Question

Give examples of two functions \[f:N\to N\]and \[g:N\to N\]such that gof is onto but f is not onto.

Answer 1

Hint: Consider example of \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}
   x-1,x>1 \\
   1,x=1 \\
\end{matrix} \right.\]. First prove that f is not onto by proving the digit formed is not a natural number as mentioned in \[f:N\to N\]. Then prove gof is onto. Find the composite of g and f \[\Rightarrow \left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]. Thus prove gof is onto.



Complete step-by-step answer:
Given to us two functions of \[f:N\to N\]and \[g:N\to N\].
We need to prove that gof is onto and f is not onto.
Onto function could be explained by considering two sets, set A and set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto.
Let us consider \[f:N\to N\]be \[f\left( x \right)=x+1\].
And \[g:N\to N\]be taken as \[g\left( x \right)=\left\{ \begin{matrix}
   x-1,x>1 \\
   1,x=1 \\
\end{matrix} \right.\].
We will first show that f is not onto.
Let \[f:N\to N\]be \[f\left( x \right)=x+1\]
Let us consider\[y=f\left( x \right)\], where \[y\in N\].
\[\begin{align}
  & f\left( x \right)=x+1\Rightarrow y=x+1 \\
 & \therefore x=y-1 \\
\end{align}\]
Let us put \[y=1\],
\[x=1-1=0\]
We get x = 0, which is not a natural number.
It is said that \[f:N\to N\]i.e. natural numbers.
But zero is not a natural number. So f is not onto and it doesn’t belong in the elements of natural numbers.
If given functions f and g with domains and target sets consisting sets consisting of real numbers, the composite of g and f is the new function gof defined as,
\[\left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]
Here, we need to find gof.
Let us consider \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}
   x-1, x>1 \\
   1, x=1 \\
\end{matrix} \right.\].
\[f\left( x \right)=x+1,g\left( x \right)=1\]
Since, \[g\left( x \right)=1\], for x = 1.
\[\begin{align}
  & g\left( f\left( x \right) \right)=1 \\
 & \therefore gof=1 \\
\end{align}\]
Since, \[g\left( x \right)=x-1,x>1\]
\[g\left( f\left( x \right) \right)=f\left( x \right)-1\], put, \[x=f\left( x \right)\].
\[\begin{align}
  & gof=\left( x+1 \right)-1 \\
 & \Rightarrow gof=x \\
 & \therefore gof=\left\{ \begin{matrix}
   x, x>1 \\
   1, x=1 \\
\end{matrix} \right. \\
\end{align}\]
Let us consider, \[gof=y\], where \[y\in N\].
\[\therefore y=\left\{ \begin{matrix}
   x ,x>1 \\
   1, x=1 \\
\end{matrix} \right.\]
Here, y is a natural number, as y = x.
So, x is also a natural number.
\[\therefore \]gof is onto, as y is a natural number and y = gof.
Hence, we proved that gof is onto and f is not onto.

Note:
Remember what is onto function. We can only prove a function is not onto, if the elements found don’t belong to it. In the case of \[f:N\to N\], the element was zero, which is not a natural number and doesn’t belong to f. If it was any number greater than zero, then it would belong to \[f:N\to N\]and f would be an onto function.