Divisible by 7, 11, 13: use specific rules or test up to √n.

Methods to find primes: Sieve of Eratosthenes: mark multiples of each prime starting from 2 to eliminate composites, remaining unmarked numbers are primes.

Short proof idea (Euclid) that primes are infinite: Assume finite list of primes p1,...,pn. Consider N = p1p2...pn + 1. N is not divisible by any pi, so it has a new prime factor, contradiction. Hence infinitely many primes.

Applications / real-life relevance: Cryptography (RSA uses large primes), digital hashing, error detection, arranging objects in equal groups (divisibility), scheduling and packing (LCM/HCF problems), gears and teeth counts where coprime counts avoid repeated alignments.

📌 Examples
  • Simple lists: Primes under 20 = {2, 3, 5, 7, 11, 13, 17, 19}; Composites under 20 = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}.
  • Prime check: Is 97 prime? Test divisors up to √97 ≈ 9.8: check 2,3,5,7 — none divide 97, so 97 is prime.
  • Prime factorization: 360 = 2^3 × 3^2 × 5 (use factor tree: 360 → 36 × 10 → 2^3 × 3^2 × 5).
  • HCF and LCM example: For 360 (2^3·3^2·5) and 420 (2^2·3·5·7): HCF = 2^min(3,2)·3^min(2,1)·5^min(1,1) = 2^2·3^1·5^1 = 60. LCM = 2^max(3,2)·3^max(2,1)·5^max(1,1)·7^max(0,1) = 2^3·3^2·5·7 = 2520. Check: HCF × LCM = 60 × 2520 = 360 × 420.
  • Sieve example (first steps): Write numbers from 2 to 30. Mark multiples of 2 (except 2), then multiples of next unmarked 3, then 5, etc. Unmarked numbers are primes: 2,3,5,7,11,13,17,19,23,29.
🧮 Formulas
  1. \[Definition: prime ⇔ number n > 1 with exactly two positive divisors: 1 and n.\]
  2. \[Fundamental Theorem of Arithmetic: n = p1^a1 × p2^a2 × ... × pk^ak (unique up to order).\]
  3. \[Check up to √n: if no divisor d (2 ≤ d ≤ ⌊√n⌋) divides n\]
    \[then n is prime.\]
  4. \[HCF (GCD) from prime factorization: product of common primes raised to minimum exponents.\]
  5. \[LCM from prime factorization: product of all primes appearing raised to maximum exponents.\]
  6. \[Relation for two numbers a and b: HCF(a,b) × LCM(a,b) = a × b.\]
🔢4

Fundamental Theorem of Arithmetic (FTA)

📐 MATHEMATICAL FORMULA / THEOREM

Fundamental Theorem of Arithmetic (FTA)

Key Point: General prime-power form: n = p1^a1 · p2^a2 · ... · pk^ak (pi distinct primes, ai ≥ 1).

Statement: Every integer greater than 1 can be written as a product of prime numbers, and this factorization is unique except for the order of the prime factors.

What it means: For any integer n > 1, there exist primes p1, p2, ..., pk and positive integers a1, a2, ..., ak such that

n = p1a1 · p2a2 · ... · pkak.

Existence (idea): Start dividing n by the smallest prime that divides it. The quotient is either 1 or a smaller integer; repeat the process. Since each division reduces the number, eventually you reach 1. The primes used along the way give a prime factorization.

Uniqueness (idea): If a number had two different prime factorizations, Euclid's lemma (a prime dividing a product divides one of the factors) forces the same primes to appear with the same exponents in both factorizations. Thus the factorization is unique up to the order of factors.

Notes: 1 is neither prime nor composite and has no prime factorization. The theorem applies only to integers > 1.

Practical algorithm (factor tree method):

