If $f\left(x\right)=\left\{\begin{array}{l}x,\text{ when }x\text{ is rational }\\ 1-x,\text{ when }x\text{ is irrational }\end{array}\right.\text{, }$then

Let a be any real number
$\begin{array}{c}\underset{x\to a}{lim} f\left(x\right)=\underset{x\to a}{lim} x=a\\ \text{ (when }x\to a\text{ through rational values) }\\ \underset{x\to a}{lim} f\left(x\right)=\underset{x\to a}{lim} (1-x)=1-a\\ \text{ (when }x\to a\text{ through irrational values) }\\ \underset{x\to a}{lim} f\left(x\right)\text{ will exist only when }a=1-a\end{array}$
i.e., when a = 1/2
Thus if x ≠1/2 , then $\underset{x\to a}{lim} f\left(x\right)$
f(x) will not exist and hence 
f(x) will be discontinuous at x = a where a ≠1/2