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
Verified
Hint: Use the conditions for surjectivity for taking the right examples.
Complete step-by-step answer: Consider examples 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. 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.
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