Prove that function is non-monotonic and is invertible

Question

We have $$f(x)=\begin{cases}\frac x2,&x\in\Bbb Q\\-\frac x2,&x\in\Bbb R\setminus\Bbb Q\end{cases}$$ Prove that function is non monotonic and is invertible.

I tried to take points from $\Bbb R\setminus\Bbb Q$ and $\Bbb Q$ and prove that is decreasing but couldn't. Can you help?

Answer 1

Note that monotonic entails both increasing and decreasing. Non-motonic means neither increasing nor decreasing.

$0<\sqrt2,f(0)=0>f(\sqrt2)=-\frac1{\sqrt2}$ so the function is not monotonically increasing.

$0<1,f(0)=0<f(1)=\frac12$ so the function is not monotonically decreasing.

---

The function is surjective over its range $\Bbb R$.

For injectivity: let $f(x_1)=f(x_2)$. Note that the function maps rational numbers to rational numbers and irrational numbers to irrational images. Thus both $f(x_1),f(x_2)$ are either rational (in which case $f(x_i)=x_i/2\implies x_1/2=x_2/2\iff x_1=x_2$) or irrational (in which case $f(x_i)=-x_i/2\implies-x_1/2=-x_2/2\iff x_1=x_2$).

Answer 2

Hint:

To prove non-monotonicity, pick $0<x_1<x_2<x_3$ such that $x_1,x_3\in\mathbb{Q}$ and $x_2\in\mathbb{R}\setminus\mathbb{Q}$.

To prove that it is invertible, show that this function is injective and surjective.

• for "injective", try $x,y$ both irrational, both rational, and exactly one rational.
• for "subjective", follow the definition.