# 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 :

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

3 years, 5 months 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 - 3 years, 5 months ago

OK...

- 3 years, 5 months ago

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

- 1 year, 5 months ago