The number of bijective functions f: Z Z such that f(x+y)=f(x)+f(y) x, y Z , is

Step-by-step solution

Let x and y be any two elements in the domain (Z) , such that f(x+y)=f(x)+f(y) (i) Differentiating above expression w.r.t ' y ', keeping x constant, we get f^ (x+y)=f^ (y) Let y=0 f^ (x+0)=f^ (0) and Assume f^ (0)=K f^ (x)=K Integrating on both sides, we get f(x)=K x+C C integration constant (ii) Now putting x=y=0 in Eq. (i), we get f(0+0)=f(0)+f(0) f(0)=2 f(0) or f(0)=0 Let x=0 , from Eq. (ii) f(0)=K 0+C or 0=0+C or C=0 f(x)=k x Case (i) If k>0 , then f(x) is strictly increasing. Case (ii) If k < 0 , then f(x) is