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Let the set A = {p, q, r}Define in extension A²Calculate Card A²Check that Card A² = (Card A) ²Card : Cardinality

User Richmond
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1 Answer

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SOLUTION

Giving The set


A=\mleft\lbrace p,q,r\mright\rbrace

Then from cartesian coordinate of a set,


A^2=A* A=\mleft\lbrace p,q,r\mright\rbrace*\mleft\lbrace p,q,r\mright\rbrace

Then we expand the set above


A^2=\mleft\lbrace(p,p\mright),(p,q),(p,r),(q,p),(q,q),(q,r),(r,p),(r,q),(r,r)\}

The extension of a set is the set itself.

Hence


\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}

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.


\begin{gathered} \text{Cardinality of A}^2=cardA^2 \\ \text{cardA}^2=9 \end{gathered}

Hence

Card A²=9

Then Card A is


\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}

Therefore


\text{cardA}^2=(\text{card)}^2=9

Therefore

cardA²=(card)²=9

User Nazgob
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