Is zero irrational? I think of the number zero as a whole number.
It can certainly be a ratio = $\frac{0}{x}, x \neq 0.$
Therefore it is rational. 
But any ratio equaling zero involves zero, or
is irrational, e.g.$\frac{x}{\infty}, x \neq 0$ is not a ratio
of integers.
Can a rational number that is rational only when the number
itself is involved still be rational?
I realize that it "should" be rational but it doesn't seem to
fit into the same category with other rational numbers.
(I'm glad mse has a soft question category for questions like
this...).
 A: One perspective on this is to regard $0$ as a symbol that actually refers to several different mathematical objects:


*

*$0$-the-natural-number, usually the first natural number defined. Let's write $0_n$ for this specifically.

*$0$-the-integer; an integer is a natural number with a sign, and so we can write $0_i = +0_n$. In fact it's also $-0_n$, but that doesn't really matter.

*$0$-the-rational-number: a rational number is an integer divided by another, so $0_r = 0_i/1_i$


From this point of view, there is nothing circular about $0 = 0/1$, because we're actually just using the same symbol to refer to two very similar objects, one an integer and one a rational number.
A: As you stated $0$ is whole number, and we know that whole numbers are subset of rational numbers. So what does that says?
A: That is a good observation, the question is not really about 0 being rational or not but that wether 0 is the only rational with infinitely many representation that can not be simplified.
Use the alternative definition of rationals :Rational can be represented by finite or repeating decimal representation. 
Using that definition there is no ambiguity trying to find two numbers that ratio will be 0. But according to that definition there are infinite numbers that their ratio is 0, where we only needed 1. So it is a rational with the property that can be expressed with infinity many different ratios (all other rational have only one ratio in its simplest terms ).
A: A real number $x$ is rational if and only if it can be written as a fraction $a/b$ with $a$ and $b$ integers and $b\neq 0$. In particular, integers are rational, and $0$ is an integer, so it is rational.
A: The division of two irrational quantities can obviously have rational results. For instance, both π and 2π are irrational; but the result of their respective division is either 2 or $\tfrac12$ , both of which are of course rational. And not just division, but other operations as well. Let's try a simple addition, for instance: both π + 1 and 1 - π are irrational; yet their sum, 2, is not...
In case you weren't aware, there is a reason why there is no symbol specifically for irrationals, as in the case of natural numbers $(\mathbb{N})$, integers $(\mathbb{Z})$, rationals $(\mathbb{Q})$, reals $(\mathbb{R})$, complex $(\mathbb{C})$, algebraics $(\mathbb{A})$, etc. This has to do with the fact that -unlike these other numerical sets- the set of irrationals is not self-contained when mixed with even the most basic arithmetic operations, like addition or multiplication, for instance. E.g., despite the fact that both $\sqrt2$ and $\sqrt2$ are irrational, their product, 2, isn't. The same observation holds for $\sqrt3$ and $^1/_\sqrt3$ , etc.
