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A student says, “For a non-zero rational number \(a\), \(a\sqrt{3}\) can be rational.” Which argument correctly explains the error in this statement?

Answer and explanation

Correct answer: If \(a\sqrt{3}\) were rational, dividing by \(a\) would make \(\sqrt{3}\) rational, which is a contradiction.

Assume \(a\sqrt{3}\) is rational, where \(a\ne0\) is rational. Then \(\frac{a\sqrt{3}}{a}=\sqrt{3}\) would be rational, contradicting its irrationality. Exam tip: always check that division is by a non-zero number.

Related tags

Number SystemsIrrational NumbersSquare Root 3Proof By ContradictionRational NumbersProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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