π Introduction to Real Numbers
Why does this chapter matter? Real Numbers is Chapter 1 of CBSE Class 10 Maths and carries 6β8 marks in the board exam every year. Mastering Euclid's Division Algorithm and the Fundamental Theorem of Arithmetic will make number theory questions feel effortless. Let's get started!
The chapter Real Numbers builds on your Class 9 knowledge of the number system. You'll learn powerful tools for finding HCF and LCM without listing all factors, prove that certain numbers like β2 and β3 are irrational, and understand the building blocks of all integers. These concepts appear not just in Class 10, but also in JEE Foundation and Olympiad problems.
Euclid's Division Lemma
Fundamental Theorem
HCF & LCM
Irrational Numbers
Rational Decimals
Prime Factorisation
π All Formulas β Real Numbers Class 10
π’ 1. Euclid's Division Lemma (EDL)
a = b Γ q + r where 0 β€ r < b
Where: a = dividend, b = divisor, q = quotient, r = remainder
Key point: For any two positive integers a and b, there exist unique whole numbers q and r satisfying this relation. The remainder r is always non-negative and less than b.
Key point: For any two positive integers a and b, there exist unique whole numbers q and r satisfying this relation. The remainder r is always non-negative and less than b.
π 2. Euclid's Division Algorithm (EDA) β HCF
HCF(a, b) β Apply EDL repeatedly until r = 0
Last non-zero remainder = HCF
Last non-zero remainder = HCF
Steps: (1) Divide a by b, get remainder rβ. (2) Now divide b by rβ, get rβ. (3) Continue until remainder = 0. The divisor at that step is HCF.
Example: HCF(135, 225) β 225 = 135Γ1 + 90 β 135 = 90Γ1 + 45 β 90 = 45Γ2 + 0 β΄ HCF = 45
Example: HCF(135, 225) β 225 = 135Γ1 + 90 β 135 = 90Γ1 + 45 β 90 = 45Γ2 + 0 β΄ HCF = 45
ποΈ 3. Fundamental Theorem of Arithmetic
Every composite number = Product of primes (unique)
Every integer greater than 1 is either prime or can be expressed as a unique product of prime numbers (order doesn't matter). This is also called Prime Factorisation.
Example: 360 = 2Β³ Γ 3Β² Γ 5
Example: 360 = 2Β³ Γ 3Β² Γ 5
π 4. HCF and LCM Relationship
HCF(a, b) Γ LCM(a, b) = a Γ b
Valid for: Any two positive integers a and b.
HCF from prime factorisation: Take lowest power of each common prime factor.
LCM from prime factorisation: Take highest power of each prime factor (common or not).
Example: a = 12, b = 18 β HCF = 6, LCM = 36 β 6 Γ 36 = 216 = 12 Γ 18 β
HCF from prime factorisation: Take lowest power of each common prime factor.
LCM from prime factorisation: Take highest power of each prime factor (common or not).
Example: a = 12, b = 18 β HCF = 6, LCM = 36 β 6 Γ 36 = 216 = 12 Γ 18 β
π 5. Irrational Numbers β Proof Technique
If p is prime and p | aΒ², then p | a
This is the key lemma used to prove β2, β3, β5, etc. are irrational. The method is Proof by Contradiction.
Steps for proving β2 is irrational:
(1) Assume β2 = p/q (in lowest terms). (2) Then 2qΒ² = pΒ², so 2 | pΒ². (3) By lemma, 2 | p, so p = 2m. (4) Then 2qΒ² = 4mΒ² β qΒ² = 2mΒ², so 2 | q. (5) Both p and q divisible by 2 β contradicts "lowest terms". β΄ β2 is irrational.
Steps for proving β2 is irrational:
(1) Assume β2 = p/q (in lowest terms). (2) Then 2qΒ² = pΒ², so 2 | pΒ². (3) By lemma, 2 | p, so p = 2m. (4) Then 2qΒ² = 4mΒ² β qΒ² = 2mΒ², so 2 | q. (5) Both p and q divisible by 2 β contradicts "lowest terms". β΄ β2 is irrational.
π 6. Decimal Expansions of Rational Numbers
p/q terminates β q = 2βΏ Γ 5α΅ (n, m β₯ 0)
A rational number p/q (in lowest terms) has a terminating decimal if and only if its denominator q has no prime factor other than 2 and 5.
Terminating: 7/8 = 7/2Β³ β 0.875 β
Non-terminating repeating: 1/3 = 0.333... (denominator has factor 3) β
Irrational: Non-terminating, non-repeating (e.g., Ο, β2)
Terminating: 7/8 = 7/2Β³ β 0.875 β
Non-terminating repeating: 1/3 = 0.333... (denominator has factor 3) β
Irrational: Non-terminating, non-repeating (e.g., Ο, β2)
π Key Theorems to Remember
Theorem 1
Euclid's Division Lemma
Given positive integers a and b, there exist unique integers q and r such that:
a = b Γ q + r (0 β€ r < b)
This guarantees the uniqueness of quotient and remainder in division.
