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While proving the irrationality of \(\sqrt{3}\) by contradiction, if \(\sqrt{3}=p/q\) where \(p\) and \(q\) are coprime, which conclusion is essential to the proof?

Answer and explanation

Correct answer: If \(p^2\) is divisible by 3, then \(p\) is divisible by 3

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisibility property of squares.

Related tags

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

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