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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.
 Fall '08
 STAFF
 Math, Calculus

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