1873_solutions

Of the collection the relations and are the same in

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Unformatted text preview: for every x S there exists one and only one member y of the set E such that x ß y. Solution: Suppose that x S. We define C to be the equivalence class of the ralation ß that contains the member x and we define y to be the member of E that lies in the equivalence class C. Since x and y belong to the same equivalence class we know that x ß y. Now, to show that this member y of E is the only member of the set E that can be related by r to x, suppose that z E and that x ß z. From the fact that x ß y and x ß z we deduce that y ß z. Therefore y and z must belong to the same equivalence class of the relation ß and, since E never contains more than one member in any one equivalence class of ß we conclude that y  z. 7. Suppose that is a family of sets and that for any two members A and B of we define A r B to mean that either A B or B A. Is r an equivalence relation in ? This relation need not be transitive. For example, if B A and B C then there is no reason to expect that one of the sets A and C should be included in the other. 8. Which of the following relations in R is an equivalence relation? a. r  x, y Yes R2 x y is an integer . b. r  No x, y R2 x y is a positive integer . c. r  x, y Yes R2 x y is an even integer . 44 d. r  No x, y R2 x y is an odd integer . e. r  x, y Yes R2 x y is rational . f. r  No R2 x y is irrrational . x, y 9. Given a subset G of R, prove that the following two conditions are equivalent: a. The relation r defined by r x, y R2 x y G is an equivalence relation in R. b. The set G is nonempty and for all numbers x and y in the set G the numbers x and x  y must also belong to G. Proof that a implies b: We assume that r is an equivalence relation in R. Since r is reflexive we know that 2 r 2 which says that 2 2 G. Therefore G is nonempty. Given any member x of the set G we know from the fact that x 0 G that x r 0 and it follows from the fact that r is symmetric that 0 r x which says that x G. Finally, suppose that x and y belong to G. Since 0 x  x G we have 0 r x and since x x  y  y G we have x r x  y. Therefore, since r is transitive we have 0 r x  y which tells us that x  y  x  y 0 G. Proof that b implies a: We assume that condition b holds. Using the fact that G we choose a member u of G. We know that u also belongs to G and that u  u G. In other words, 0 G. Given any number x we deduce from the fact that x x  0 G that x r x and so r is reflexive. Given any numbers x and y, if x y G then y x G and so the condition x r y implies that y r x and we conclude that r is symmetric. Finally, suppose that x and y and z are given numbers and that x r y and y r z. Since x y G and y z G we have x y  y z G which tells us that x r z and we conclude that r is transitive. 10. Suppose that G R and that the relation r x, y R2 x y G is an equivalence relation in R. Suppose that in every equivalence class C of r we have chosen a number and named it x C . Prove that every real number x can be written in one and only on...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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