f,g:N−{1}→N defined as f(a)=α , where α is the maximum power of those primes p such that pα divides a. g(a)=a+1 Now, f(2)=1,‌‌g(2)=3‌‌⇒‌‌(f+g)(2)=4 f(3)=1,‌‌g(3)=4‌‌⇒‌‌(f+g)(3)=5 f(4)=2,‌‌g(4)=5⇒(f+g)(4)=7 f(5)=1,‌‌g(5)=6⇒(f+g)(5)=7 ∵(f+g)(5)=(f+g)(4) ∴f+g is not one-one Now, ∵fmin=1,gmin=3 So, there does not exist any x∈N−{1} such that (f+g)(x)=1,2,3 ∴f+g is not onto