MHB Operations on Sets: Explained & Examples

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The discussion clarifies the concept of the Cartesian product of sets, specifically when set A contains a single element. It explains that if A has one element and B has multiple elements, the Cartesian product A × B results in a set of ordered pairs where the first element is from A and the second from B. For example, with A = {a} and B = {b1, b2, b3}, the product A × B yields {(a, b1), (a, b2), (a, b3)}. The thread emphasizes that the total number of pairs is the product of the cardinalities of the sets, confirming that a single element in A leads to as many pairs as there are elements in B. Understanding this operation is essential for grasping more complex mathematical concepts involving sets.
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please help me understand what my book says:

If set A has only one element a, then $\displaystyle A\,x\,B\,=\, \{\left(a,\, b\right)\,|\,b\,\epsilon\,B\}$, then there is exactly one such element for each element from B.

can you explain what it means and give some examples. thanks! :)
 
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Re: Operations on set

From Wikipedia:

In mathematics, a Cartesian product is a mathematical operation which returns a set (or product set) from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

For example, if the cardinality of set $A$ is one, where $A=\{a\}$ and we have a set $B$ of cardinality $n$, i.e., $B=\{b_1,b_2,b_3,\cdots,b_n\}$ then:

$$A\,\times\,B=\{(a,b_1),(a,b_2),(a,b_3),\cdots,(a,b_n),\}$$
 
Cartesian products get their name from the prototypical example, the Cartesian plane, which is the Cartesian product of two orthogonal lines.

It's easier to see what is going on if we consider a Cartesian product of two finite sets, say:

A = a bag of red marbles,
B = a bag of green marbles.

Suppose we want "all possible pairs" of marbles, and A has 3 marbles, and B has 4 marbles. We can label these r1,r2,r3 (for the red marbles) and: g1,g2,g3,g4 (for the green marbles). Then the set of all possible pairs looks like this:

(r1,g1) (r1,g2) (r1,g3) (r1,g4)

(r2,g1) (r2,g2) (r2,g3) (r2,g4)

(r3,g1) (r3,g2) (r3,g3) (r3,g4)

Laid out like this, it's clear we have 3*4 = 12 pairs in all. And, in general:

[math]|A \times B| = |A|\cdot|B|[/math]

so, if A and B are sets of 1 element each, their Cartesian product has 1*1 = 1 element (only one possible choice for the "first coordinate", and only one possible choice for the "second coordinate").
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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