  • Pick any composite number n.
  • Write n as a product of two factors (preferably using a small prime factor).
  • If a factor is composite, split it further; if it's prime, stop splitting it.
  • Continue until all leaves are primes; multiply them (with exponents) to get the prime-power representation.
📌 Examples
  • Example 1 — Factorize 72: 72 = 2 × 36 = 2 × 2 × 18 = 2 × 2 × 2 × 9 = 2^3 × 3^2.
  • Example 2 — Factorize 360: 360 = 2 × 180 = 2^2 × 90 = 2^3 × 45 = 2^3 × 3 × 15 = 2^3 × 3^2 × 5.
  • Example 3 — Use prime factorization to find HCF and LCM of 12 and 18: 12 = 2^2 × 3^1, 18 = 2^1 × 3^2. HCF = 2^min(2,1) × 3^min(1,2) = 2^1 × 3^1 = 6. LCM = 2^max(2,1) × 3^max(1,2) = 2^2 × 3^2 = 36.
  • Real-life example — Scheduling: To find when two events repeat together (e.g., bus A every 12 min and bus B every 18 min), compute LCM(12,18)=36 minutes. Prime factorization is a fast method to compute LCM.
  • Real-life example — Tiling/packing: To tile a rectangle with identical square tiles without cutting, prime factorization can help find the largest possible square tile (side = gcd of rectangle sides).
  • Connection — Cryptography: Modern RSA cryptography relies on properties of prime numbers and difficulty of factoring large integers into primes (an application of prime factorization idea).
🧮 Formulas
  1. \[General prime-power form: n = p1^a1 · p2^a2 · ... · pk^ak (pi distinct primes\]
    \[ai ≥ 1).\]
  2. \[HCF (GCD) from prime powers: gcd(n,m) = ∏ pi^min(ai,bi).\]
  3. \[LCM from prime powers: lcm(n,m) = ∏ pi^max(ai,bi).\]
  4. \[Number of positive divisors (τ function): if n = ∏ pi^ai then τ(n) = ∏ (ai + 1).\]
  5. \[Product relation: n · m = gcd(n,m) · lcm(n,m) for positive integers n,m (can be checked via prime powers).\]
🔢5

Euclid's Lemma

📐 MATHEMATICAL FORMULA / THEOREM

Euclid's Lemma

Key Point: Euclid's Lemma: If p is prime and p | (a·b) then p | a or p | b.

Statement: If p is a prime number and p divides the product ab (written p | ab), then p divides a or p divides b (or both).

Why it matters: Euclid's Lemma is a key step in proving the Fundamental Theorem of Arithmetic (unique prime factorisation). It also has a useful generalisation: if p divides a1·a2·...·an then p divides at least one of the factors.

Proof (using gcd and Bezout's identity):

  • Assume p is prime and p | ab.
  • If p divides a, we are done. Otherwise gcd(a,p)=1 because p has no divisors other than 1 and itself.
  • By Bezout's identity there exist integers x,y such that ax + py = 1.
  • Multiply both sides by b: abx + pyb = b. The left side is divisible by p (abx is, because p | ab; pyb is obviously), so p divides b.
  • Thus if p does not divide a, it must divide b. This proves the lemma.

Alternative explanation (prime-factor view): In the prime factorisation of ab every prime factor comes from the factors a or b. If p appears in the prime factors of ab, it must appear in the factorisation of a or b. Hence p divides a or p divides b.

Generalisation: If gcd(a,c)=1 and a | bc, then a | c. This follows by writing a as product of primes and applying Euclid's Lemma to each prime factor.

Important remark (why primality is needed): The lemma fails if the divisor is composite. Example: 6 divides 2·3 = 6, but 6 divides neither 2 nor 3.

📌 Examples
  • Example 1: Let p = 3, a = 6, b = 10. Then 3 | (6·10) = 60. Since 3 | 6, Euclid's Lemma holds.
  • Example 2: Let p = 7, a = 14, b = 15. 7 | (14·15). Here 7 | 14, so the lemma holds.
  • Example 3 (multiple factors): If 5 divides 2·3·5·7, then 5 divides one of those factors (here it divides 5).
  • Example 4 (gcd-version): gcd(4,3)=1 and 4 | (3·8) = 24, so by the generalisation 4 | 8.
  • Counterexample (non-prime divisor): 6 | (2·3) = 6 but 6 does not divide 2 and does not divide 3. This shows primality is necessary.
🧮 Formulas
  1. \[Euclid's Lemma: If p is prime and p | (a·b) then p | a or p | b.\]
  2. \[Generalisation: If p prime and p | (a1·a2·...·an) then p divides at least one ai.\]
  3. \[GCD version: If gcd(a,b)=1 and a | (b·c) then a | c.\]
  4. \[Bezout identity (used in proof): If gcd(a,p)=1 then there exist integers x,y with ax + py = 1.\]
  5. \[Prime-factor form: If a = ∏ p_i^{α_i}\]
    \[b = ∏ p_i^{β_i}\]
    \[then a·b = ∏ p_i^{α_i+β_i}\]
    \[any prime appearing in a·b appears in a or b.\]
✖️6

Highest Common Factor (HCF) and Lowest Common Multiple (LCM)

📐 MATHEMATICAL FORMULA / THEOREM

Highest Common Factor (HCF) and Lowest Common Multiple (LCM)

Key Point: HCF(a,b) × LCM(a,b) = |a × b| (for two nonzero integers a and b).

Definitions

HCF (Highest Common Factor), also called GCD (Greatest Common Divisor), of two or more integers is the largest positive integer that divides each of them exactly. LCM (Lowest Common Multiple) of two or more integers is the smallest positive integer that is a multiple of each of them.

