Cartesian Product!

We know that given two non - void sets, say AA and BB, their Cartesian Product\textbf{Cartesian Product} is defined as A×B={(a,b):aA,bB}A\,\times\,B\,=\,\{(a,b)\,:\,a\,\in\,A\,,\,b\,\in\,B\}. For eg. If A={1,3}A\,=\,\{1,3\} and B={2,4,9}B\,=\,\{2,4,9\}, then their Cartesian product is defined as {(1,2),(1,4),(1,9),(3,2),(3,4),(3,9)}\{(1,2),(1,4),(1,9),(3,2),(3,4),(3,9)\}.

I was wondering what if the set AA itself consists of ordered pairs, like what if A={(1,5),(1,9),(2,5)}A\,=\,\{(1,5),(1,9),(2,5)\} and say B={3,7}B\,=\,\{3,7\}, then how do we define the A×BA\,\times B?

One thing that I've observed is that ,in case, if the set that consists of ordered pairs ( n-tuples in general) can be expressed as Cartesian product of simple sets (sets that consists of numbers only), then the overall cartesian product can be found. What I mean is

Say, A={(3,4),(3,7),(2,4),(2,7)}A\,=\,\{(3,4),(3,7),(2,4),(2,7)\} and B={2,5}B\,=\,\{2,5\} and A×BA\times B is to be found, then one can proceed as follows :

A×B={(3,4),(3,7),(2,4),(2,7)}×{2,5}={3,2}×{4,7}×{2,5}={(2,3,4),(2,3,7),(2,2,4),(2,2,7),(5,3,4),(5,3,7),(5,2,4),(5,2,7)}A\times B\,=\,\{(3,4),(3,7),(2,4),(2,7)\}\,\times\,\{2,5\}\,\,=\,\{3,2\}\,\times\,\{4,7\}\,\times\,\{2,5\}\,\,=\,\{(2,3,4),(2,3,7),(2,2,4),(2,2,7),(5,3,4),(5,3,7),(5,2,4),(5,2,7)\}.

Note by Aditya Sky
2 years, 3 months ago

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Do the same thing. A×B={(a,b)aA,bB} A \times B = \{ ( a, b) | a \in A, b \in B \} . It doesn't matter what the sets AA and BB are.

E.g. you can determine { elephant, zinc}×{12,(1,2)} \{ \text{ elephant, zinc} \} \times \{ \frac{1}{2} , ( 1, 2) \} .

Calvin Lin Staff - 2 years, 3 months ago

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OK...

Aditya Sky - 2 years, 3 months ago

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Infact, this is the method followed to generate n-ary relations.......

Aaghaz Mahajan - 3 months, 2 weeks ago

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