Definition of XY: Equivalence Relation for x,y ∈ Q

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The discussion defines the product "xy" for rational numbers x and y, represented as x = n/m and y = p/q, leading to the formula xy = (np)/(mq) where m and q are nonzero. It emphasizes that this definition is independent of the specific representations of x and y, ensuring well-definedness. The conversation also touches on equivalence relations, illustrating how rational numbers can be viewed as equivalence classes derived from pairs of integers. An equivalence relation is established on the Cartesian product of integers and positive integers, defined by (a, b) ~ (c, d) if ad = bc. Ultimately, the multiplication of rational numbers is shown to be well-defined by confirming that different representatives yield the same result.
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Definition of "xy"

What is "the definition of 'xy' for x,y element of Q"?

For a reference point, this is listed within the subject of equivalence relations and well-definedness.
 
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You write x = n/m, y = p/q, where m,q are nonzero and define xy = (np)/(mq). You check that this definition doesn't depend on the representation of x and y. (I.e. a/b = n/m iff am = nb.)
 
Since you are talking about "equivalence relations" you might be using a more fundamental definition of "rational numbers". If X= IxN, the Cartesian product of the set of integers and the set of counting numbers (positive integers), then we can define an equivalence relation on X by (a, b)~ (c, d) if and only if ad= bc. It's easy to show that is an equivalence relation and so partitions X into equivalence classes. We can define the rational numbers to be that set of equivalence classes. (Then if (a,b) is in an equivalence class, that equivalence class corresponds to the fraction a/b).

Multiplication is then defined by "If x and y are such equivalence classes, choose one "representative", (a,b), from the class x and one "representative", (c, d), from the class y. xy is the class containing (ac, bd)." Of course, you have to prove that this is "well defined"- that is, that if you were to choose different "representatives" from the same classes, the result would be the same. (That's the same as rudinreader's "(I.e. a/b = n/m iff am = nb)".)
 

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