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Which condition guarantees that the square root of a natural number \(n\) is rational?

Answer and explanation

Correct answer: \(n\) is a perfect square

If \(n=k^2\) for an integer \(k\), then \(\sqrt{n}=k\), which is rational. Being even or composite alone is not enough; for example, \(\sqrt{6}\) is irrational. Exam tip: in prime factorisation, all exponents must be even for a perfect square.

Related tags

Number SystemsIrrational NumbersSquare RootsPerfect SquaresProof Of IrrationalityProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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