Echoing what other people have already said: no $i= \sqrt{-1}$ is not a rational number.
You have $ \begin{align} &\mathbb{C} \;\text{ the complex numbers}\\ &\cup \\ &\mathbb{R} \;\text{ the real numbers}\\ &\cup \\ &\mathbb{Q} \;\text{ the rational numbers}\\ &\cup \\ &\mathbb{Z} \;\text{ the integers}\\ &\cup \\ &\mathbb{N} \;\text{ the natural numbers} \end{align} $ Here the $\cup$ denotes that the lower is contained in the upper. So for example all real numbers are complex numbers. And: an integer is a real number. Note that for example not all complex numbers are real numbers. Not all complex numbers are rational numbers. Not all integers are natural numbers.
So the question now is whether $i = \sqrt{-1}$ (which is a complex number) is a rational number. And here we first note that the rational numbers are those numbers that can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers (so belongs to $\mathbb{Z}$) ($b\neq 0$). So is $i = \frac{a}{b}$ for any integers $a$, and $b$? As provided in Austin's fine answer, one can show that indeed the answer is no.
To have something to think about, maybe you can answer the question: Is $\sqrt{2}$ a rational number? (You can find the answer here on M.SE).