Question

Prove that the function f : N → N, defined by f(x) = x² + x + 1, is one-one but not onto.

Answer 1

f : N → N, defined by f(x) = x² + x + 1

Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
$$\begin{gathered} \Rightarrow x^2+x+1=y^2+y+1 \\ \Rightarrow \left(x^2-y^2\right)+\left(x-y\right)=0 \\ \Rightarrow \left(x+y\right)\left(x-y\right)+\left(x-y\right)=0 \\ \Rightarrow \left(x-y\right)\left(x+y+1\right)=0 \\ \Rightarrow x-y=0\left[\left(\text{x+y+1}\right)\text{cannot be zero because}\mathit{\text{x}}\text{and}\mathit{\text{y}}\text{are natural numbers}\right] \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
$$\begin{gathered} \text{The minimum number in}N\text{is 1.} \\ \text{When}x\text{=1,} \\ {x}^2+x+1=1+1+1=3 \\ \Rightarrow x^2+x+1\ge 3,\text{for every}x\text{in}N\text{.} \\ \text{⇒}f\left(x\right)\text{will not assume the values 1 and 2.} \\ \text{So,}f\text{is not onto.} \end{gathered}$$