Why they matter

HCF is useful for simplifying fractions and dividing objects into maximum equal groups. LCM is used to find when repeating events coincide (schedules, cycles), or to find a common denominator for adding/subtracting fractions.

Methods to find HCF and LCM

  • Prime factorization method: Express each number as product of primes. For HCF multiply common primes with their lowest powers. For LCM multiply all primes that appear with their highest powers.
  • Division (short division) method: Repeatedly divide numbers by common prime divisors until no common divisor > 1 remains. Product of divisors gives HCF (for the divisor method focused on HCF) or combine appropriately for LCM when using combined division tables.
  • Euclid's algorithm (efficient for HCF/GCD): For two numbers a and b (a > b), compute remainder r = a mod b, then replace a by b and b by r, repeat until remainder is 0. Last nonzero remainder is HCF. Then use HCF to get LCM by LCM = |a*b| / HCF.

Relationship between HCF and LCM (for two numbers)

For two nonzero integers a and b: HCF(a, b) × LCM(a, b) = |a × b|.

Special notes

  • HCF of numbers that are pairwise coprime may still be >1 only if a common factor exists in all; otherwise HCF = 1.
  • For more than two numbers, you can iteratively use HCF and LCM: HCF(a,b,c) = HCF(HCF(a,b), c); LCM(a,b,c) = LCM(LCM(a,b), c).
📌 Examples
  • Example 1 (prime factorization + relation): Find HCF and LCM of 48 and 180. 48 = 2^4 × 3; 180 = 2^2 × 3^2 × 5. HCF = 2^(min(4,2)) × 3^(min(1,2)) = 2^2 × 3 = 4 × 3 = 12. LCM = 2^(max(4,2)) × 3^(max(1,2)) × 5^(max(0,1)) = 2^4 × 3^2 × 5 = 16 × 9 × 5 = 720. Check: HCF × LCM = 12 × 720 = 8640 = 48 × 180.
  • Example 2 (Euclid's algorithm): Find HCF of 180 and 48. 180 ÷ 48 = 3 remainder 36 → replace (180,48) by (48,36). 48 ÷ 36 = 1 remainder 12 → replace by (36,12). 36 ÷ 12 = 3 remainder 0 → HCF = 12. Then LCM = |180 × 48| / 12 = 8640 / 12 = 720 (same as above).
  • Example 3 (three numbers): Find HCF and LCM of 12, 15 and 20. 12 = 2^2 × 3; 15 = 3 × 5; 20 = 2^2 × 5. HCF = product of common primes with lowest powers: no prime common to all three → HCF = 1. LCM = 2^2 × 3 × 5 = 4 × 3 × 5 = 60. Real-life interpretation: if three events repeat every 12, 15 and 20 minutes, they will coincide every 60 minutes.
  • Real-life HCF example: Two classes of sizes 30 and 45 students want to be arranged in equal rows with same number of students per row and no mixing. Maximum students per row = HCF(30,45) = 15. So rows of 15: one class makes 2 rows, the other 3 rows.
  • Real-life LCM example: A traffic light cycles every 40 s and another nearby cycles every 60 s. They will turn green together every LCM(40,60) = 120 seconds (2 minutes).
🧮 Formulas
  1. \[HCF(a,b) × LCM(a,b) = |a × b| (for two nonzero integers a and b).\]
  2. \[LCM(a,b) = |a × b| / HCF(a,b).\]
  3. \[Prime-factor method: If a = ∏ p_i^{α_i} and b = ∏ p_i^{β_i}\]
    \[then HCF(a,b) = ∏ p_i^{min(α_i,β_i)}\]
    \[LCM(a,b) = ∏ p_i^{max(α_i,β_i)}.\]
  4. \[Euclid's algorithm (for HCF): HCF(a,b) = HCF(b\]
    \[a mod b) repeated until remainder 0\]
    \[last nonzero remainder is the HCF.\]
  5. \[For more than two numbers: HCF(a,b,c,...) = HCF(HCF(a,b),c,...)\]
    \[LCM(a,b,c,...) = LCM(LCM(a,b),c,...).\]
🔢7

Decimal Representation of Rational Numbers

📐 MATHEMATICAL FORMULA / THEOREM

Decimal Representation of Rational Numbers

Key Point: Decimal of p/q (in lowest terms) terminates iff q = 2^a * 5^b for some nonnegative integers a, b.

Rational numbers are numbers that can be expressed as a fraction p/q where p and q are integers and q ≠ 0. Every rational number has a decimal representation obtained by dividing p by q. There are two possible types of decimal expansions for rational numbers:

  • Terminating decimals: The decimal stops after a finite number of digits (for example 0.75).
  • Repeating (recurring) decimals: After some point a block of digits repeats indefinitely (for example 0.333..., 0.1818...).

