Real Numbers (Class 10): Notes और MCQ Practice

Class 10 Mathematics • Class 10

आसान भाषा में

इस चैप्टर में हम मुख्य रूप से Composite numbers का Prime Factorisation करना और √2, √3, √5 जैसे numbers की Irrationality prove करना सीखते हैं। यह पूरा चैप्टर Fundamental Theorem of Arithmetic पर टिका है, जो हमें Prime Factors की मदद से HCF और LCM निकालने का आसान तरीका सिखाता है।

Key Formulas & Definitions

  1. Fundamental Theorem of Arithmetic: प्रत्येक Composite number को Primes के product के रूप में uniquely express किया जा सकता है (factors के order को छोड़कर)।
  2. Important Relation (केवल 2 numbers के लिए):

HCF(a, b) × LCM(a, b) = a × b

  1. Divisibility Rule for Primes: यदि p एक prime number है और p, a² को divide करता है, तो p, a को भी divide करेगा (जहाँ a एक positive integer है)।
  2. Irrational Numbers: जिन्हें p/q form में नहीं लिखा जा सकता (q ≠ 0, p और q integers हैं), जैसे √2, √5, 3 + 2√5।

2 Memory Tricks

  • HCF vs LCM Powers:
  • HCF में नाम High है, लेकिन काम उल्टा: केवल common prime factors की Smallest power लेनी है।
  • LCM में नाम Low है, लेकिन हर prime factor की Greatest power लेनी है।
  • Two-Body Rule: याद रखें "HL = P" (HCF × LCM = Product)। यह rule केवल 2 numbers के लिए ही valid है, तीसरे को entry नहीं मिलती!

2 Common Board Exam Mistakes

  1. 3 Numbers पर Formula लगाना: छात्र अक्सर HCF(a, b, c) × LCM(a, b, c) = a × b × c लिख देते हैं, जो कि बिल्कुल गलत है। यह rule 3 numbers के लिए काम नहीं करता।
  2. Co-prime mention न करना: √2 या √3 को irrational prove करते समय 'Let a and b are co-prime integers (HCF = 1)' लिखना भूल जाना। इसके बिना पूरे steps के marks कट जाते हैं!

Real Numbers - Class 10 NCERT Study Notes

1. Chapter in Short

In Class 10, the chapter Real Numbers focuses on the foundational properties of positive integers and real numbers. The entire chapter revolves around two central themes:

  1. The Fundamental Theorem of Arithmetic: Understanding how composite numbers are uniquely built from prime numbers, and using prime factorization to find the Highest Common Factor (HCF) and Least Common Multiple (LCM).
  2. Revisiting Irrational Numbers: Proving that numbers like √2, √3, √5, and their combinations (like 3 + 2√5) are irrational using the powerful mathematical technique called Proof by Contradiction.

2. Key Concepts Explained Simply

A. Prime and Composite Numbers

  • Prime Numbers: Numbers greater than 1 that have only two factors: 1 and the number itself (e.g., 2, 3, 5, 7, 11).
  • Composite Numbers: Numbers that have more than two factors (e.g., 4, 6, 8, 9, 12).
  • Note: 1 is neither prime nor composite. 2 is the only even prime number.

B. The Fundamental Theorem of Arithmetic

  • Idea: Think of prime numbers as basic building blocks (like LEGO bricks). Any composite number can be broken down into a unique set of prime blocks.
  • Everyday Example: Think of a smoothie recipe. If you blend 2 bananas and 3 strawberries, it does not matter if you throw in the bananas first or the strawberries first; the final drink is uniquely determined by those ingredients. Similarly, 12 is always 2 × 2 × 3 = 2² × 3, regardless of the order in which you write the factors.

C. HCF and LCM by Prime Factorization

  • HCF (Highest Common Factor): Product of the smallest power of each common prime factor involved in the numbers.
  • LCM (Least Common Multiple): Product of the greatest power of each prime factor involved in the numbers.

