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For $a,b \in \mathbb{R}$ define $a \sim b$ if $a - b \in \mathbb{Z}$

How would you find the equivalence class of 5. In other words what I'm trying to describe is the set $[5]$ = {$y : 5 \sim y$}. And $[5]$ is just the name of the set.

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You show how would one go to prove the relation is an equivalence one. How does this serve us to find $\;[5]\;$ ? –  DonAntonio Oct 29 '13 at 5:09

2 Answers 2

The answers to your previous question contain a description of the equivalence classes:

$a \sim b$ iff $a$ and $b$ have the same fractional part.

The fractional part of $5$ is $0$. Which other real numbers have zero fractional part?

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HINT: By definition $[5]=\{x\in\Bbb R:5\sim x\}=\{x\in\Bbb R:5-x\in\Bbb Z\}$. $5-x\in\Bbb Z$ if and only if $5-x$ is an integer, i.e., there is an $n\in\Bbb Z$ such that $5-x=n$. By elementary algebra this is equivalent to saying that $x=5-n$ for some integer $n$, so $[5]=\{5-n:n\in\Bbb Z\}$. What real numbers can be written in the form $5-n$ for some integer $n$? This set has a very simple description.

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So would the real numbers be 0,1,2,3,4? –  micheal wal Oct 29 '13 at 16:04
    
@micheal: That would be the case if $[5]$ were $\{x\in\Bbb R:5-x\in\Bbb Z^+\}$, but $\Bbb Z$ is the set of all integers, positive, negative, and zero. For example, $5-7=-2$, and $-2\in\Bbb Z$, so $7\in[5]$. So what all is in $[5]$? –  Brian M. Scott Oct 29 '13 at 16:07
    
Oh, so it would be all numbers -infinity to 5. Or would it be -infinity to infinity since the set includes negatives and such. –  micheal wal Oct 29 '13 at 16:22
    
@micheal: The latter: every integer is $5-n$ for some integer $n$. $100$ is $5-(-95)$; $-100$ is $5-105$; and you can easily see that every integer can be written in this way. Thus, $[5]=\Bbb Z$. Similarly, $[n]=\Bbb Z$ no matter what integer $n$ is. –  Brian M. Scott Oct 29 '13 at 16:25
    
Ok that makes sense. So if on the other hand let's say you were the trying to find the equivalence class of something like [3.2]. Would the class be empty then since 3.2 is not an integer? –  micheal wal Oct 29 '13 at 16:31

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