Question

Let $f,g:N\to N$ such that $f(n+1)=f(n)+f(1)$ $\forall n\in N$ and $g$ be any arbitrary function. Which of the following statements is $NOT$ true?

• A. If fog is one-one, then $g$ is one-one
• B. If $f$ is onto, then $f(n)=n\forall n\in N$
• C. $f$ is one-one
• D. If $g$ is onto, then fog is one-one

Answer 1

Answer: (D)

$f(n+1)-f(n)=f(1)$
$\Rightarrow f(n)=nf(1)$
$\Rightarrow f$ is one-one
Now, Let $f\left(g\left(x_2\right)\right)=f\left(g\left(x_1\right)\right)$
$\Rightarrow g\left(x_2\right)=g\left(x_1\right)$ (as $f$ is one-one )
$\Rightarrow x_1=x_2$ ( as fog is one-one )
$\Rightarrow g$ is one-one
Now, $f(g(n))=g(n)f(1)$
may be many-one if
$g(n)$ is many-one