D. Important Relationship Between HCF and LCM

For any two positive integers a and b:

  • HCF(a, b) × LCM(a, b) = a × b

(Warning: This rule works only for two numbers, never for three or more numbers!)

E. Revisiting Irrational Numbers

  • A number is called irrational if it cannot be written in the form p ÷ q, where p and q are integers and q ≠ 0.
  • When written as decimals, irrational numbers are non-terminating and non-repeating (e.g., √2 = 1.4142135...).
  • Method of Contradiction: To prove a statement is true, we assume the exact opposite is true. We then follow logical mathematical steps until we arrive at an impossible situation (a contradiction). This forces us to conclude that our initial assumption was wrong.

3. Important Definitions, Formulas, and Facts

  • Fundamental Theorem of Arithmetic: 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.
  • Core Divisibility Theorem: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
  • Co-prime Integers: Two integers a and b are called co-prime if their HCF is 1 (they share no common factor other than 1).
  • HCF and LCM Formula (For 2 numbers a and b):
  • HCF(a, b) × LCM(a, b) = a × b
  • LCM(a, b) = (a × b) ÷ HCF(a, b)
  • HCF(a, b) = (a × b) ÷ LCM(a, b)
  • Properties of Rational and Irrational Operations:
  • Rational + Irrational = Irrational
  • Rational − Irrational = Irrational
  • Non-zero Rational × Irrational = Irrational
  • Non-zero Rational ÷ Irrational = Irrational

4. Solved Board-Exam Style Examples

Example 1: Finding HCF and LCM and Verifying the Formula

Question: Find the HCF and LCM of 12 and 18 using prime factorisation. Verify that HCF × LCM = Product of the two numbers.

Solution:

  • Prime factorisation of 12: 12 = 2 × 2 × 3 = 2² × 3¹
  • Prime factorisation of 18: 18 = 2 × 3 × 3 = 2¹ × 3²
  • HCF(12, 18) = Smallest powers of common factors = 2¹ × 3¹ = 6
  • LCM(12, 18) = Highest powers of all prime factors = 2² × 3² = 4 × 9 = 36
  • Verification:
  • HCF × LCM = 6 × 36 = 216
  • Product of numbers = 12 × 18 = 216
  • Since HCF × LCM = Product of numbers (216 = 216), the relationship is verified.

Example 2: Using the HCF-LCM Relationship

Question: Given that HCF(306, 657) = 9, find LCM(306, 657).

Solution:

We know that for any two numbers a and b:

LCM(a, b) = (a × b) ÷ HCF(a, b)

Here, a = 306, b = 657, and HCF = 9.

  • LCM(306, 657) = (306 × 657) ÷ 9
  • Divide 306 by 9: 306 ÷ 9 = 34
  • LCM = 34 × 657 = 22338

Answer: LCM(306, 657) = 22338

Example 3: Proving a Number is Irrational

Question: Prove that √5 is an irrational number.

Solution:

  1. Let us assume, to the contrary, that √5 is rational.
  2. Therefore, we can find co-prime integers a and b (where b ≠ 0) such that:

√5 = a ÷ b

  1. Rearranging, we get:

a = b√5

  1. Squaring both sides:

a² = 5b² ... (Equation 1)

  1. This means 5 divides a². By the theorem, if 5 divides a², then 5 divides a.
  2. So, we can write a = 5c for some integer c.
  3. Substituting a = 5c into Equation 1:

(5c)² = 5b²

25c² = 5b²

5c² = b² (Dividing both sides by 5)

  1. This means 5 divides b². Therefore, 5 divides b.
  2. From steps 5 and 8, both a and b have at least 5 as a common factor.
  3. But this contradicts our initial fact that a and b are co-prime (having no common factor other than 1).
  4. This contradiction arose because of our incorrect assumption that √5 is rational.
  5. Hence, √5 is irrational.

