MP Board Class 10 Maths Chapter 1: Real Numbers (वास्तविक…
MP Board Class 10 Maths Chapter 1: Real Numbers (वास्तविक संख्याएँ) — This chapter lays the foundation for number theory in Class 10 Mathematics. It covers Euclid’s Division Lemma and Algorithm, the Fundamental Theorem of Arithmetic, properties of prime and composite numbers, HCF and LCM calculations, proofs of irrationality of numbers like √2, √3, and √5, and the concept of terminating and non-terminating repeating decimals. In the MP Board Class 10 board exam, this chapter carries 6–8 marks and typically appears as short answer questions (2–3 marks each) and one long answer question (4 marks).
📑 Table of Contents
- 1. Euclid’s Division Lemma and Algorithm
- 2. The Fundamental Theorem of Arithmetic
- 3. HCF and LCM Using Prime Factorisation
- 4. Proving Irrational Numbers
- 5. Decimal Expansions of Rational Numbers
- 6. Important Theorems and Proofs
- 7. Solved Examples from MP Board Exams
- 📋 Previous Year Questions (2017–2026)
- ❓ Frequently Asked Questions
🔢 1. Euclid’s Division Lemma and Algorithm
Euclid’s Division Lemma
Statement: For any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b
- a = Dividend
- b = Divisor
- q = Quotient
- r = Remainder
This lemma is the foundation of Euclid’s Division Algorithm. It states that when we divide any integer by another positive integer, we get a unique quotient and remainder. The remainder is always less than the divisor and greater than or equal to zero.
Euclid’s Division Algorithm
Euclid’s Division Algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. It involves repeatedly applying the division lemma until the remainder becomes zero.
Steps to find HCF using Euclid’s Algorithm:
- Apply Euclid’s division lemma to the larger number c and the smaller number d: c = dq + r, where 0 ≤ r < d
- If r = 0, then HCF(c, d) = d
- If r ≠ 0, apply the lemma to d and r: d = rq₁ + r₁
- Continue this process until the remainder becomes zero
- The divisor at this stage is the HCF
📘 Example: Find HCF of 135 and 225 using Euclid’s algorithm.
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
HCF(135, 225) = 45
Since the remainder becomes zero at the last step, the divisor at that stage (45) is the HCF.
🧮 2. The Fundamental Theorem of Arithmetic
Statement of the Theorem
The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
In other words, for any composite number n, there is exactly one way (ignoring the order of factors) to write it as:
n = p₁ᵃ × p₂ᵇ × p₃ᶜ × … × pₖᶻ
where p₁, p₂, p₃, …, pₖ are prime numbers and a, b, c, …, z are positive integers (their powers).
📘 Examples of Prime Factorisation:
12 = 2 × 2 × 3 = 2² × 3
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
5005 = 5 × 7 × 11 × 13
156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
Notice that each number has a unique prime factorisation — no two composite numbers have the same set of prime factors with the same powers.
Applications of the Fundamental Theorem
- Finding HCF: Take the product of the smallest power of each common prime factor
- Finding LCM: Take the product of the greatest power of each prime factor involved
- Proving irrationality: Used in contradiction proofs for √2, √3, √5
- Classifying decimal expansions: Determining if a rational number has terminating or non-terminating decimal expansion
📊 3. HCF and LCM Using Prime Factorisation
Finding HCF by Prime Factorisation
To find the HCF of two or more numbers using prime factorisation:
- Express each number as a product of primes
- Identify the common prime factors
- Take the smallest exponent for each common prime
- Multiply these together
Finding LCM by Prime Factorisation
To find the LCM of two or more numbers using prime factorisation:
- Express each number as a product of primes
- Consider ALL prime factors that appear in any number
- Take the greatest exponent for each prime
- Multiply these together
Relationship Between HCF and LCM
For any two positive integers a and b:
HCF(a, b) × LCM(a, b) = a × b
📐 4. Proving Irrational Numbers
What is an Irrational Number?
An irrational number is a number that cannot be expressed in the form p/q, where p and q are integers and q ≠ 0. In other words, it is a non-terminating, non-repeating decimal. Examples: √2, √3, √5, π, e.
Proof that √2 is Irrational
This is one of the most important proofs in Class 10 Mathematics. It uses the method of contradiction:
- Assume that √2 is rational. Then √2 = p/q, where p and q are co-prime integers (q ≠ 0)
- Squaring both sides: 2 = p²/q² → p² = 2q²
- Since p² = 2q², p² is divisible by 2. Therefore p is divisible by 2 (by the Fundamental Theorem of Arithmetic)
- Let p = 2r for some integer r. Then (2r)² = 2q² → 4r² = 2q² → 2r² = q²
- This means q² is divisible by 2, so q is also divisible by 2
- But p and q are both divisible by 2, which contradicts our assumption that p and q are co-prime!
- Therefore, our assumption was wrong, and √2 is irrational.
Proving 3 + 2√5 is Irrational
This is a common variation asked in MP Board exams:
- Assume 3 + 2√5 is rational = a/b (where a, b are co-prime, b ≠ 0)
- Then 2√5 = a/b − 3 = (a − 3b)/b
- So √5 = (a − 3b)/(2b)
- The right-hand side is a rational number (difference/product of integers divided by integers)
- But we know √5 is irrational — contradiction!
- Therefore, 3 + 2√5 is irrational.
