Let $f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$ be functions defined by $f(a)=\alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$, and $g(a)=a+1$, for all $a \in \mathbb{N}-\{1\}$. Then, the function $f+g$ is

<p>$f,g:N - \{ 1\} \to N$ defined as</p> <p>$f(a) = \alpha$, where $\alpha$ is the maximum power of those primes p such that p^$\alpha$ divides a.</p> <p>$g(a) = a + 1$,</p> <p>Now,</p> <p>$$\matrix{ {f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr {f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \cr {f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \cr {f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \cr } $$</p> <p>$\because$ $(f + g)\,(5) = (f + g)\,(4)$</p> <p>$\therefore$ $f + g$ is not one-one</p> <p>Now, $\because$ ${f_{\min }} = 1,\,{g_{\min }} = 3$</p> <p>So, there does not exist any $x \in N - \{ 1\}$ such that $(f + g)(x) = 1,2,3$</p> <p>$\therefore$ $f + g$ is not onto</p>