Question

If $A=\{0,1\}$ and$B=\{1,2,3\}$ show that $AxB\ne BxA$

Answer 1

Solve for the required values and compare them.

Given,

$A=\{0,1\}$

$B=\{1,2,3\}$

$$\begin{gathered} \Rightarrow AxB=\left\{\right(0,1),(0,2),(0,3),(1,1),(1,2),(1,3\left)\right\} \\ \Rightarrow BxA=\left\{\right(1,0),(2,0),(3,0),(1,1),(2,1),(3,1)\} \end{gathered}$$

By the definition of equality of ordered pair,

The pair$(0,1)$ in$AxB$ is not equal to the pair$(1,0)$ in$BxA$

So, $AxB\ne BxA$

Hence proved.