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Reema says, “\(\sqrt{3}\approx1.732\); therefore, \(\sqrt{3}\) is rational because 1.732 is rational.” What is the correct evaluation of Reema’s argument?

Answer and explanation

Correct answer: 1.732 is an approximation; its rationality does not prove that \(\sqrt{3}\) is rational

1.732 is a terminating rational approximation, not \(\sqrt{3}\) itself. Check: \(1.732^2=2.999824\), not 3. Exam tip: distinguish an approximate value from an exact value.

Related tags

Irrational NumbersSquare Root 3Decimal ApproximationNumber SystemsMathematical ReasoningProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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