Subject GRE question - set of points of discontinuity I was just working on a Math Subject GRE practice test, and I got the following problem wrong:
Let $f$ be the function defined on the real line by
$\displaystyle f(x) = \begin{cases} \displaystyle \frac{x}{2} &\mbox{if } x \text{ is rational}, \\ 
\displaystyle \frac{x}{3} & \mbox{if } x \text{ is irrational}. \end{cases}$
If $D$ is the set of points of discontinuity of $f$, then $D$ is the 
$(A)\, \text{Empty set} \\
(B)\, \text{set of rational numbers}\\
(C)\, \text{set of irrational numbers}\\
(D)\, \text{set of nonzero real numbers}\\
(E)\, \text{set of real numbers}\\ $
I chose $(E)$, the set of real numbers, since this set is clearly discontinuous at both every rational number and every irrational number.  However, the answer key told me the answer is $(D)$, the set of nonzero real numbers. 
I find this very confusing because, as far as I know, $0$ is a rational number, since it can be written in the form $\displaystyle \frac{0}{q}$, for any integer $q$. Can $0$ also be irrational?  I've been looking online to figure that out, and everywhere I've looked has told me that $0$ is only rational.  Are they all wrong, or is there some other reason why $0$ is not a point of discontinuity?
I'm very confused :(
 A: Your initial reasoning is mostly correct; however, you are missing a subtlety here:  $f$ is continuous at $0$ because, loosely speaking, the behavior of $x/2$ and $x/3$ in a sufficiently small neighborhood of $0$ are such that they tend to the same limit.  If the function had been $$f(x) = \begin{cases} x/2 & x \in \mathbb Q, \\ x/2 + 1 & x \not\in \mathbb Q, \end{cases}$$ then this function is not continuous anywhere on the real line.
A: Let $\epsilon > 0$ be given. To prove continuity at $a$, we must find $\delta(\epsilon) > 0$ such that 
$$|x-a| < \delta(\epsilon) \implies |f(x) - f(a)| < \epsilon$$
for $x \in V_{\delta}(a)$, where $V_{\delta}(a) = (x-a, x+a)$. 
To prove continuity at $a = 0$, notice for nonzero $k$,
$$|x-0| = |x| = \left|\dfrac{x}{k}\right|(k)=k\left|\dfrac{x}{k} - \dfrac{0}{2}\right|\text{.}$$
(I'm going by the assumption that $0$ is rational.) 
Assume $x$ were rational. Then take $k = 2$ and 
$$2\left|\dfrac{x}{2}-\dfrac{0}{2}\right| = 2\left|f(x)-f(0)\right| < \delta(\epsilon) = \dfrac{\epsilon}{2}$$
so that $|f(x) - f(0)| < \epsilon$. If $x$ were irrational, take $k = 3$ and notice similarly that 
$$3\left|\dfrac{x}{3}-\dfrac{0}{2}\right| = 3\left|f(x)-f(0)\right| < \delta(\epsilon) = \dfrac{\epsilon}{3}$$
so that $|f(x) - f(0)| < \epsilon$ and hence, $f$ is continuous at $0$. (I didn't touch on this too much, but the most logical $\delta(\epsilon) = \min\left(\dfrac{\epsilon}{2}, \dfrac{\epsilon}{3}\right)$.)
A: Let $a \in \mathbb R$ s.t. rational sequence $b_n \to a$ and irrational sequence $c_n \to a$
$$f \ \text{is continuous at} \  x=a$$
$$\iff$$
$$\lim f(b_n) = \lim f(c_n)$$
$$\iff$$
$$\lim b_n/2 = \lim c_n/3$$
$$\iff$$
$$a/2=a/3$$
$$\iff$$
$$a=0$$