Key criterion (in lowest terms p/q): the decimal expansion of p/q terminates iff the denominator q has no prime factors other than 2 and/or 5. If q has any other prime factor (e.g., 3, 7, 11), the decimal expansion is repeating.

How to get the decimal from a fraction: perform long division of numerator by denominator. If remainders become 0 at some step, the division terminates. If a remainder repeats, the digits between two occurrences of the same remainder form the repeating block.

How to convert repeating decimals back to fractions (methods):

  • Pure repeating (only repeating block): if x = 0.(R) where R is an r-digit repeating block, then x = integer(R)/(10^r - 1). Example: 0.(3) = 3/9 = 1/3.
  • Mixed decimal (non-repeating part followed by a repeating block): if x has n non-repeating digits followed by r repeating digits, then x = (integer formed by first n+r digits - integer formed by first n digits)/(10^{n+r} - 10^n).

Length of the repeating block (repetend): For a reduced fraction p/q where q' is q with all factors 2 and 5 removed, the length of the repetend is the smallest positive integer k such that 10^k ≡ 1 (mod q'). Thus the repetend length depends on the multiplicative order of 10 modulo q'.

All rational numbers have either terminating or eventually periodic decimal expansions; irrational numbers have non-terminating, non-repeating decimals.

📌 Examples
  • 1) 1/2 = 0.5 (terminating).
  • 2) 3/4 = 0.75 (terminating).
  • 3) 1/3 = 0.(3) = 0.333... (pure repeating).
  • 4) 2/11 = 0.(18) = 0.1818... (pure repeating of 2 digits).
  • 5) 7/12 = 0.58(3) = 0.58333... (mixed: two non-repeating digits 58 then repeating 3).
  • 6) Convert 0.125 to a fraction: 0.125 = 125/1000 = 1/8 (terminating).
🧮 Formulas
  1. \[Decimal of p/q (in lowest terms) terminates iff q = 2^a * 5^b for some nonnegative integers a\]
    \[b.\]
  2. \[Pure repeating decimal 0.(R) with r-digit block R: value = R / (10^r - 1)\]
    \[Example: 0.(142857) = 142857 / 999999.\]
  3. \[Mixed decimal with n non-repeating digits followed by r repeating digits: value = (integer formed by first n+r digits - integer formed by first n digits) / (10^{n+r} - 10^n).\]
  4. \[Length of repetend k: smallest k such that 10^k ≡ 1 (mod q')\]
    \[where q' = denominator after removing all factors 2 and 5.\]
  5. \[Procedure for converting terminating decimal D with d decimal places to fraction: D = integer(D * 10^d) / 10^d and then reduce.\]
🔢8

Irrational Numbers

📐 MATHEMATICAL FORMULA / THEOREM

Irrational Numbers

Key Point: Real numbers: R = Q ∪ (R \ Q), where Q = set of rational numbers and R \ Q = set of irrational numbers

Definition: An irrational number is a real number that cannot be expressed as a ratio of two integers (i.e., it cannot be written as p/q where p and q are integers and q ≠ 0). In decimal form an irrational number has a non-terminating, non-repeating expansion.

Key points:

  • Irrational numbers together with rational numbers make up the set of real numbers: R = Q ∪ (R\Q).
  • Common examples: π = 3.14159..., e = 2.71828..., √2 = 1.4142135..., √3 = 1.73205....
  • Decimal property: irrational decimals never terminate and never repeat a fixed pattern of digits.
  • Density: Between any two distinct real numbers there are infinitely many irrational numbers (and also infinitely many rational numbers).

Simple proof idea (√2 is irrational): Suppose √2 = p/q in lowest terms with integers p,q. Then p² = 2q², so p² is even ⇒ p is even. Write p = 2k; then 4k² = 2q² ⇒ q² = 2k² ⇒ q is even. Both p and q are even, contradicting that p/q was in lowest terms. So √2 is not rational.

Properties (useful at Class 10 level):

  • The sum or difference of two irrational numbers may be rational or irrational (e.g., √2 + (1−√2)=1 is rational; √2+√3 is irrational).
  • Product or quotient of two irrational numbers may be rational or irrational (e.g., √2·√2=2 is rational; √2·√3 is irrational).
  • If r is a nonzero rational number and x is irrational, then r·x is irrational. (If r·x were rational then x would be rational.)
  • Adding a rational number to an irrational number gives an irrational number (except in the special case when the rational cancels the irrational part).

How to recognize some irrationals quickly: √n is irrational whenever n is a positive integer that is not a perfect square. Numbers that arise from geometry (diagonals, circumferences) or certain limits (π, e) are typically irrational.

