Is the square root of -1 rational? This is not a deep question, but if there is a definite answer then here is the place where I will find it.
Is it justified to say that $i =\sqrt{-1}$  is rational?
The origin of this question lies in a regular discussion I have over this t-shirt of mine:

While obvious $\pi$'s comment is completely legit, $\sqrt{-1}$'s might be hypocritical, if the rationality of $\sqrt{-1}$ is questionable.
 A: $\sqrt{-1} = i$ is an imaginary number, not lying anywhere on the real number line.  Therefore as others have said, it is neither rational nor irrational in the usual senses of those words.  
To comment on the specific grammar of the shirt: $\pi$'s imperative is 'Get real' which I would say has the connotation of 'join [the speaker's] group'.  Clearly this makes sense, as $\pi$ is a real number.  $i$, on the other hand, says only 'be rational' which I would say lacks the connotation of joining the speakers group; instead, $i$ is only imploring $\pi$ to join the rationals, with no implication of $i$'s membership.   
A: Rather $\sqrt{-1}$ is the algebraic number, but not rational.
A: It is a Gaussian rational number, but it is not rational in the conventional sense of the word because rational numbers are real.
A: 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).
A: The number $\sqrt{-1}$ is not real. Since the rationals are just a particular type of real number, it cannot be rational, either.
Another way to look at it: Were there integers $a$ and $b$ such that $\sqrt{-1} = \frac{a}{b}$  then
$$
-1 = \frac{a^2}{b^2},
$$
and so
$$
a^2 = -b^2.
$$
Since $b^2$ is certainly positive, that means $a^2$ is certainly negative, which is impossible.
