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


Prove that the function $f: N \rightarrow N$, defined by


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


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

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




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

$\Rightarrow x=y$

 So, is one-one.


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

When $x=1$,


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

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

So, $f$ is not onto.


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