A student says, “\(\sqrt{3}=\frac{6}{\sqrt{12}}\), so \(\sqrt{3}\) is rational.” What is the error in the argument?

Correct answer: To establish rationality, both the numerator and denominator must be integers; here \(\sqrt{12}\) is not an integer.

The equality is correct because \(\sqrt{12}=2\sqrt{3}\) and \(\frac{6}{2\sqrt{3}}=\sqrt{3}\). However, a rational number must be expressed as \(a/b\) with integers \(a,b\); the denominator here is irrational. Check the denominator in such claims.