Which option shows the correct logical chain in the proof of (√3)?
Answer and explanation
Correct answer: (p²=3q²), (p) divisible by (3), (p=3k), (q) divisible by (3)
For the contradiction proof of √3, assume √3 = p/q in lowest terms, with q not equal to zero. Squaring gives p² = 3q². Therefore p² is divisible by the prime 3, and the prime-divisibility property implies that p is divisible by 3. Write p = 3k. Substituting and simplifying gives 3k² = q², so q², and hence q, is also divisible by 3. Both p and q then have a common factor 3, contradicting lowest terms. Option C states this correct chain. Option A uses the factor 2 and belongs to √2; the remaining options do not establish the contradiction.
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What is the correct answer to this question?
(p²=3q²), (p) divisible by (3), (p=3k), (q) divisible by (3)
Why is this the correct answer?
For the contradiction proof of √3, assume √3 = p/q in lowest terms, with q not equal to zero. Squaring gives p² = 3q². Therefore p² is divisible by the prime 3, and the prime-divisibility property implies that p is divisible by 3. Write p = 3k. Substituting and simplifying gives 3k² = q², so q², and hence q, is also divisible by 3. Both p and q then have a common factor 3, contradicting lowest terms. Option C states this correct chain. Option A uses the factor 2 and belongs to √2; the remaining options do not establish the contradiction.
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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