$\text{ If the functions }f\left(x\right)\text{ and }g\left(x\right)\text{ are defined on }R\to R\text{ such that }f\left(x\right)=\left\{\begin{array}{l}0,x\in \text{ rational }\\ x,x\in \text{ irrational }\end{array}\right.$

$\text{ and }g\left(x\right)=\left\{\begin{array}{c}0,x\in \text{ irrational }\\ x,x\in \text{ rational }\end{array}\right.\text{ then }(f-g)\left(x\right)\text{ is }$

We are given that
$\begin{array}{c}f:R\to R,f\left(x\right)=\left\{\begin{array}{c}0,x\in \text{ rational }\\ x,x\in \text{ irrational }\end{array}\right.\\ g:R\to R,g\left(x\right)=\left\{\begin{array}{c}0,x\in \text{ irrational }\\ x,x\in \text{ rational }\end{array}\right.\\ \therefore (f-g):R\to R\text{ such that }\\ (f-g)\left(x\right)=\left\{\begin{array}{l}-x,\text{ if }x\in \text{ rational }\\ x,\text{ if }x\in \text{ irrational }\end{array}\right.\end{array}$

$\text{Since }(f-g):R\to R\text{ for any }x,\text{ there is only one value of ( }f\left(x\right)-g\left(x\right))\text{ whether }x\text{ is rational or irrational. Moreover, }$

$\text{ as }x\in R,f\left(x\right)-g\left(x\right)\text{ also belongs to }R\text{ . Therefore, }(f-g)\text{ is one-one and onto.}$