Classify the following function as injection, surjection or bijection:

    \[\mathbf{f}:~\mathbf{Q}~-\text{ }\left\{ \mathbf{3} \right\}\text{ }\to ~\mathbf{Q}\]

, defined by

    \[\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }\left( \mathbf{2x}\text{ }+\mathbf{3} \right)/\left( \mathbf{x}-\mathbf{3} \right)\]

Classify the following function as injection, surjection or bijection:

    \[\mathbf{f}:~\mathbf{Q}~-\text{ }\left\{ \mathbf{3} \right\}\text{ }\to ~\mathbf{Q}\]

, defined by

    \[\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }\left( \mathbf{2x}\text{ }+\mathbf{3} \right)/\left( \mathbf{x}-\mathbf{3} \right)\]

Given

    \[f:~Q~-\text{ }\left\{ 3 \right\}\text{ }\to ~Q\]

, defined by

    \[f\text{ }\left( x \right)\text{ }=\text{ }\left( 2x\text{ }+3 \right)/\left( x-3 \right)\]

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (Q − {

    \[3\]

}), such that f(x) = f(y).

f(x) = f(y)

    \[\left( 2x\text{ }+\text{ }3 \right)/\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }\left( 2y\text{ }+\text{ }3 \right)/\left( y\text{ }\text{ }3 \right)\]

    \[\left( 2x\text{ }+\text{ }3 \right)~\left( y\text{ }-\text{ }3 \right)\text{ }=~\left( 2y\text{ }+\text{ }3 \right)~\left( x\text{ }-\text{ }3 \right)\]

    \[2xy~-~6x\text{ }+~3y\text{ }-\text{ }9~=~2xy~-~6y\text{ }+\text{ }3x\text{ }-\text{ }9\]

    \[9x~=~9y\]

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {

    \[3\]

}), such that f(x) = y for some element x in Q (domain).

f(x) = y

    \[\left( 2x\text{ }+\text{ }3 \right)/\left( x\text{ }\text{ }3 \right)\text{ }=\text{ }y\]

    \[2x\text{ }+\text{ }3\text{ }=\text{ }x\text{ }y\text{ }-\text{ }3y\]

    \[2x\text{ }\text{ }x\text{ }y\text{ }=\text{ }-3y\text{ }-\text{ }3\]

    \[x~\left( 2-y \right)\text{ }=\text{ }-3~\left( y\text{ }+\text{ }1 \right)\]

    \[x\text{ }=\text{ }-3\left( y\text{ }+\text{ }1 \right)/\left( 2\text{ }\text{ }y \right)~\]

which is not defined at y =

    \[2\]

.

So, f is not a surjection and f is not a bijection.