Let [f, g: N-1 N] be functions defined by [f(a)=], where [] is the maximum of the powers o…

Question

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

Accepted Answer

$$f,g:N - \{ 1\}  \to N$$ defined as

$$f(a) = \alpha $$, where $$\alpha$$ is the maximum power of those primes p such that p$$\alpha$$ divides a.

$g(a) = a + 1$,

Now,

$$\begin{pmatrix} {f(2) = 1,} & {g(2) = 3} &  \Rightarrow  & {(f + g)\,(2) = 4}  \
   {f(3) = 1,} & {g(3) = 4} &  \Rightarrow  & {(f + g)\,(3) = 5}  \
   {f(4) = 2,} & {g(4) = 5} &  \Rightarrow  & {(f + g)\,(4) = 7}  \
   {f(5) = 1,} & {g(5) = 6} &  \Rightarrow  & {(f + g)\,(5) = 7}  \

\end{pmatrix} $$

$$\because$$ $$(f + g)\,(5) = (f + g)\,(4)$$

$$\therefore$$ $f + g$ is not one-one

Now, $$\because$$ $${f_{\min }} = 1,\,{g_{\min }} = 3$$

So, there does not exist any $$x \in N - \{ 1\} $$ such that $(f + g)(x) = 1,2,3$

$$\therefore$$ $f + g$ is not onto

Solutions · 1