1. The problem asks whether $\sqrt{9}$ is an irrational number. 2. Recall that an irrational number is a number that cannot be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. 3. To determine if $\sqrt{9}$ is irrational, first calculate $\sqrt{9}$. 4. Since $9 = 3^2$, $\sqrt{9} = 3$. 5. The number $3$ is an integer and can be expressed as $\frac{3}{1}$, which is a ratio of two integers. 6. Therefore, $\sqrt{9}$ is a rational number, not an irrational number. Final answer: $\sqrt{9}$ is not an irrational number; it is a rational number equal to $3$.