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

Last updated date: 30th Mar 2023

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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.

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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