In the proof of √3, after taking m = 3k, which step from m² = 3n² proves n divisible by 3?
Answer and explanation
Correct answer: 9k² = 3n² ⇒ n² = 3k²
Starting from m² = 3n² and using the earlier conclusion that 3 divides m, write m = 3k. Substitution gives (3k)² = 3n², so 9k² = 3n². Dividing both sides by 3 yields 3k² = n², or n² = 3k². Thus n² is divisible by 3. Since 3 is prime, Euclid’s lemma implies that n itself is divisible by 3; write n = 3l if continuing the proof. This creates a common factor 3 in m and n, contradicting their assumed coprimality. Option D contains the necessary algebraic step. The other choices either assert an unsupported equality or give an irrelevant value.
Frequently asked questions
What is the correct answer to this question?
9k² = 3n² ⇒ n² = 3k²
Why is this the correct answer?
Starting from m² = 3n² and using the earlier conclusion that 3 divides m, write m = 3k. Substitution gives (3k)² = 3n², so 9k² = 3n². Dividing both sides by 3 yields 3k² = n², or n² = 3k². Thus n² is divisible by 3. Since 3 is prime, Euclid’s lemma implies that n itself is divisible by 3; write n = 3l if continuing the proof. This creates a common factor 3 in m and n, contradicting their assumed coprimality. Option D contains the necessary algebraic step. The other choices either assert an unsupported equality or give an irrelevant value.
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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