Real-life connections: The number π appears in formulas for circumference and area of circles (C = 2πr, A = πr²). √2 appears as the diagonal length of a unit square (important in construction and geometry). e appears in continuous growth models such as continuously compounded interest and some natural processes.

📌 Examples
  • √2 (diagonal of a unit square) ≈ 1.4142135... — irrational
  • π (ratio of circumference to diameter) ≈ 3.14159265... — irrational
  • e (base of natural logarithm, appears in growth/compound interest) ≈ 2.7182818... — irrational
  • √3 (diagonal space in unit cube projection) ≈ 1.7320508... — irrational
  • Cube root of 2, ∛2 ≈ 1.259921... — irrational
🧮 Formulas
  1. \[Real numbers: R = Q ∪ (R \ Q)\]
    \[where Q = set of rational numbers and R \ Q = set of irrational numbers\]
  2. \[Irrational decimal expansion: non-terminating\]
    \[non-repeating (example: π = 3.1415926535... no repeating block)\]
  3. \[√n is irrational if n ∈ N and n is not a perfect square (e.g., √2, √3, √5 ...)\]
  4. \[If r ∈ Q\{0} and x ∉ Q (x irrational) then r·x is irrational\]
  5. \[Circumference of circle: C = 2πr (π is irrational) — shows an everyday appearance of an irrational constant\]
  6. \[Area of circle: A = πr² (π is irrational)\]
🔢9

Real Numbers on the Number Line

📐 MATHEMATICAL FORMULA / THEOREM

Real Numbers on the Number Line

Key Point: Rational number form: any rational number can be written as p/q, where p ∈ Z, q ∈ N, gcd(p,q)=1.

What are real numbers? Real numbers are all numbers that can be placed as points on a continuous number line. They include natural numbers, whole numbers, integers, rational numbers (fractions and terminating/repeating decimals) and irrational numbers (non‑terminating, non‑repeating decimals).

Classification

  • Natural numbers: 1, 2, 3, ...
  • Whole numbers: 0, 1, 2, ...
  • Integers: ..., -2, -1, 0, 1, 2, ...
  • Rational numbers: numbers expressible as p/q (q ≠ 0). Their decimal expansion is either terminating or repeating.
  • Irrational numbers: numbers that cannot be written as p/q. Their decimal expansion is non‑terminating and non‑repeating (examples: √2, π).

Representation on the number line

  • Every real number corresponds to exactly one point on the number line and every point on the number line corresponds to exactly one real number.
  • To locate a rational number p/q: perform the division p ÷ q to get its decimal expansion and mark that point. If the simplified denominator has only prime factors 2 and/or 5, the decimal terminates; otherwise it repeats.
  • To locate an irrational number (for example √2): use successive approximations—find increasingly narrow intervals that contain the number (interval nesting) or construct geometrically (√2 is the diagonal of a unit square) and project its length onto the number line. Decimal approximations (1.414, 1.4142, 1.41421, ...) give increasingly accurate locations.

Important properties

  • Density: Between any two distinct real numbers there exist infinitely many rational numbers and infinitely many irrational numbers.
  • Ordering and distance: Given real numbers a and b, a point representing a is to the left of b if a < b; the distance between them is |a - b|.
  • Archimedean property (useful for locating and approximating): For any real x, there exists a natural number n such that n > x. Equivalently, for any ε > 0 there exists n in N with 1/n < ε.

Why this matters

The number line gives a geometric picture of numbers and their order, and it is the basis for measuring distances, comparing magnitudes, and understanding continuity and approximation. Placing rationals and irrationals on the same line shows the completeness of real numbers: there are no “gaps” left unrepresented.

📌 Examples
  • Locate -3, 0, and 4 on the number line: mark equally spaced ticks and place the points at respective positions left (negative) or right (positive) of zero.
  • Represent 2/5 on the number line: divide the segment from 0 to 1 into 5 equal parts and take the 2nd tick (0.4). As a decimal 2/5 = 0.4 (terminating).
  • Represent 1/3 on the number line: divide 0 to 1 into 3 equal parts; 1/3 = 0.333... (repeating). Mark the point at 0.333... using the geometric third or an approximation like 0.333 or 0.3333.
  • Locate √2: note 1 < √2 < 2. Check 1.4^2 = 1.96 (less than 2) and 1.5^2 = 2.25 (greater than 2), so 1.4 < √2 < 1.5. Refine by testing 1.41, 1.414 etc., and mark the point at the resulting decimal approximation.
  • Show density: between 1 and 2 there are numbers like 1.1 (rational), 1.414 (approximation to an irrational), and infinite others. You can always take midpoint (a + b)/2 to get another number between any two a and b.
🧮 Formulas
  1. \[Rational number form: any rational number can be written as p/q\]
    \[where p ∈ Z\]
    \[q ∈ N\]
    \[gcd(p,q)=1.\]
  2. \[Decimal termination condition: a reduced fraction p/q (q>0) has a terminating decimal expansion iff q has no prime factors other than 2 and/or 5.\]
  3. \[Distance between two real numbers a and b: |a - b| (absolute value).\]
  4. \[Midpoint of two real numbers a and b: (a + b) / 2\]
    \[which lies strictly between a and b if a ≠ b.\]
  5. \[Archimedean property (practical form): ∀ε > 0, ∃n ∈ N such that 1/n < ε (used for arbitrarily fine approximations).\]
  6. \[Density statement (informal): Between any two real numbers there exist infinitely many rationals and infinitely many irrationals.\]
🔢10

