let f:z-z:f(n)=3n and let g:z-z defined by
g(n)=n/3 if n multiple of 3
0 if n is not a multiple of 3
show that gof=Iz and fog not =Iz
Given, f : Z → Z
f(n) = 3n
g : Z → Z
f : Z → Z and g : Z → Z
∴ gof : Z → Z and fog : Z → Z
For any n∈Z, we have
For any n∈Z, we have
fog (n) = f(g(n))
So, fog (n) ≠ n
Thus, fog (n) ≠ I Z