## Calculus (3rd Edition)

We are given the function: $f(x)=\begin{cases} 1,\text{ for }x \text{ rational}\\ -1,\text{ for }x \text{ irrational} \end{cases}$ Determine $[f(x)]^2$: $[f(x)]^2=\begin{cases} 1^2,\text{ for }x \text{ rational}\\ (-1)^2,\text{ for }x \text{ irrational} \end{cases}$ $[f(x)]^2=\begin{cases} 1,\text{ for }x \text{ rational}\\ 1,\text{ for }x \text{ irrational} \end{cases}$ Let $x_0$ be a rational number. There are infinitely many irrational numbers to the left and to the right of $x_0$, so $f^(x)$ is continuous in $x_0$. The function takes the same value for rational and irrational numbers. Therefore $f^2(x)$ is continuous on $\mathbb{R}$.