Which statement about exponents of prime factors in perfect squares is useful in the proof of √3?
Answer and explanation
Correct answer: In a perfect square, the exponent of 3 must be even
A perfect square has an even exponent for every prime in its prime factorisation. This applies to the prime 3 as well. In the usual contradiction proof, suppose √3 = p/q in lowest terms. Squaring gives p² = 3q². Because the right side is divisible by 3, p is divisible by 3; write p = 3k. Substitution gives q² = 3k², so q is also divisible by 3. The resulting common factor contradicts the assumption that p/q was in lowest terms. Equivalently, the square p² must have an even exponent of 3, while the factor 3q² introduces an odd exponent. Therefore option C is the precise useful statement; the remaining options are false.
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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?
A perfect square has an even exponent for every prime in its prime factorisation. This applies to the prime 3 as well. In the usual contradiction proof, suppose √3 = p/q in lowest terms. Squaring gives p² = 3q². Because the right side is divisible by 3, p is divisible by 3; write p = 3k. Substitution gives q² = 3k², so q is also divisible by 3. The resulting common factor contradicts the assumption that p/q was in lowest terms. Equivalently, the square p² must have an even exponent of 3, while the factor 3q² introduces an odd exponent. Therefore option C is the precise useful statement; the remaining options are false.
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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