**REAL NUMBERS - COMPREHENSIVE CHEAT SHEET**
**FUNDAMENTAL THEOREM OF ARITHMETIC**
• Definition: 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.
• Key Point: The prime factorisation of a natural number is unique → only ONE way to write it as a product of primes (ignoring order)
• Standard Form: For composite number x: x = p₁^a₁ × p₂^a₂ × ... × pₙ^aₙ where p₁ < p₂ < ... < pₙ are primes and a₁, a₂, ... aₙ are positive integers
• Example: 32760 = 2³ × 3² × 5 × 7 × 13 (same as 13 × 7 × 5 × 3² × 2³)
• Why It Matters: Guarantees that no other prime factors exist beyond those found → used to prove irrationality and decimal expansion properties
**APPLICATION 1: FINDING HCF AND LCM USING PRIME FACTORISATION**
• Prime Factorisation Method Steps:
• Example: For 6 = 2¹ × 3¹ and 20 = 2² × 5¹
• Critical Formula: HCF(a, b) × LCM(a, b) = a × b
**APPLICATION 2: DETERMINING IF A NUMBER ENDS WITH ZERO**
• Rule: A number ends with zero ↔ it is divisible by both 2 AND 5 ↔ prime factorisation contains at least one 2 AND one 5
• Method: Find prime factorisation → check if both 2 and 5 appear → if not, number cannot end with zero
• Example: Does 4ⁿ ever end with zero? 4ⁿ = (2²)ⁿ = 2^(2n) → only prime factor is 2 → NO, by uniqueness theorem, 5 cannot appear
**EUCLID'S DIVISION ALGORITHM**
• Definition: For any two positive integers a and b (a > b), there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b
• Key Concept: The remainder r is ALWAYS smaller than the divisor b
• Geometric Meaning: This is the standard long division process you use in arithmetic
• Main Application: Finding HCF(Greatest Common Divisor) of two numbers
• Algorithm to Find HCF(a, b):
• Example: Find HCF(135, 225)
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
**CONNECTION BETWEEN BOTH ALGORITHMS**
• HCF Using Prime Factorisation: Find smallest power of common primes
• HCF Using Euclid's Algorithm: Repeated division process
• Both methods give the SAME answer but Euclid's is more efficient for large numbers
**APPLICATIONS TO RATIONAL NUMBERS AND DECIMAL EXPANSION**
• Terminating Decimal Expansion: p/q has terminating decimal ↔ Denominator q (in lowest terms) has prime factorisation of form 2ᵐ × 5ⁿ
• Non-Terminating Repeating Decimal: If q has prime factors OTHER than 2 and 5 → decimal is non-terminating and repeating
**APPLICATION 3: PROVING IRRATIONALITY**
• Proof Strategy for √p (p is prime): Assume √p = a/b (rational), reduce to lowest terms, apply uniqueness of prime factorisation to derive contradiction
• Key Insight: If a² = pb² → prime p divides a² → p divides a → a = pc for some integer c → substituting leads to p dividing b too → contradicts lowest terms assumption
• Proven Irrational: √2, √3, √5, and √p for any prime p
**COMMON MISTAKES TO AVOID**
⚠ Mistake 1: Confusing HCF and LCM formulas
⚠ Mistake 2: Forgetting that HCF × LCM = a × b
⚠ Mistake 3: Not reducing fraction to lowest terms before checking decimal expansion
⚠ Mistake 4: In Euclid's algorithm, stopping when remainder is small instead of when it's ZERO
⚠ Mistake 5: Assuming uniqueness of prime factorisation means numbers CAN'T have common factors
⚠ Mistake 6: Forgetting that prime factorisation must contain ONLY prime numbers
**TYPES OF EXAM PROBLEMS**
1. Prime Factorisation: Find unique prime factorisation of given number
2. HCF/LCM Calculation: Use prime factorisation or Euclid's algorithm
3. Decimal Expansion: Determine if p/q is terminating or repeating
4. Irrationality Proofs: Prove √n is irrational using contradiction
5. Divisibility Problems: Use FTA to prove/disprove ending digits, divisibility properties
6. Euclid's Algorithm: Apply repeated division to find HCF
7. Combination Problems: Find LCM using HCF × LCM = a × b formula
**MEMORY AIDS**
• FTA = Fundamental Theorem of Arithmetic (composite → unique prime factors)
• HCF = Highest Common Factor (use SMALLEST power of common primes)
• LCM = Least Common Multiple (use GREATEST power of ALL primes)
• Euclid = Long division method (divide until remainder = 0)
• Terminating Decimal ↔ Denominator = 2ᵐ × 5ⁿ only
Q1. A student claims that since every composite number can be expressed as a product of primes, and this expression is unique, then the number 60 can be written as 2² × 3 × 5 in only one way. Using the Fundamental Theorem of Arithmetic, this claim is —
Answer: A — The FTA guarantees uniqueness apart from the order of factors; uniqueness means 2² × 3 × 5 and 5 × 3 × 2² represent the same factorization, not different ones.
Q2. Consider the numbers 6n where n is a natural number. Based on the Fundamental Theorem of Arithmetic, which statement correctly explains why 6n can never end with the digit zero?
Answer: B — Ending in zero requires both prime factors 2 and 5 in the factorization; 6n = (2 × 3)ⁿ guarantees 2 is present but 5 never appears, so by FTA's uniqueness, no such number exists.
