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# Cartesian Product!

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

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

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

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

5 months, 2 weeks ago

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

E.g. you can determine $$\{ \text{ elephant, zinc} \} \times \{ \frac{1}{2} , ( 1, 2) \}$$. Staff · 5 months, 1 week ago