In the proof of √3, after which sequence is n proved divisible by 3?
Answer and explanation
Correct answer: m² = 3n² ⇒ m is divisible by 3 ⇒ m = 3k ⇒ n² = 3k²
From m² = 3n², the right side is divisible by 3, so m² and therefore m are divisible by 3. Write m = 3k. Substitution gives 9k² = 3n²; after division by 3, n² = 3k². Hence n² is divisible by 3, and because 3 is prime, n is divisible by 3. This complete chain is given only in option A. It is important that the conclusion about n comes after introducing m = 3k and simplifying the substituted equation; it does not follow merely from the original equation without those steps. Once both m and n are divisible by 3, their assumed coprimality is contradicted.
Frequently asked questions
What is the correct answer to this question?
m² = 3n² ⇒ m is divisible by 3 ⇒ m = 3k ⇒ n² = 3k²
Why is this the correct answer?
From m² = 3n², the right side is divisible by 3, so m² and therefore m are divisible by 3. Write m = 3k. Substitution gives 9k² = 3n²; after division by 3, n² = 3k². Hence n² is divisible by 3, and because 3 is prime, n is divisible by 3. This complete chain is given only in option A. It is important that the conclusion about n comes after introducing m = 3k and simplifying the substituted equation; it does not follow merely from the original equation without those steps. Once both m and n are divisible by 3, their assumed coprimality is contradicted.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Proof of irrationality of square root 2 and square root 3.
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