Respuesta :
SOLUTION
Giving The set
[tex]A=\mleft\lbrace p,q,r\mright\rbrace[/tex]Then from cartesian coordinate of a set,
[tex]A^2=A\times A=\mleft\lbrace p,q,r\mright\rbrace\times\mleft\lbrace p,q,r\mright\rbrace[/tex]Then we expand the set above
[tex]A^2=\mleft\lbrace(p,p\mright),(p,q),(p,r),(q,p),(q,q),(q,r),(r,p),(r,q),(r,r)\}[/tex]The extension of a set is the set itself.
Hence
[tex]\begin{gathered} \text{extension A}^2=A^2 \\ \text{extension A}^2=\lbrace(p,p),(p,q),(p,r),(q,p),(q,q),(q,r),(r,p),(r,q),(r,r)\} \end{gathered}[/tex]Extension A² = {(p,p),(p,q),(p,r),(q,p),(q,q),(q,r),(r,p),(r,q),(r,r)}
The cardinality of a set is the number of elements in the set.
[tex]\begin{gathered} \text{Cardinality of A}^2=cardA^2 \\ \text{cardA}^2=9 \end{gathered}[/tex]Hence
Card A²=9
Then Card A is
[tex]\begin{gathered} \sin ce\text{ A=}\mleft\lbrace p,q,r\mright\rbrace \\ \text{cardinality of A=CardA=3} \\ \text{Hence } \\ (CardA)=3^2=9^{} \end{gathered}[/tex]Therefore
[tex]\text{cardA}^2=(\text{card)}^2=9[/tex]Therefore
cardA²=(card)²=9
