Question

8) show that AXB is not equal to
BX A​

Answer 1

Answer:

A & B are two non- empty subsets and we have to prove AxB=BxA iff A=B. Proof: ... So, you can observe A×B not equal B×A.

Answer 2

Answer:

$A \times B \neq B \times A \text { unless } A=B$

Step-by-step explanation:

If we assume that $A \times B=B \times A$ and if we can then derive that $A=B$, then we have shown that $A \times B \neq B \times A$ unless $A=B$

Given:

$\begin{gathered}A \times B=B \times A \\A \neq \emptyset \\B \neq \emptyset\end{gathered}$

To prove: $A=B$

PROOF

FIRST PART Let $x$ be an element of $A:$

$x \in A$

Then for every element $b in B$, we know that $(x, b)$ is in the Cartesian product $A \times B$:

$\forall b \in B:(x, b) \in A \times B$

Use $A \times B=B \times A$ :

$\forall b \in B:(x, b) \in B \times A$

By the definition of the Cartesian product, we then know that $x$ is an element of $B and b$is an element in $A$ :

$x \in B$

We have then derived that every element in $A$ is also an element in $B$:

$A \subseteq B$

SECOND PART Let $x$ be an element of $B$ :

$x \in B$

Then for every element $a in A$, we know that $(a, x)$is in the Cartesian product $A \times B$ :

$\forall a \in A:(a, x) \in A \times B$

Use $A \times B=B \times A$:

$\forall a \in A:(a, x) \in B \times A$

By the definition of the Cartesian product, we then know that $a$ is an element of $B and x$ is an element in $A$ :

$x \in A$

We have then derived that every element in $B$ is also an element in $A$ :

$B \subseteq A$

Since $A \subseteq B$ and $B \subseteq A$, the two sets have to be equal

$A=B$

$A \times B \neq B \times A \text { unless } A=B$