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

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