The number $\sqrt{3}$ is

### Concept & Logic The square root of any non-perfect square is an irrational number. The defining characteristic of all irrational numbers is that their decimal form expands infinitely without forming a repeating sequence (an **infinite non-recurring decimal**). ### Step-by-Step Solution **Deduction:** 1. The given number is $\sqrt{3}$. 2. Check if the integer $3$ is a perfect square. It is not. Therefore, by definition, $\sqrt{3}$ is an irrational number. 3. Next, map the properties of rational vs. irrational numbers to decimal forms. Rational numbers have decimals that either terminate (finite) or repeat (infinite recurring). 4. Irrational numbers strictly have decimals that never terminate and never repeat in a block pattern. 5. Therefore, the decimal expansion of $\sqrt{3}$ must inherently be infinite and non-recurring. ### Exam Strategy & Shortcut Recognize synonymous terminology: "Irrational" is perfectly synonymous with "infinite non-recurring decimal." Once you know $\sqrt{3}$ is irrational, the specific decimal descriptor is locked in without calculation. ### Common Pitfall Selecting "equal to $1.732$" (Option C). While $1.732$ is routinely used for practical calculations in geometry, it is a rounded truncation. The actual value continues forever without repeating, meaning it is not strictly *equal* to $1.732$. ### Final Answer Therefore, the correct answer is **an infinite non-recurring decimal**.