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Exercise 1.1 · Q10

Q.If A×AA\times A has 16 elements, S={(a,b)∈A×A:a<b}S=\{(a,b)\in A\times A : a<b\}; (−1,2)(-1,2) and (0,1)(0,1) are two elements of SS, then find the remaining elements of SS.

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Step 1. n(A×A)=16⇒n(A)=4n(A\times A)=16\Rightarrow n(A)=4.

Step 2. (−1,2)∈S(-1,2)\in S and (0,1)∈S(0,1)\in S tell us −1,2,0,1∈A-1,2,0,1\in A -- that's already 4 distinct elements, so A={−1,0,1,2}A=\{-1,0,1,2\}.

Step 3. Sort AA ascending: −1<0<1<2-1<0<1<2. S={(a,b)∈A×A:a<b}S=\{(a,b)\in A\times A:a<b\} lists every pair in this increasing order: (−1,0),(−1,1),(−1,2),(0,1),(0,2),(1,2)(-1,0),(-1,1),(-1,2),(0,1),(0,2),(1,2) -- six pairs total. …

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