Q3. A mathematician states: 'If two different composite numbers have identical sets of prime factors (ignoring powers), they must be equal.' Based on the Fundamental Theorem of Arithmetic, this statement is —
Answer: B — Uniqueness of FTA refers to the exponents AND primes combined; identical sets of primes with different exponents yield different numbers (12 ≠ 18), so the statement is false.
Q4. Why does Euclid's division algorithm guarantee that we can always find the HCF of two positive integers a and b (where a > b) by repeatedly applying division?
Answer: A — The algorithm works by replacing the pair (a,b) with (b, remainder) repeatedly; since remainders strictly decrease and are bounded below by zero, the process terminates with HCF as the last non-zero remainder.
Q5. Assertion (A): For any natural number n, if 2ⁿ × 5ⁿ is the complete prime factorization of a number, then that number must end with the digit zero. Reason (R): A number ends in zero if and only if it is divisible by 10, which requires exactly one factor of 2 and one factor of 5. Choose the correct option:
Answer: A — A is true: 2ⁿ × 5ⁿ = (2×5)ⁿ × 2^(n-1) × 5^(n-1) or simplified, 10ⁿ × (other factors), always divisible by 10; R is true and correctly explains why A is true.
Q6. Assertion (A): The prime factorization of 32760 is unique. Reason (R): The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes in exactly one way, except for the order of factors. Choose the correct option:
Answer: A — A is true (32760 = 2³ × 3² × 5 × 7 × 13 uniquely); R is true and directly explains why A must be true—FTA guarantees uniqueness for every composite number.
Q7. Assertion (A): It is impossible to find a natural number n such that 4ⁿ ends with the digit 5. Reason (R): Numbers ending in 5 must have 5 as a prime factor, but 4ⁿ = 2^(2n) contains only 2 as its prime factor. Choose the correct option:
Answer: A — A is true by the contrapositive logic; R is true and correctly explains A: FTA's uniqueness means no prime factor 5 can appear in 4ⁿ, so it cannot end in 5.
Q8. A student tries to verify whether 3³ × 5² × 7 and 3² × 5³ × 7 represent the same prime factorization by checking if they have the same set of prime divisors {3, 5, 7}. The student's approach is —
Answer: B — The student confuses 'same prime factors' with 'same prime factorization'; FTA demands unique exponent combinations (3³ × 5² × 7 ≠ 3² × 5³ × 7), so checking only the set of primes is insufficient.
Q9. Why does the statement 'if a number is divisible by both 6 and 9, it must be divisible by 54' fail to be always true?
Answer: A — Using FTA: LCM(6, 9) = 2 × 3² = 18 (taking the highest power of each prime); the statement claims LCM = 54 = 2 × 3³, which wrongly adds exponents instead of taking maximum exponents.
Q10. A number N has the property that 'whenever it divides the product of two numbers, it must divide at least one of them.' Which of the following must be true about N's prime factorization?
Answer: A — This is the definition of a prime in terms of divisibility (Euclid's lemma); if N were composite (e.g., 6 = 2 × 3), then 6 | (2 × 3) but 6 ∤ 2 and 6 ∤ 3, violating the property—only primes satisfy it.
What is the Fundamental Theorem of Arithmetic?
Every composite number can be expressed uniquely as a product of primes, regardless of the order in which the primes are written.
How do you find HCF using prime factorization?
HCF is the product of the smallest powers of all common prime factors in the given numbers.
How do you find LCM using prime factorization?
LCM is the product of the greatest powers of all prime factors involved in the given numbers.
What is the relationship between HCF and LCM?
For any two positive integers a and b: HCF(a,b) × LCM(a,b) = a × b.
Prime factorize 32760.
32760 = 2³ × 3² × 5 × 7 × 13.
Can 4^n ever end with digit zero for any natural number n?
No, because 4^n = 2^(2n) contains only prime factor 2, but a number ending in zero must be divisible by both 2 and 5.
Find HCF and LCM of 6 and 20.
6 = 2 × 3, 20 = 2² × 5; HCF = 2, LCM = 60.
Is the prime factorization of a composite number unique?
Yes, the prime factorization is unique apart from the order in which the prime factors occur.
What does 'ascending order' mean in prime factorization?
Writing prime factors from smallest to largest: p₁ ≤ p₂ ≤ ... ≤ pₙ.
Why is 2 × 3 × 5 considered the same as 5 × 3 × 2?
Because the Fundamental Theorem states uniqueness is independent of order; both represent the same unique prime factorization.
State the Fundamental Theorem of Arithmetic and verify it for the number 84. [2 marks]
State that every composite number has unique prime factorization; then find prime factors of 84 by division: 84 = 2² × 3 × 7.
Find the HCF and LCM of 12 and 18 using the prime factorization method. Verify that HCF × LCM = product of the numbers. [3 marks]
Factor both: 12 = 2² × 3, 18 = 2 × 3²; HCF = 2 × 3 = 6, LCM = 2² × 3² = 36; verify: 6 × 36 = 216 = 12 × 18.
Prove that 4ⁿ can never end with the digit 0 for any natural number n. Explain using the Fundamental Theorem of Arithmetic. [5 marks]
Show 4ⁿ = (2²)ⁿ = 2²ⁿ contains only prime factor 2; for a number to end in 0, it must be divisible by both 2 and 5; since 5 never appears in 2²ⁿ by Fundamental Theorem's uniqueness, 4ⁿ cannot end in 0.
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