# 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?

1. $$x$$ is rational
2. $$x$$ is irrational
3. $$x = 1$$
4. $$x = \frac 2 3$$
5. $$x = \frac 3 2$$

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?

• Do you mean $\dfrac 43$ is irrational, or the solution to $\dfrac {4x}{6x^2} = \dfrac 43$, $x = \dfrac 12$, is irrational? Oct 1 '20 at 14:28
• MathJax, please Oct 1 '20 at 14:28

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.

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 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.