Example 4: Proving a Combination is Irrational

Question: Prove that 3 + 2√5 is irrational, given that √5 is irrational.

Solution:

  1. Let us assume, to the contrary, that 3 + 2√5 is rational.
  2. Then we can write 3 + 2√5 = a ÷ b, where a and b are integers, b ≠ 0, and a, b are co-prime.
  3. Rearranging the equation to isolate √5:

2√5 = (a ÷ b) − 3

2√5 = (a − 3b) ÷ b

√5 = (a − 3b) ÷ 2b

  1. Since a and b are integers, (a − 3b) ÷ 2b is a rational number.
  2. This means √5 must be rational.
  3. But this contradicts the known fact that √5 is irrational.
  4. Therefore, our assumption was incorrect. Hence, 3 + 2√5 is irrational.

Example 5: Testing End Digits (Application of Fundamental Theorem)

Question: Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution:

  1. If any number ends with the digit 0, it must be divisible by 10.
  2. Since 10 = 2 × 5, a number ending in 0 must have both 2 and 5 in its prime factorisation.
  3. Prime factorisation of 6ⁿ:

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

  1. The only prime factors of 6ⁿ are 2 and 3. The prime 5 does not appear.
  2. By the Fundamental Theorem of Arithmetic, the prime factorisation of any number is unique.
  3. Hence, 6ⁿ can never end with the digit 0 for any natural number n.

5. Common Mistakes and Quick Revision Points

Common Mistakes

  • Applying the formula to 3 numbers: Students often mistakenly write HCF(a, b, c) × LCM(a, b, c) = a × b × c. This is false! The formula holds true strictly for two numbers.
  • Forgetting the word 'co-prime': In contradiction proofs for irrational numbers, failing to define a and b as co-prime integers leads to marks deduction.
  • Mixing up powers for HCF and LCM:
  • HCF uses the lowest/minimum power of common primes.
  • LCM uses the highest/maximum power of all primes involved.
  • Skipping reason statements: In proofs, always state the theorem: "If a prime p divides a², then p divides a."

Quick Revision Points

  • A rational number can always be written as a fraction p/q; an irrational number cannot.
  • Any composite number can be uniquely factored into primes.
  • To find if a number like 4ⁿ or 6ⁿ ends with 0, check if both 2 and 5 are factors.
  • HCF of two prime numbers is always 1, and their LCM is their product.

Practice MCQ (15)

  1. प्राकृतिक संख्या 140 का अभाज्य गुणनखंडन क्या है?

    • (A) 2 × 5 × 7
    • (B) 2² × 5 × 7
    • (C) 2 × 5² × 7
    • (D) 2² × 5² × 7

    उत्तर: (B)

  2. यदि p एक परिमेय संख्या है और √q एक अपरिमेय संख्या है, तो योगफल p + √q सदैव क्या होगा?

    • (A) An integer
    • (B) A rational number
    • (C) An irrational number
    • (D) A natural number

    उत्तर: (C)

  3. यदि दो धनात्मक पूर्णांकों a और b का HCF 12 है और उनका गुणनफल 1800 है, तो उनका LCM क्या होगा?

    • (A) 120
    • (B) 150
    • (C) 180
    • (D) 216

    उत्तर: (B)

  4. अंकगणित की आधारभूत प्रमेय के अनुसार, प्रत्येक भाज्य संख्या को किसके गुणनफल के रूप में अद्वितीय रूप से गुणनखंडित किया जा सकता है?

    • (A) Odd numbers only
    • (B) Co-prime numbers only
    • (C) Composite numbers only
    • (D) Prime numbers, apart from the order in which the factors occur

    उत्तर: (D)

  5. निम्नलिखित में से कौन-सी संख्या अपरिमेय (irrational) है?

    • (A) √4
    • (B) 3 + √9
    • (C) 2 − √5
    • (D) (√3)²

    उत्तर: (C)

  6. 196 के अभाज्य गुणनखंडन में अभाज्य गुणनखंडों के घातों का योग कितना है?

