Question

Show that \(A \times B \ne B \times A\), when A and B are non empty, unless \(A = B\)

Short answer

\(A \times B \ne B \times A\) unless \(A = B\)

Step-by-step solution

Cartesian product

The cartesian product of A and B denoted by\({\bf{A \times B}}\), is the set of all ordered pairs (a,b).

Therefore, \(A \times B = \left\{ {\left( {a,b} \right)\left| {a \in A\Lambda b \in B} \right.} \right\}\)

To show that \(A \times B \ne B \times A\) unless \(A{\bf{ = }}B\).\(\) 

Here,

The cartesian product of a set A and B is same as cartesian product of a set B and A.

\(A \times B \ne B \times A\)

Also, \(A \ne \phi \) and \(B \ne \phi \)

Proof:

Part A:

Let x be an element of A, i.e., \(x \in A\)

Now, Cartesian product is \(A \times B\) for every element B in b

Therefore, \(\forall b \in B:\left( {x,b} \right) \in A \times B\)

Also, \(A \times B = B \times A\)

Hence, it can be written as

\(\forall b \in B:\left( {x,b} \right) \in B \times A\)

By definition of Cartesian product, x is an element of B and B is an element in A.

So, \(x \in B\)

Now, A is an element of B.

So, \(A \subseteq B\)

Part B:

Let x be an element of B, i.e., \(x \in B\)

Now, Cartesian product is \(A \times B\) for every element B in a

Therefore, \(\forall a \in B:\left( {x,b} \right) \in A \times B\)

Also, \(A \times B = B \times A\)

Hence, it can be written as

\(\forall a \in B:\left( {x,b} \right) \in B \times A\)

By definition of Cartesian product, x is an element of A and A is an element in B.

So, \(x \in A\)

Now, A is an element of B.

So, \(B \subseteq A\)

Now, since \(A \subseteq B\)and \(B \subseteq A\), the two sets have to be equal.

\(A = B\)

Therefore, it is proved that \(A \times B \ne B \times A\) unless, \(A = B\).