Question

which number is irrational?
\\(\frac{\sqrt3{-27}}{3}\\)
\\(\frac{25}{9}\\)
\\(\frac{\pi}{\sqrt3{27}}\\)
\\(4.\overline{12}\\)

Explanation:

Step1: Analyze $\frac{\sqrt[3]{-27}}{3}$

First, calculate the cube root: $\sqrt[3]{-27} = -3$ (since $(-3)^3 = -27$). Then, $\frac{-3}{3} = -1$, which is an integer (rational).

Step2: Analyze $\frac{25}{9}$

This is a fraction of two integers, so it is a rational number (by definition of rational numbers: $\frac{p}{q}$, $q
eq0$, $p,q\in\mathbb{Z}$).

Step3: Analyze $\frac{\pi}{\sqrt[3]{27}}$

First, calculate the cube root: $\sqrt[3]{27} = 3$ (since $3^3 = 27$). So the expression becomes $\frac{\pi}{3}$. $\pi$ is an irrational number, and dividing an irrational number by a non - zero rational number ($3$ is rational) results in an irrational number.

Step4: Analyze $4.\overline{12}$

A repeating decimal can be expressed as a fraction of two integers. Let $x = 4.\overline{12}=4.121212\cdots$. Then $100x = 412.121212\cdots$. Subtract $x$ from $100x$: $100x - x=412.1212\cdots - 4.1212\cdots$, which gives $99x = 408$, so $x=\frac{408}{99}=\frac{136}{33}$, a rational number.

Answer:

$\frac{\pi}{\sqrt[3]{27}}$ (the option with $\frac{\pi}{\sqrt[3]{27}}$)