    • (A) 2
    • (B) 3
    • (C) 4
    • (D) 5

    उत्तर: (C)

  7. किसी भी प्राकृतिक संख्या n के लिए, संख्या 6ⁿ अंक 0 पर केवल तभी समाप्त होती है जब इसके अभाज्य गुणनखंडन में निम्नलिखित अभाज्य गुणनखंड शामिल हों:

    • (A) 2 and 3
    • (B) 3 and 5
    • (C) 5 and 7
    • (D) 2 and 5

    उत्तर: (D)

  8. यदि दो धनात्मक पूर्णांकों x और y को अभाज्य संख्याओं p और q के रूप में x = p³q² और y = p²q⁴ द्वारा व्यक्त किया जा सकता है, तो HCF(x, y) क्या होगा?

    • (A) pq
    • (B) p²q²
    • (C) p³q⁴
    • (D) p⁵q⁶

    उत्तर: (B)

  9. दो धनात्मक पूर्णांक a और b को a = x²y³ और b = x³y के रूप में लिखा जाता है, जहाँ x और y अभाज्य संख्याएँ हैं। तो LCM(a, b) क्या होगा?

    • (A) xy
    • (B) x²y
    • (C) x⁵y⁴
    • (D) x³y³

    उत्तर: (D)

  10. एक मिठाई विक्रेता के पास 420 काजू की बर्फ़ियाँ और 130 बादाम की बर्फ़ियाँ हैं। वह उनकी ऐसी ढेरियाँ बनाना चाहती है कि प्रत्येक ढेरी में बर्फ़ियों की संख्या समान हो तथा ये ट्रे में न्यूनतम स्थान घेरें। प्रत्येक ढेरी में रखी जा सकने वाली बर्फ़ियों की अधिकतम संख्या क्या है?

    • (A) 10
    • (B) 15
    • (C) 20
    • (D) 26

    उत्तर: (A)

  11. सोनिया और रवि एक वृत्ताकार खेल के मैदान के चारों ओर गाड़ी चलाते हैं। सोनिया को एक चक्कर पूरा करने में 18 मिनट लगते हैं, जबकि रवि को इसी के लिए 12 मिनट लगते हैं। यदि वे दोनों एक ही बिंदु से, एक ही समय पर चलना प्रारंभ करते हैं और एक ही दिशा में चलते हैं, तो वे कितने मिनट बाद प्रारंभिक बिंदु पर पुनः मिलेंगे?

    • (A) 6
    • (B) 24
    • (C) 36
    • (D) 72

    उत्तर: (C)

  12. तीन घंटियाँ क्रमशः 4, 7 और 12 मिनट के अंतराल पर बजती हैं। यदि वे सभी एक साथ सुबह 6:00 AM पर बजी हों, तो वे अगली बार एक साथ किस समय बजेंगी?

    • (A) 6:42 AM
    • (B) 7:12 AM
    • (C) 7:24 AM
    • (D) 7:48 AM

    उत्तर: (C)

  13. वह सबसे बड़ी संख्या कौन-सी है जिससे 70 और 125 को विभाजित करने पर क्रमशः 5 और 8 शेषफल प्राप्त होता है?

    • (A) 13
    • (B) 65
    • (C) 875
    • (D) 1755

    उत्तर: (A)

  14. यदि दो संख्याएं a = 2³ × 3² × 5 और b = 2² × 3³ × 7 हैं, तो LCM(a, b) का HCF(a, b) से अनुपात क्या है?

    • (A) 105 : 1
    • (B) 210 : 1
    • (C) 420 : 1
    • (D) 35 : 1

    उत्तर: (B)

  15. यदि p और q दो भिन्न अभाज्य संख्याएं हैं, तो HCF(p + q, p × q) किसके बराबर होगा?

    • (A) 1
    • (B) 2
    • (C) p
    • (D) q

    उत्तर: (A)