Applications and Problem Solving

📐 MATHEMATICAL FORMULA / THEOREM

Applications and Problem Solving

Key Point: Euclid's division lemma: a = bq + r, 0 ≤ r < b (a, b integers, b>0).

Overview: The chapter Real Numbers introduces tools (Euclid's division lemma, Euclid's algorithm, prime factorization and the Fundamental Theorem of Arithmetic) that are used to solve practical problems involving highest common factor (HCF/GCD), least common multiple (LCM), classification of numbers (rational/irrational), and repeating decimals. "Applications and Problem Solving" means using these tools to model and solve everyday partitioning, scheduling, remainder and representation problems.

Key methods:

  • Euclid's division lemma: For integers a and b (b > 0) there exist unique integers q and r such that a = bq + r, 0 ≤ r < b. This is the basis of the Euclid (GCD) algorithm.
  • Euclid's algorithm: Repeatedly apply the division lemma to find the HCF/GCD quickly. If a = bq + r, then gcd(a,b) = gcd(b,r) and continue until remainder 0.
  • Prime factorization and Fundamental Theorem of Arithmetic: Every integer > 1 can be written uniquely (up to order) as product of primes. Use prime powers to compute HCF and LCM: HCF uses minimum exponents, LCM uses maximum exponents.
  • Converting repeating decimals to rational numbers: Use algebraic manipulation: if x has repeating part, form equations multiplying by powers of 10 to eliminate repeat and solve for x.

Typical applications:

  • Grouping or partition problems (use HCF): e.g., divide objects into the largest possible identical groups, or place items in equal rows with maximum elements per row.
  • Scheduling / synchronization problems (use LCM): e.g., find when two repeating events coincide (buses, bells, lights).
  • Remainder problems: use modular reasoning and sometimes LCM/HCF to find numbers leaving specified remainders on division by different divisors.
  • Decimal representation: classify numbers as terminating/recurring (rational) or non-terminating non-repeating (irrational); convert recurring decimals into fractions for exact answers.

Strategy for problem solving:

  • Model the situation with integers (counts, minutes, objects).
  • Decide whether HCF (equal largest groups), LCM (first common occurrence) or prime-factor techniques are needed.
  • Apply Euclid's algorithm for large numbers or prime-factorization for transparency.
  • Check answers in the context (units, positive integers, remainders range).
📌 Examples
  • 1) HCF and LCM by prime factorization: Find HCF and LCM of 84 and 120. 84 = 2^2 × 3 × 7; 120 = 2^3 × 3 × 5. HCF = product of common primes with minimum exponents = 2^2 × 3 = 12. LCM = product of all primes with maximum exponents = 2^3 × 3 × 5 × 7 = 840.
  • 2) Euclid's algorithm example: Find gcd(544,119). 544 = 119×4 + 68 119 = 68×1 + 51 68 = 51×1 + 17 51 = 17×3 + 0 ⇒ gcd = 17. (Euclid's algorithm quickly reduces large numbers by repeated division.)
  • 3) Grouping (HCF application): You have 120 red and 84 blue chairs and want identical rows each containing only one colour and each row having same number of chairs. Maximum chairs per row = HCF(120,84) = 12. So make 10 rows of red (120/12) and 7 rows of blue (84/12).
  • 4) Scheduling (LCM application): Two lamps flash every 12 s and 18 s. They flash together every LCM(12,18) = 36 seconds. So after both flash together, next simultaneous flash is after 36 s.
  • 5) Converting recurring decimal: x = 0.2\overline{34} = 0.2343434... Let x = 0.2343434... Multiply by 10 to move non-repeating digit: 10x = 2.343434... Multiply by 1000 (10 × 100) to align repeats: 1000x = 234.343434... Subtract: 1000x - 10x = 234.343... - 2.343... ⇒ 990x = 232 ⇒ x = 232/990 = 116/495 (in lowest terms).
  • 6) Remainder problem: Find smallest positive N that leaves remainder 2 when divided by 3 and remainder 3 when divided by 5. Model: N ≡ 2 (mod 3), N ≡ 3 (mod 5). Check numbers ≡ 3 (mod 5): 3,8,13,18,23,... Test mod 3: 8 ≡ 2 (mod 3). So N = 8 is smallest. More systematic method uses CRT-like search or LCM steps when moduli coprime.
🧮 Formulas
  1. \[Euclid's division lemma: a = bq + r, 0 ≤ r < b (a\]
    \[b integers\]
    \[b>0).\]
  2. \[Euclid's algorithm: gcd(a,b) = gcd(b,r) where r is remainder when a is divided by b\]
    \[iterate until remainder 0.\]
  3. \[Prime-factor representation: n = p1^{a1} × p2^{a2} × ... × pk^{ak} (unique for n>1).\]
  4. \[HCF (GCD) using prime powers: gcd = ∏ p_i^{min(exponent in a\]
    \[exponent in b)}.\]
  5. \[LCM using prime powers: lcm = ∏ p_i^{max(exponent in a\]
    \[exponent in b)}.\]
  6. \[Relation for two numbers: a × b = gcd(a,b) × lcm(a,b).\]

