Let f, g: N → N such that f(n+1)=f(n)+f(1)
Question:

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

  1. If fog is one-one, then $g$ is one-one

  2. If $\mathrm{f}$ is onto, then $\mathrm{f}(\mathrm{n})=\mathrm{n} \forall \mathrm{n} \in \mathrm{N}$

  3. $\mathrm{f}$ is one-one

  4. If $g$ is onto, then fog is one-one


Correct Option: , 4

Solution:

$f(n+1)-f(n)=f(1)$

$\Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{nf}(1)$

$\Rightarrow \mathrm{f}$ is one-one

Now, Let $\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{2}\right)\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{1}\right)\right)$

$\Rightarrow \mathrm{g}\left(\mathrm{x}_{2}\right)=\mathrm{g}\left(\mathrm{x}_{1}\right)$ (as $\mathrm{f}$ is one-one)

$\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$ (as fog is one-one)

$\Rightarrow \mathrm{g}$ is one-one

Now, $\mathrm{f}(\mathrm{g}(\mathrm{n}))=\mathrm{g}(\mathrm{n}) \mathrm{f}(\mathrm{l})$

may be many-one if

$\mathrm{g}(\mathrm{n})$ is many-one

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