Proving Periodic Functions

A function denoted by:

$$f(x)=\{\begin{array}{ll}1& ifx\in Q\\ 0& ifx\notin Q\end{array}$$

$f(x)$ is periodic

Proof

WTS: $\exists p>0$ such that $f(x)=f(x+p)$ Choose $p=1$ Case 1: $x\in Q$ $f(x)=1$ $f(x+p)=f(x+1)$ Note that $x+1\in Q$ (by properties of Real Number) Then, $f(x+1)=1$ SInce both the results are 1, this means that $f(x)=f(x+p)$ for $x\in Q$ Case 2: $x\notin Q$ $f(x)=0$ $f(x+p)=f(x+1)$ Note that $x+1\notin Q$ (by properties of Irrational Number. See proof Rational + Irrational = Irrational) Then, $f(x+1)=0$. SInce both the results are 0, this means that $f(x)=f(x+p)$ for $x\notin Q$. As both cases are correct, we can conclude that the function is periodic for all inputs set by the function. $\square$