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Q. 6 Give examples of two functions $f : N \rightarrow Z$ and $g : Z \rightarrow Z$ such that $gof$ is
injective but g is not injective.

(Hint : Consider $f (x) = x$ and $g (x) = | x |$ ).

Answers (1)

best_answer

$f : N \rightarrow Z$

$g : Z \rightarrow Z$

$f (x) = x$

$g (x) = | x |$

One - one:

Since $g (x) = | x |$

$f(1)=\left | 1 \right | = 1$

$f(-1)=\left |- 1 \right | = 1$

As we can see $f(1)=f(-1)$ but $1\neq -1$ so $g(x)$ is not one-one.

Thus , g(x) is not injective.

$gof : N \rightarrow Z$

$gof(x) = g(f(x)) = g(x) =$ $\left | x \right |$

Let $gof(x)=gof(y)\, \, \, \, \, x,y \in N$

$\left | x \right |=\left | y \right |$

Since, $x,y \in N$ so x and y are both positive.

$\therefore x=y$

Hence, gof is injective.

Posted by

seema garhwal

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