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

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SUMMARY

The discussion defines the multiplication of rational numbers as an equivalence relation, specifically for x, y ∈ Q. It establishes that if x = n/m and y = p/q, where m and q are nonzero, then xy is defined as (np)/(mq). The definition ensures that the multiplication is well-defined by demonstrating that the result does not depend on the representation of x and y. Furthermore, it introduces the concept of equivalence classes in the Cartesian product of integers and positive integers, showing how these classes correspond to rational numbers.

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  • Knowledge of Cartesian products in set theory
  • Basic concepts of well-definedness in mathematics
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Mathematicians, educators, and students interested in abstract algebra, particularly those focusing on rational numbers and equivalence relations.

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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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