Key Concepts

Natural number
Positive integers used for counting (1, 2, 3, ...).
Whole number
Natural numbers together with 0 (0, 1, 2, 3, ...).
Integer
Whole numbers and their negatives (..., −2, −1, 0, 1, 2, ...).
Rational number
A number expressible as p/q where p and q are integers and q ≠ 0.
Irrational number
A real number that cannot be written as a ratio of two integers; its decimal expansion is non-terminating and non-repeating.
Real number
All rational and irrational numbers; points on the number line.
Prime number
A natural number greater than 1 that has no positive divisors other than 1 and itself.
Composite number
A natural number greater than 1 that has at least one divisor other than 1 and itself.
Unit
The number 1, which has exactly one positive divisor and is neither prime nor composite.
Co-prime (Relatively prime)
Two integers whose greatest common divisor is 1.
Prime factorization
Expression of a number as a product of prime numbers.
Fundamental Theorem of Arithmetic
Every integer greater than 1 can be expressed uniquely as a product of primes, up to the order of factors.
Euclid's division lemma
For integers a and b (b > 0), there exist unique integers q and r such that a = bq + r with 0 ≤ r < b.
Euclid's algorithm
A method using repeated division (remainder steps) to compute the gcd of two integers.
Highest Common Factor (HCF) / Greatest Common Divisor (GCD)
The largest positive integer that divides two or more integers without remainder.
Least Common Multiple (LCM)
The smallest positive integer that is a multiple of two or more integers.
Terminating decimal
A decimal representation that has a finite number of non-zero digits after the decimal point.
Recurring (Repeating) decimal
A decimal in which a finite block of digits repeats indefinitely.
Decimal expansion
Representation of a real number in base 10 as an integer part and a decimal part (may terminate, repeat, or be non-repeating).
Irreducible fraction (Lowest terms)
A fraction p/q is in lowest terms if gcd(p,q) = 1; it cannot be simplified further.

Practice Questions

  1. State Euclid's Division Lemma and apply it to express 25 divided by 7. / यूक्लिड का विभाजन प्रमेयिका बताएं और इसे 7 से विभाजित 25 को व्यक्त करने पर लागू करें।
    Show answer

    For integers a and b with b > 0, there exist unique integers q and r such that a = bq + r, with 0 ≤ r < b. For 25 and 7: 25 = 7×3 + 4, so q = 3 and r = 4, where 0 ≤ 4 < 7. / पूर्णांकों a और b के लिए जहाँ b > 0, अद्वितीय पूर्णांक q और r होते हैं ताकि a = bq + r, जहाँ 0 ≤ r < b। 25 और 7 के लिए: 25 = 7×3 + 4, अतः q = 3 और r = 4, जहाँ 0 ≤ 4 < 7।

  2. Find the HCF of 252 and 105 using Euclid's algorithm, showing each step. / यूक्लिड एल्गोरिथम का उपयोग करके 252 और 105 का HCF ज्ञात करें, प्रत्येक चरण दिखाएं।
    Show answer

    252 = 105×2 + 42; 105 = 42×2 + 21; 42 = 21×2 + 0. Since the remainder is now 0, the last nonzero remainder 21 is the HCF, so HCF(252,105) = 21. / 252 = 105×2 + 42; 105 = 42×2 + 21; 42 = 21×2 + 0। चूँकि शेषफल अब 0 है, अंतिम शून्येतर शेषफल 21 HCF है, अतः HCF(252,105) = 21।

  3. State the Fundamental Theorem of Arithmetic and use it to find the HCF and LCM of 12 and 18. / अंकगणित की आधारभूत प्रमेय बताएं और इसका उपयोग करके 12 और 18 का HCF और LCM ज्ञात करें।
    Show answer

