
Let A and B be sets. Show that $f:A \times B \to B \times A$ such that $f\left( {a,b} \right) = \left( {b,a} \right)$ is a bijective function.
Answer
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Hint: In this question, we want to prove that the given function is bijective. The function will be bijective if the function will be one-one and the function will be onto. Based on the property of the ordered pair we can prove that the function is one-one. Based on the pre-image or matching element, we can prove that the function is onto.
Complete step-by-step solution:
In this question,
$f:A \times B \to B \times A$ is defined as $f\left( {a,b} \right) = \left( {b,a} \right)$.
To show the given function is bijective:
First, let us prove the function is one-one.
Here, two elements $\left( {{a_1},{b_1}} \right)$ and $\left( {{a_2},{b_2}} \right)$ belongs to $A \times B$.
Therefore,
$\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right) \in A \times B$ such that $f\left( {{a_1},{b_1}} \right) = f\left( {{a_2},{b_2}} \right)$
From the property of ordered pairs and based on function definition.
$ \Rightarrow ({b_1},{a_1}) = ({b_2},{a_2})$
If two ordered pairs are equal then the first element should be equal and the second element should also be equal.
$ \Rightarrow {b_1} = {b_2}$ and ${a_1} = {a_2}$
So, we can write,
$ \Rightarrow ({a_1},{b_1}) = ({a_2},{b_2})$
Hence, the function f is one-one.
Now, let us prove that the function f is onto.
Take $\left( {b,a} \right) \in B \times A$ be any element.
$ \Rightarrow b \in B$ and $a \in A$
Here, $\left( {a,b} \right) = \left( {b,a} \right)$ is a pre-image if $\left( {a,b} \right) \in A \times B$ because $b \in B$ and $a \in A$
Then there exists $\left( {a,b} \right) \in A \times B$
We also know$f\left( {a,b} \right) = \left( {b,a} \right)$ from the function definition.
Hence, the function f is onto.
Here, the function f is one-one and onto.
Therefore, this is a bijective function.
Note: An ordered pair consists of two elements that are written in the fixed order. The pair of elements that occur in a particular order and are enclosed in brackets is called a set of ordered pairs. Ordered pair is not a set consisting of two elements.
Complete step-by-step solution:
In this question,
$f:A \times B \to B \times A$ is defined as $f\left( {a,b} \right) = \left( {b,a} \right)$.
To show the given function is bijective:
First, let us prove the function is one-one.
Here, two elements $\left( {{a_1},{b_1}} \right)$ and $\left( {{a_2},{b_2}} \right)$ belongs to $A \times B$.
Therefore,
$\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right) \in A \times B$ such that $f\left( {{a_1},{b_1}} \right) = f\left( {{a_2},{b_2}} \right)$
From the property of ordered pairs and based on function definition.
$ \Rightarrow ({b_1},{a_1}) = ({b_2},{a_2})$
If two ordered pairs are equal then the first element should be equal and the second element should also be equal.
$ \Rightarrow {b_1} = {b_2}$ and ${a_1} = {a_2}$
So, we can write,
$ \Rightarrow ({a_1},{b_1}) = ({a_2},{b_2})$
Hence, the function f is one-one.
Now, let us prove that the function f is onto.
Take $\left( {b,a} \right) \in B \times A$ be any element.
$ \Rightarrow b \in B$ and $a \in A$
Here, $\left( {a,b} \right) = \left( {b,a} \right)$ is a pre-image if $\left( {a,b} \right) \in A \times B$ because $b \in B$ and $a \in A$
Then there exists $\left( {a,b} \right) \in A \times B$
We also know$f\left( {a,b} \right) = \left( {b,a} \right)$ from the function definition.
Hence, the function f is onto.
Here, the function f is one-one and onto.
Therefore, this is a bijective function.
Note: An ordered pair consists of two elements that are written in the fixed order. The pair of elements that occur in a particular order and are enclosed in brackets is called a set of ordered pairs. Ordered pair is not a set consisting of two elements.
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