Prove that the function $f: N ightarrow N$, defined by

Solution:

$f: N \rightarrow N$, defined by $f(x)=x^{2}+x+1$

Injectivity:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

$\Rightarrow x^{2}+x+1=y^{2}+y+1$

$\Rightarrow\left(x^{2}-y^{2}\right)+(x-y)=0$

$\Rightarrow(x+y)(x-y)+(x-y)=0$

$\Rightarrow(x-y)(x+y+1)=0$

$\Rightarrow x-y=0 \quad[(\mathrm{x}+\mathrm{y}+1)$ cannot be zero because $x$ and $y$ are natural numbers $]$

$\Rightarrow x=y$

 So, f is one-one.

Surjectivity:

The minimum number in $N$ is $1 .$

When $x=1$,

$x^{2}+x+1=1+1+1=3$

$\Rightarrow x^{2}+x+1 \geq 3$, for every $x$ in $N$.

$\Rightarrow f(x)$ will not assume the values 1 and 2 .

So, $f$ is not onto.