Solutions

We are given

$$g(x)=x^2$$

and

$$h(x)= \begin{cases} 0, & x \text{ rational},\\ 1, & x \text{ irrational}. \end{cases}$$

i) Let us judiciously consider the following partitions of $y$:

• For $y<0$, $y<0\le h(y)$, so the inequality does not hold.
• For $y=0$, $h(0)=0=y$, so it holds.
• For rational numbers in $0<y\le 1$, $h(y)=0<y$, so it holds.
• For irrational numbers in $0<y<1$, $y<1=h(y)$, so it does not hold.
• For $y>1$, $h(y)\le 1<y$, so it holds.

ii) For all rational numbers $y$, it holds because

$$h(y)=0\le y^2=g(y).$$

For irrational numbers in $-1<y<1$,

$$g(y)=y^2<1=h(y),$$

so it does not hold.

For irrational numbers $y$ such that $|y|>1$,

$$h(y)=1<|y|<y^2=g(y),$$

so it holds.

iii) We have

$$g(h(z))-h(z)=h(z)^2-h(z).$$

Notice that $h(z)$ is either $0$ or $1$. Thus, $h(z)=h(z)^2$. Hence,

$$g(h(z))-h(z)=h(z)-h(z)=0.$$

iv) For $|w|>1$, we have

$$g(w)=w^2>w,$$

so the inequality does not hold.

• For $0\le w\le 1$, $g(w)=w^2\le w$, so it holds.
• For $w=-1$, $g(-1)=1$, so it holds also.
• For $-1<w<0$,

$$g(w)=w^2>0>w,$$

so the inequality does not hold.

v) Notice that

$$\begin{align*} g(g(\varepsilon))=g(\varepsilon) &\implies g(\varepsilon^2)=\varepsilon^2 \\ &\implies \varepsilon^4=\varepsilon^2 \\ &\implies \varepsilon^2(\varepsilon^2-1)=0 \\ &\implies \varepsilon^2(\varepsilon+1)(\varepsilon-1)=0. \end{align*}$$

So the equality holds for $\varepsilon=0$, $\varepsilon=1$, and $\varepsilon=-1$.