Prove that f: N → N defined as f ( x )= a x + b ( a, b ∈ N ) is one-one but not onto.

Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{~N}$ (Domain) such that $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)$ \[\begin{aligned} & \Rightarrow \mathrm{ax}_{1}+\mathrm{b}=\mathrm{ax}_{2}+\mathrm{b} \\ & \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2} \end{aligned} \] Therefore, f is one-one. Let $\displaystyle \mathrm{y} \in \mathrm{N}$ (codomain). Then $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{y}$ if, $\displaystyle \mathrm{ax}+\mathrm{b}=\mathrm{y}$ i.e., if, $\displaystyle \mathrm{x}=\frac{\mathrm{y}-\mathrm{b}}{\mathrm{a}}$, which may not belong to N (domain) Therefore, f is not onto.