The Cartesian product is an essential notion to know in set theory. Let's discover its definition together.
Definition of the Cartesian product
We call Cartesian product of two sets A and B (it is denoted A x B) the set of couples (a;b) where a is an element of A and b an element of B. Mathematically, this is written:
A \times B = \{(a,b) ,a \in A, b\in B\}
When A and B are finite, the cardinality of A x B is the product of the cardinality of A and the cardinality of B. So if the cardinality of A is p, that of B is q then the cardinality of A x B is pq.
\#(A\times B) = \#A \times \#B
Here is some additional information:
- The Cartesian product is not commutative, E x F is generally different from F x E, they are also different when F is different from E
- If F = E, we can replace the notation E x E by E2
- We can define a Cartesian product with n terms:
A_1 \times \ldots \times A_n = \{ (a_1,\ldots,a_n), a_1 \in A_1, \ldots, a_n \in A_n \}
- And if all the A'si are equal to A, we also have the following notation: A1 x … x An = An which is much more condensed.
Examples
- If A = {a,b} and B = {1,2,3} then the set A x B resulting from the Cartesian product is made up of the following 6 elements: A x B = {(a,1); (a,2); (a,3); (b,1); (b,2); (b,3)}. Note that B x A is quite different from A x B and equals {(1,a); (2,a); (3,a); (1,b); (2,b); (3,b)}
- If A and B are equal to the set of real numbers, we define
\mathbb{R}^2
by
\mathbb{R}^2 = \{(a,b), a \in \R, b\in \R\}
The Cartesian Product in SQL
To make the connection with the SQL, the joint cross join corresponds exactly to the Cartesian product. If we have a first table of k rows and a second table of l rows then the cross join between these two tables gives a table of kxl rows corresponding to the Cartesian product of the 2 rows.
Hello,
Why is E^nx E^n \= E^(2n)
Hello,
And thank you for your question! The answer is quite simple:
We define E^n as E x E^(n-1). By induction we get that it is E x E x … x E n times
E^(2n) so it's E x E x … x E 2n times
We therefore have: E^nx E^n = E x E x .. x E 2n times = E^(2n)
Does that answer your question?
I'm sorry but the \= was for \not=
It was my high school teacher who told me that and I didn't really understand. I understand that they are equal up to an isomorphism but not why they are not equal. Thanks in advance and have a good evening.
Here is my answer :
On one side it's 2n-uplets, on the other a 2n-upets
These sets are therefore clearly isomorphic but not equal unless E is empty
However, that seems more conventional than anything else.
As I had a doubt, I asked the question on Twitter, you can read the answers to form an opinion. Here is the direct link: https://twitter.com/Progressermaths/status/1526326467654733827
thank you so much!!