If the functions f(x) and g(x) are defined on R→R such  that f(x)=0,x∈ rational x,x∈ irrational  and g(x)=0,x∈ irrational x,x∈ rational  then (f−g)(x) is

We are given thatf:R→R,f(x)=0,x∈ rational x,x∈ irrational g:R→R,g(x)=0,x∈ irrational x,x∈ rational ∴ (f−g):R→R such that (f−g)(x)=−x, if x∈ rational x, if x∈ irrational  Since (f−g):R→R for any x, there is only one value of  ( f(x)−g(x)) whether x is rational or irrational. Moreover,  as x∈R,f(x)−g(x) also belongs to R . Therefore, (f−g) is one-one and onto.