What is the idea of prime factors of a perfect square in the proof of √3?
Answer and explanation
Correct answer: In a perfect square, the exponent of 3 must be even
The governing principle is that the exponent of every prime in the factorization of a perfect square is even. Assume, for contradiction, that √3 = m/n in lowest terms. Squaring gives m² = 3n², so 3 divides m. Put m = 3r; then 9r² = 3n², which simplifies to n² = 3r². Hence 3 also divides n. The numerator and denominator are therefore both divisible by 3, contradicting the fact that the fraction was in lowest terms. Option B expresses the exact prime-exponent idea behind this contradiction. Option A is too broad, option C confuses a root with the number under the radical, and option D is unrelated to rational representation.
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What is the correct answer to this question?
In a perfect square, the exponent of 3 must be even
Why is this the correct answer?
The governing principle is that the exponent of every prime in the factorization of a perfect square is even. Assume, for contradiction, that √3 = m/n in lowest terms. Squaring gives m² = 3n², so 3 divides m. Put m = 3r; then 9r² = 3n², which simplifies to n² = 3r². Hence 3 also divides n. The numerator and denominator are therefore both divisible by 3, contradicting the fact that the fraction was in lowest terms. Option B expresses the exact prime-exponent idea behind this contradiction. Option A is too broad, option C confuses a root with the number under the radical, and option D is unrelated to rational representation.
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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