1. The problem is to evaluate the square root of 3, written as $\sqrt{3}$.\n\n2. The square root function $\sqrt{x}$ gives the number which, when multiplied by itself, equals $x$.\n\n3. For $\sqrt{3}$, we want a number $y$ such that $y^2 = 3$.\n\n4. Since 3 is not a perfect square, $\sqrt{3}$ is an irrational number approximately equal to 1.732.\n\n5. Therefore, $\sqrt{3} \approx 1.732$.