Given a fraction with $ x $ is a real number, judge the scope of $ x $. From an ACT Math test:

Suppose that $ x $ is a real number and $ \frac { 4 x } { 6 x ^ 2 } $ is a rational number. Which of the following statements about $ x $ must be true?

*

*$ x $ is rational

*$ x $ is irrational

*$ x = 1 $

*$ x = \frac 2 3 $

*$ x = \frac 3 2 $

The answer says it must be a rational number, but how about an irrational number, say, $ \frac 4 3 $, which can also satisfy $ \frac { 4 x } { 6 x ^ 2 } $ a rational number?
 A: A rational number is any number that can be expressed as a ratio of two integers $\frac{p}{q}$ with $q\neq 0$. That is $\mathbb Q:= \{\frac{p}{q}|p\in\mathbb Z,q\in\mathbb Z, q\neq0\}.$ Moreover $\frac{4}{3}$ is a rational number.
Since $x$ is a positive real number we have $$\frac{4x}{6x^2}=\frac{4}{6x}=\frac{2}{3x}$$
which we are told is a rational number. Then $\frac{2}{3x}$ is of the form $\frac{p}{q}$, where $p=2$ and $q=3x$ (and we know $x\neq 0$).
So statement $A$ must be true.
A: Simply $\dfrac{4x}{6x^2}=\dfrac{4}{6x}$ can be irrational, but doesn't have to -- for example let's take $x=\dfrac{10}{3}.$
Then we have
$\dfrac{4}{6x}=\dfrac{2}{3x}=\dfrac{2}{3\cdot\frac{10}{3}}=\dfrac{2}{10}.$
It isn't irrational number definitely.
By the way the question says which statement must be true.
A: A number is rational if it can be written as a fraction with integer numerator and denominator.  $\frac43$ is not irrational, it's an improper fraction.  If $\frac{4x}{6x^2}=\frac{2}{3x}=\frac ab$, then $x=\frac{2b}{3a}$.  If $a$ and $b$ are integers, so are $2b$ and $3a$.  Therefore, $x$ is rational.