Theorem 2
Fundamental Theorem of Arithmetic
Every composite number can be factorised as a product of primes, and this factorisation is unique apart from the order of the prime factors.
n = pβα΅ Γ pβα΅ Γ pβαΆ Γ ... (where pβ, pβ, pβ ... are primes)
Theorem 3
Irrationality of βp (where p is prime)
If p is a prime number, then βp is irrational. This is proven by contradiction using the key lemma:
If p | aΒ², then p | a (for prime p)
Theorem 4
Terminating Decimal Condition
Let p/q be a rational number in lowest terms. Its decimal expansion terminates if and only if q is of the form:
q = 2βΏ Γ 5α΅ (n, m are non-negative integers)
π Quick Reference Table
| Concept | Key Formula / Condition | Example |
|---|---|---|
| Euclid's Lemma | a = bq + r, 0 β€ r < b | 135 = 8Γ16 + 7 |
| HCF (by EDA) | Last non-zero remainder | HCF(48,18) = 6 |
| LCM (by PF) | Highest power of all primes | LCM(12,18) = 36 |
| HCF Γ LCM | HCF Γ LCM = a Γ b | 6 Γ 36 = 12 Γ 18 β |
| Terminating decimal | q = 2βΏ Γ 5α΅ | 7/8 β 0.875 β |
| Non-terminating | q has prime β 2, 5 | 1/6 = 0.1666... |
| Irrational number | βprime, sum of irrationals | β2, β3, β5, Ο |
| Rational + Irrational | Always irrational | 2 + β3 is irrational |
π Important Questions & Solutions
Q1. Use Euclid's division algorithm to find HCF of 867 and 255.
2 marks+
Solution:
Step 1: 867 = 255 Γ 3 + 102
Step 2: 255 = 102 Γ 2 + 51
Step 3: 102 = 51 Γ 2 + 0
Since remainder = 0, HCF(867, 255) = 51
Step 1: 867 = 255 Γ 3 + 102
Step 2: 255 = 102 Γ 2 + 51
Step 3: 102 = 51 Γ 2 + 0
Since remainder = 0, HCF(867, 255) = 51
Q2. Prove that β5 is irrational.
3 marks+
Proof by Contradiction:
Assume β5 is rational β β5 = a/b (in lowest terms, HCF(a,b) = 1)
β 5 = aΒ²/bΒ² β aΒ² = 5bΒ² β 5 | aΒ² β 5 | a (by key lemma)
Let a = 5c β aΒ² = 25cΒ² β 5bΒ² = 25cΒ² β bΒ² = 5cΒ² β 5 | b
Both a and b are divisible by 5. This contradicts HCF(a,b) = 1.
β΄ Our assumption was wrong. β5 is irrational. β‘
Assume β5 is rational β β5 = a/b (in lowest terms, HCF(a,b) = 1)
β 5 = aΒ²/bΒ² β aΒ² = 5bΒ² β 5 | aΒ² β 5 | a (by key lemma)
Let a = 5c β aΒ² = 25cΒ² β 5bΒ² = 25cΒ² β bΒ² = 5cΒ² β 5 | b
Both a and b are divisible by 5. This contradicts HCF(a,b) = 1.
β΄ Our assumption was wrong. β5 is irrational. β‘
Q3. Find HCF and LCM of 96 and 404 by prime factorisation. Verify HCF Γ LCM = a Γ b.
3 marks+
96 = 2β΅ Γ 3 | 404 = 2Β² Γ 101
HCF = 2Β² = 4
LCM = 2β΅ Γ 3 Γ 101 = 9696
Verification: HCF Γ LCM = 4 Γ 9696 = 38784
a Γ b = 96 Γ 404 = 38784 β
HCF = 2Β² = 4
LCM = 2β΅ Γ 3 Γ 101 = 9696
Verification: HCF Γ LCM = 4 Γ 9696 = 38784
a Γ b = 96 Γ 404 = 38784 β
Q4. Without actually performing the long division, state whether 17/8 has a terminating or non-terminating decimal expansion.
1 mark+
Denominator = 8 = 2Β³. Since 8 = 2Β³ Γ 5β° (form 2βΏ Γ 5α΅), the decimal expansion of 17/8 is terminating.
17/8 = 2.125
17/8 = 2.125
Q5. Prove that 3 + 2β5 is irrational.
3 marks+
Assume 3 + 2β5 is rational. Then 3 + 2β5 = a/b (where a, b are integers, b β 0).
β 2β5 = a/b β 3 = (a β 3b)/b
β β5 = (a β 3b)/(2b)
Since a, b are integers, (a β 3b)/(2b) is rational β β5 is rational.
