Question

Prove that the function $F:N\to N$, defined by $f\left( x \right)={{x}^{2}}+x+1$ is one-one but not onto.

Answer 1

Hint: For a function to be one-one the condition is, if f (x) = f (y), then x = y. For a function to be onto, the condition is each pre-image has its own unique image. We will also use the identity, $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ to solve this question. By using these concepts, we will prove that the function is one-one but not onto.

Complete step by step solution:
It is given in the question that we have to prove that the function $F:N\to N$, defined by $f\left( x \right)={{x}^{2}}+x+1$ is one-one but not onto.
We will first prove that the function is one-one.
So, the given function is $f\left( x \right)={{x}^{2}}+x+1,x\in N$. Now, for a function to be one-one, the condition is that, if f (x) = f (y), then x = y.
Let us assume $x,y\in N$ such that f (x) = f (y). So, we get,
${{x}^{2}}+x+1={{y}^{2}}+y+1$
On cancelling the like terms on both the sides, we get,
${{x}^{2}}+x={{y}^{2}}+y$
Now, on transposing ${{y}^{2}}$ from RHS to the LHS and x from LHS to the RHS, we get,
${{x}^{2}}-{{y}^{2}}=y-x$
Now, we know that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, so we can write,

& \left( x+y \right)\left( x-y \right)=\left( y-x \right) \\\ & \left( x+y \right)\left( x-y \right)=-\left( x-y \right) \\\ & \left( x+y \right)\left( x-y \right)+\left( x-y \right)=0 \\\ \end{aligned}$$ On taking (x-y) common from both the terms, we get, $\begin{aligned} & \left( x-y \right)\left[ \left( x+y \right)+1 \right]=0 \\\ & \left( x-y \right)\left( x+y+1 \right)=0 \\\ \end{aligned}$ Now, (x + y + 1) cannot be equal to 0, because it is greater than 0, so it is neglected. Hence, we get, (x - y) = 0 x = y Therefore, we have proved that the function is one-one. Now, we will consider the function and show that it is not onto. We know that a function is onto when each pre-image has its own unique image. Here, we have $f\left( x \right)={{x}^{2}}+x+1$.  When we put x = 1 in f (x), we get, ${{x}^{2}}+x+1\Rightarrow 1+1+1\Rightarrow 3$ and if we put x = 2 in f (x), we get, ${{x}^{2}}+x+1\Rightarrow 4+2+1\Rightarrow 7$. So, it shows that $\left( {{x}^{2}}+x+1 \right)\ge 3$ for every x in N. It means that f (x) will not have the pre-image of value 1 and 2. Therefore the function is not onto. Hence we have proved that the function $F:N\to N$, defined by $f\left( x \right)={{x}^{2}}+x+1$ is one-one but not onto. Note: While solving this question, the students must not skip any steps, like some students write that f (x) = f (y), therefore ${{x}^{2}}+x+1={{y}^{2}}+y+1$ and x = y. But then the solution will be incomplete, so the students must solve each step and show how they reached the conclusion x = y. Also some students tend to get confused between the conditions of one-one and onto function and interchange them, so this must be avoided.