    Every integer greater than 1 can be expressed uniquely as a product of primes (up to order). Here 12 = 2²×3 and 18 = 2×3²; HCF = 2^min(2,1)×3^min(1,2) = 2×3 = 6 and LCM = 2^max(2,1)×3^max(1,2) = 2²×3² = 36. / 1 से बड़े प्रत्येक पूर्णांक को अभाज्य संख्याओं के गुणनफल के रूप में अद्वितीय रूप से (क्रम को छोड़कर) व्यक्त किया जा सकता है। यहाँ 12 = 2²×3 और 18 = 2×3²; HCF = 2^min(2,1)×3^min(1,2) = 2×3 = 6 और LCM = 2^max(2,1)×3^max(1,2) = 2²×3² = 36।

  4. Prove that √2 is irrational. / सिद्ध करें कि √2 अपरिमेय है।
    Show answer

    Suppose √2 = p/q in lowest terms; then p² = 2q², so p² is even, hence p is even. Writing p = 2k gives 4k² = 2q², so q² = 2k², making q even too. Then p and q share factor 2, contradicting lowest terms, so √2 is irrational. / मान लें √2 = p/q निम्नतम पद में है; तब p² = 2q², अतः p² सम है, इसलिए p सम है। p = 2k लिखने पर 4k² = 2q², अतः q² = 2k², जिससे q भी सम है। तब p और q में 2 उभयनिष्ठ गुणनखंड है, जो निम्नतम पद का खंडन करता है, अतः √2 अपरिमेय है।

  5. Without long division, state whether 7/12 has a terminating or recurring decimal expansion and justify. / दीर्घ विभाजन के बिना बताएं कि 7/12 का दशमलव प्रसार सांत है या आवर्ती और इसे न्यायसंगत ठहराएं।
    Show answer

    A fraction p/q in lowest terms terminates only if q has no prime factor other than 2 and 5. Here 12 = 2²×3 contains the prime 3, so 7/12 has a recurring (non-terminating repeating) decimal, namely 0.58(3). / निम्नतम पद में भिन्न p/q तभी सांत होती है जब q में 2 और 5 के अलावा कोई अभाज्य गुणनखंड न हो। यहाँ 12 = 2²×3 में अभाज्य 3 है, अतः 7/12 का दशमलव आवर्ती (अनवसानी आवर्ती) है, अर्थात् 0.58(3)।

  6. Two lamps flash every 12 seconds and 18 seconds respectively. If they flash together now, after how many seconds will they next flash together? / दो लैंप क्रमशः हर 12 सेकंड और 18 सेकंड में चमकते हैं। यदि वे अभी एक साथ चमकते हैं, तो कितने सेकंड बाद वे अगली बार एक साथ चमकेंगे?
    Show answer

    They flash together at the LCM of 12 and 18. Since 12 = 2²×3 and 18 = 2×3², LCM = 2²×3² = 36, so they next flash together after 36 seconds. / वे 12 और 18 के LCM पर एक साथ चमकते हैं। चूँकि 12 = 2²×3 और 18 = 2×3², LCM = 2²×3² = 36, अतः वे 36 सेकंड बाद अगली बार एक साथ चमकेंगे।

  7. Convert the recurring decimal 0.2̄34̄ (= 0.2343434...) into a rational number in lowest terms. / आवर्ती दशमलव 0.2̄34̄ (= 0.2343434...) को निम्नतम पद में परिमेय संख्या में बदलें।
    Show answer

    Let x = 0.2343434...; then 10x = 2.343434... and 1000x = 234.343434...; subtracting gives 990x = 232, so x = 232/990 = 116/495 in lowest terms. / मान लें x = 0.2343434...; तब 10x = 2.343434... और 1000x = 234.343434...; घटाने पर 990x = 232, अतः x = 232/990 = 116/495 निम्नतम पद में।

  8. Explain why the product of a non-zero rational number and an irrational number is always irrational. / समझाएं कि एक शून्येतर परिमेय संख्या और एक अपरिमेय संख्या का गुणनफल हमेशा अपरिमेय क्यों होता है।
    Show answer

    Let r be a non-zero rational and x irrational. If r·x were rational, then x = (r·x)/r would be a quotient of two rationals, hence rational, contradicting that x is irrational. Therefore r·x must be irrational. / मान लें r एक शून्येतर परिमेय और x अपरिमेय है। यदि r·x परिमेय होता, तो x = (r·x)/r दो परिमेय संख्याओं का भागफल होता, अतः परिमेय होता, जो x के अपरिमेय होने का खंडन करता है। इसलिए r·x अपरिमेय होना चाहिए।

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