But this contradicts the fact that β5 is irrational.
β΄ 3 + 2β5 is irrational. β‘
β 2β5 = a/b β 3 = (a β 3b)/b
β β5 = (a β 3b)/(2b)
Since a, b are integers, (a β 3b)/(2b) is rational β β5 is rational.
But this contradicts the fact that β5 is irrational.
β΄ 3 + 2β5 is irrational. β‘
Q6. The HCF of two numbers is 23 and their LCM is 1449. If one number is 161, find the other.
2 marks+
Using: HCF Γ LCM = a Γ b
23 Γ 1449 = 161 Γ b
b = (23 Γ 1449)/161 = 33327/161 = 207
23 Γ 1449 = 161 Γ b
b = (23 Γ 1449)/161 = 33327/161 = 207
β οΈ Common Mistakes Students Make
β Wrong
HCF Γ LCM = a Γ b applies to THREE numbers too
β
Correct
HCF Γ LCM = a Γ b is valid ONLY for TWO numbers
β Wrong
Writing LCM as product of lowest powers of prime factors
β
Correct
LCM = product of HIGHEST powers; HCF = LOWEST powers
β Wrong
Stopping Euclid's algorithm when q = 0 (quotient)
β
Correct
Stop when REMAINDER r = 0; the divisor at that step is HCF
β Wrong
1/7 has terminating decimal (7 is not 2 or 5, but ignoring)
β
Correct
1/7 = 0.142857... non-terminating (7 β 2βΏ Γ 5α΅ form)
β Wrong
In proof of irrationality, forgetting to state "HCF(a,b) = 1"
β
Correct
Always assume p/q in lowest terms (i.e., HCF = 1) at start of proof
π― Exam Tips from Manish Sir
-
Master Euclid's Algorithm first It's guaranteed in every board paper. Practice 10 examples until you can do it in under 3 minutes.
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Write every step in proof questions Irrationality proofs are 3-markers. Board checkers award marks for each logical step β never skip!
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Always simplify p/q first Before checking for terminating decimals, reduce p/q to lowest terms. HCF might cancel a problematic factor.
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Memorise the key lemma "If p is prime and p | aΒ², then p | a." This single line unlocks every irrational number proof.
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This chapter = easy 7+ marks Chapter 1 questions are predictable. Focus here before moving to harder chapters β guaranteed returns!
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Verify HCF Γ LCM = a Γ b In 3-mark questions asking for both HCF and LCM, always add a verification line β examiners love it!
β Frequently Asked Questions
What is Euclid's Division Lemma in Class 10 Maths?
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 = b Γ q + r, where 0 β€ r < b. It is the foundation of Euclid's Division Algorithm used to find HCF.
What is the difference between HCF and LCM in prime factorisation?
In prime factorisation method: HCF = product of the LOWEST (minimum) powers of all COMMON prime factors. LCM = product of the HIGHEST (maximum) powers of ALL prime factors (common and non-common). And always: HCF Γ LCM = Product of the two numbers.
How do I prove a number is irrational in Class 10?
Use Proof by Contradiction: (1) Assume the number is rational, i.e., equal to p/q in lowest terms (HCF = 1). (2) Square both sides and manipulate to show a prime divides both p and q. (3) This contradicts HCF(p,q) = 1. Therefore the number must be irrational. This method works for β2, β3, β5 and their combinations.
How to check if a rational number has a terminating decimal?
Reduce the fraction p/q to its lowest terms. Then check the denominator q. If q can be written as 2βΏ Γ 5α΅ (only powers of 2 and 5), the decimal is terminating. If q has any other prime factor (like 3, 7, 11, etc.), the decimal is non-terminating repeating.
Can HCF be greater than LCM?
No! HCF is always less than or equal to LCM. HCF = LCM only when both numbers are equal. For different numbers, HCF < LCM always. Also, HCF always divides LCM exactly.
Is Ο (pi) a rational or irrational number?
Ο is an irrational number. Its decimal expansion is non-terminating and non-repeating (3.14159265...). Note that 22/7 is only an approximation of Ο β it is NOT equal to Ο. 22/7 β 3.142857... while Ο β 3.14159...
How many marks does Real Numbers carry in CBSE Class 10 board exam?
Chapter 1 Real Numbers typically carries 6 to 8 marks in the CBSE Class 10 Maths board paper. This usually includes one 1-mark MCQ, one 2-mark question (HCF by algorithm), and one 3-mark question (either LCM+HCF by factorisation or an irrationality proof).
What is the Fundamental Theorem of Arithmetic?
The Fundamental Theorem of Arithmetic states that every composite number greater than 1 can be expressed as a product of prime numbers in a unique way (ignoring order). For example: 360 = 2Β³ Γ 3Β² Γ 5. This uniqueness is what makes prime factorisation so powerful for finding HCF and LCM.
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