Using exponents of prime factors in perfect squares, which idea is correct in the proof of √3?
Answer and explanation
Correct answer: In a perfect square, the exponent of 3 must be even
The correct idea is that every prime has an even exponent in the factorisation of a perfect square. In particular, the exponent of 3 must be even. Suppose √3 = a/b in lowest terms. Then a² = 3b². Since 3 divides a², it divides a, so let a = 3k. Substitution gives b² = 3k², which similarly forces 3 to divide b. This contradicts the assumption that a and b have no common factor. The exponent viewpoint expresses the same conflict: a square has an even exponent of 3, but multiplication by one additional factor 3 changes the relevant parity. Consequently option C is correct. The other choices make unjustified universal claims or give a false numerical equality.
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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 correct idea is that every prime has an even exponent in the factorisation of a perfect square. In particular, the exponent of 3 must be even. Suppose √3 = a/b in lowest terms. Then a² = 3b². Since 3 divides a², it divides a, so let a = 3k. Substitution gives b² = 3k², which similarly forces 3 to divide b. This contradicts the assumption that a and b have no common factor. The exponent viewpoint expresses the same conflict: a square has an even exponent of 3, but multiplication by one additional factor 3 changes the relevant parity. Consequently option C is correct. The other choices make unjustified universal claims or give a false numerical equality.
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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