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A student assumes that \(\sqrt{3}=\frac{p}{q}\), where \(p\) and \(q\) are coprime integers. After obtaining \(p^2=3q^2\), the student says, “\(p\) is divisible by 3, but \(q\) need not be divisible by 3.” Which statement correctly identifies the error?

Answer and explanation

Correct answer: On putting \(p=3k\), \(q\) is also proved divisible by 3, contradicting the coprimality of \(p\) and \(q\).

Putting \(p=3k\) gives \(9k^2=3q^2\), so \(q^2=3k^2\) and hence \(q\) is also divisible by 3. This contradicts coprimality. In exams, track prime factors in squares carefully.

Related tags

Number SystemsIrrational NumbersSquare Root Of 3Proof By ContradictionCoprime IntegersProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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