🔄 5. Decimal Expansions of Rational Numbers
Terminating vs Non-Terminating Decimals
Let x = p/q be a rational number in its simplest form (p and q are co-prime). The decimal expansion of x is:
- Terminating if the prime factorisation of q has only 2 and/or 5 as prime factors
- Non-terminating repeating if q has a prime factor other than 2 or 5
Classification Table
- If q = 2ᵐ × 5ⁿ → Terminating decimal expansion
- If q has any other prime factor → Non-terminating repeating
📝 6. Important Theorems and Proofs
Theorem 1.1: Euclid’s Division Lemma
For any positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.
Theorem 1.2: Fundamental Theorem of Arithmetic
Every composite number can be expressed uniquely as a product of primes, apart from the order of factors.
Theorem 1.3: Irrationality of √p
Let p be a prime number. If p divides a², then p divides a (where a is a positive integer). Using this, we prove that √p is irrational.
Theorem 1.4: Decimal Expansion Classification
A rational number p/q (in its simplest form) has a terminating decimal expansion if and only if the denominator q is of the form 2ᵐ × 5ⁿ, where m and n are non-negative integers.
✏️ 7. Solved Examples from MP Board Exams
Example 1: Using Euclid’s Algorithm
Question: Find the HCF of 867 and 255 using Euclid’s division algorithm.
Solution:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Therefore, HCF(867, 255) = 51
Example 2: HCF and LCM
Question: Find the HCF and LCM of 510 and 92 and verify that HCF × LCM = product of the two numbers.
Solution:
510 = 2 × 3 × 5 × 17
92 = 2² × 23
HCF(510, 92) = 2
LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 4 × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920
Product of numbers = 510 × 92 = 46920 ✓
Example 3: Irrationality Proof (MP Board 2024)
Question: Prove that 7√5 is irrational.
Solution:
Assume 7√5 is rational. Then 7√5 = a/b (where a, b are co-prime integers, b ≠ 0).
So √5 = a/(7b). The RHS is rational (ratio of integers).
But √5 is irrational (proved in Theorem 1.3). Contradiction!
Therefore, 7√5 is irrational.
Example 4: Decimal Expansions (MP Board 2025)
Question: Without actually performing the long division, state whether 17/125 has a terminating or non-terminating repeating decimal expansion. Write the decimal expansion if terminating.
Solution:
125 = 5³ (only prime factor 5)
Since the denominator is of the form 2ᵐ × 5ⁿ, this has a terminating decimal expansion.
17/125 = (17 × 8)/(125 × 8) = 136/1000 = 0.136
📋 Previous Year Questions (2017–2026)
❓ Frequently Asked Questions
Q1: What is the difference between Euclid’s Division Lemma and Euclid’s Division Algorithm?
Answer: Euclid’s Division Lemma is a statement about the existence of unique quotient and remainder (a = bq + r, 0 ≤ r < b). Euclid’s Division Algorithm is the step-by-step process of repeatedly applying the lemma to find the HCF of two numbers.
Q2: What is the Fundamental Theorem of Arithmetic used for?
Answer: It is used to: (a) Find HCF and LCM of numbers through prime factorisation, (b) Prove the irrationality of numbers like √2, √3, and √5, (c) Determine whether a rational number has a terminating or non-terminating decimal expansion.
Q3: How do I check if a rational number has a terminating decimal expansion?
Answer: Write the rational number in its simplest form (p/q where p and q are co-prime). Factorise q. If q has only 2 and/or 5 as prime factors, it is terminating. If any other prime appears, it is non-terminating repeating.
Q4: Can 4ⁿ end with the digit 0 for any natural number n?
Answer: No. For a number to end with 0, it must have 5 as a factor. 4ⁿ = (2²)ⁿ = 2²ⁿ, which only has 2 as a prime factor. Since 5 never divides 4ⁿ, it cannot end with 0 for any natural number n.
Q5: Is the product of two irrational numbers always irrational?
Answer: No. √2 is irrational and √2 × √2 = 2, which is rational. However, √2 × √3 = √6, which is irrational. It depends on the specific numbers.
Q6: What is the maximum number of columns in which 616 members and 32 officers can march?
Answer: This is an HCF problem. HCF(616, 32) = ? 616 = 32 × 19 + 8, 32 = 8 × 4 + 0. So HCF = 8. Maximum 8 columns. (This is NCERT Example 2.)
Q7: What is the smallest composite number?
Answer: 4 = 2 × 2 is the smallest composite number. Note that 1 is neither prime nor composite.
Q8: What is the meaning of co-prime numbers?
Answer: Two numbers are said to be co-prime (or relatively prime) if their HCF is 1. For example, (2, 3), (4, 9), (15, 28) are co-prime pairs. Note that co-prime numbers need not themselves be prime — 4 and 9 are co-prime but both are composite.
Q9: How many prime factors does 100 have?
Answer: 100 = 2² × 5². The distinct prime factors are 2 and 5 (two distinct primes). The total number of prime factors (counting multiplicities) is 4 (2 + 2).
Q10: Explain why 7 × 11 × 13 + 13 is a composite number.
Answer: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 3 × 13 = 2 × 3 × 13². Since it has prime factors 2, 3, and 13, it is a composite number.
Q11: What is the HCF of smallest prime number and smallest composite number?
Answer: Smallest prime number = 2, smallest composite number = 4. HCF(2, 4) = 2.
Q12: Can two numbers have 16 as LCM and 380 as product?
Answer: No. If two numbers have LCM = 16, then their product = LCM × HCF = 16 × HCF. The product must be divisible by 16. 380 ÷ 16 = 23.75, which is not an integer. Therefore, two numbers cannot have LCM = 16 and product = 380.