# Equivalence Relation on the set R of all Real Numbers

Show that == (where == is the equivalence relation defined below) is an equivalence on A, and find a (well-defined) bijection %: A== -> B, where

(a) A = R (the set of all real numbers)

(b) B={x: x is an element of R and 0 <= x < 1}

(c) for real numbers x and y, "x==y" (x is equivalent to y) if and only if x - y is an element of Z (the set of all integers)

(d) "A==" denotes the set of all equivalence classes

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#### Solution Preview

The following are the steps in the solution of this problem:

1. To show that == is an equivalence relation, we need to show that == is reflexive, symmetric, and transitive.

a. To show that == is reflexive, we show that for every real number x: x == x (i.e., x - x is in Z).

If x is any real number, then x - x = 0. Since 0 is in Z, we have x == x.

b. To show that == is symmetric, we show that for any real numbers x, y: if x == y (i.e., x - y is in Z), then y == x (i.e., y - x is in Z).

Let x and y be real numbers such that x==y. Then x - y is an integer. Let i = x - y. Then y - x = -(x - y) = -i. Since i is an integer, -i is also an integer, so y==x.

c. To show that == is transitive, we show that for any real numbers x, y, z: if x == y (i.e., x - y is in Z) and y == z (i.e., y - z is in Z), then x == z (i.e., x - z is in ...

#### Solution Summary

The solution consists of a review of what an equivalence relation is, a step-by-step proof that the given binary relation is indeed an equivalence relation, the definition of a mapping from the set of equivalence classes (for the given equivalence relation) to the set {x: x is a real number and 0 <= x < 1}, and a step-by-step proof that that mapping is a well-defined bijection.