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Attempt.

Part one

Proof

Define a relation $\sim$ on $\mathbb{Q}$ as $r\sim s$ iff $r-s\in\mathbb{Z}$. To prove that $\sim$ is an equivalence relation, we must verify that $\sim$ is reflexive, symmetric, and transitive.

1) Suppose $r$ is a rational number. Then $r\sim r$ or $r-r=0$ is an integer. So $\sim$ is reflexive.

2) Suppose $r,s$ are rational numbers. Then $r\sim s$ or $r-s$ is an integer. Now, $-(r-s)$ or $s-r$ is also an integer so $s\sim r$. Therefore, $\sim$ is symmetric.

3) Suppose $r,s,t$ are rational numbers. Then $r\sim s$ and $s\sim t$ or $r-s$ is an integer and $s-t$ is an integer. Now, adding $r-s$ and $s-t$ to each other gives $r-t$, which is also an integer and we get $r\sim t$. Therefore, $\sim$ is transitive.

Hence, since $\sim$ is reflexive, symmetric, and transitive, $\sim$ is an equivalence relation.

Part Two

Now, assuming this is correct, I'm having a little difficulty with describing the partition of this equivalence relation. I'd be grateful for any suggestions on how to proceed.

Thanks for your time and attention and I hope this work has not been too painful (I apologize for the poor math and reasoning on my part) too read. Live long and prosper.

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  • $\begingroup$ There's no point to introducing the letters $a$, $b$, $c$, $d$, etc., in any of your proofs in Part One. Indeed you never even use these letters after introducing them. $\endgroup$ – symplectomorphic Mar 16 '17 at 7:12
  • $\begingroup$ In part 2 I wouldn't bother multiplying by -1. Just stating s-r=-(r-s) in Z. Is enough. $\endgroup$ – fleablood Mar 16 '17 at 7:17
  • $\begingroup$ Pick a number. What is [1.37]? If a~ 1.37 then a-1.37=k in Z so a= k+1.37 =m+.37. And if b = n+.37 then a-b=1.37 - n-.37=1-n. So a~b. So [1.37] ={r|r=k+.37;k1 in Z}. Are you familiar with the concept {x}= "the fractional part" of x? $\endgroup$ – fleablood Mar 16 '17 at 7:26
  • $\begingroup$ @ fleablood "Are you familiar with the concept {x}= "the fractional part" of x?" I'm not. Could you elaborate on this concept please? Thanks. Also, I'm grateful for the feedback that all of you have given me: I have tried to edit my proof for part one accordingly. I also edited out part two, because I really didn't know what I was doing, and it seemed like irrelevant prattle on my part. Thanks again. $\endgroup$ – Kernel Sohcahtoa Mar 16 '17 at 7:40
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Your proof of part one is O.K. ($a,b,c, ...$ are not needed !)

Part two: let us denote the equivalence class of $r$ by $[r]$. Then:

$x \in [r] \iff x \sim r \iff x=r+k $ for some $k \in \mathbb Z$

Hence: $[r]=\{r+k : k \in \mathbb Z\}$

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I'm going to try and address the second part in an intuitive way, as I feel this might be helpful in addition to knowing how to shuffle the symbols around properly.

I would start concretely, maybe by asking "what are all the things that are equivalent to 1?" 1 is equivalent to all rational numbers who are an integer distance away from one. So $$\ldots, -2, -1, 0, 1, 2\ldots$$ are all equivalent to 1. If I take something like $\frac{1}{2}$, all the things that are equivalent would be $$\ldots, -1+\frac{1}{2},\ 0+\frac{1}{2},\ 1+\frac{1}{2},\ 2+\frac{1}{2},\ 3+\frac{1}{2},\ldots$$ You can do this starting with any rational number.

What you should notice is that if you take an arbitrary equivalence class, you'll find that the distinguishing feature of that class is the fractional part of all its elements. The integer part doesn't matter, and this is what the relation $r-s\in\mathbf{Z}$ is trying to capture.

Another way to say it is this: We're defining a new type of equality on the rationals. Two rationals are the "same" if they have the same fractional part and they are "different" if their fractional parts are different.

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  • $\begingroup$ Thanks for your post, Daenerys Naharis. IMO, it complimented Fred's post nicely. So, assuming that I'm understanding correctly, suppose we take 3/4. Then all the things equivalent would be ...,-1+3/4,0+3/4,1+3/4, 2+3/4. Now, if we take any two rational numbers, say r and s, in the equivalence class 3/4, then subtracting r-s will produce an integer, as each rational number is equivalent to each other. On the other hand, if we take an r that is in the equivalence class 1/2 and an s that in the equivalence class 3/4, then r-s will not be an integer as 1/2 and 3/4 are not equivalent to each other $\endgroup$ – Kernel Sohcahtoa Mar 16 '17 at 15:31
  • $\begingroup$ Also, I noticed that if we take the equivalence class of 1/2, then 1/2 is equivalent to all rational numbers that are an integer distance away from 1/2, and the same observation could be made for any class. Also, if we continued on with all possible equivalence classes, then we would produce classes that are disjoint, and taking the union of all classes would produce the set of rational numbers. So ∼ is a partition on the set of rational numbers. Have I understood correctly? $\endgroup$ – Kernel Sohcahtoa Mar 16 '17 at 15:32
  • $\begingroup$ @KernelSohcahtoa Yep, that's it! This kind of construction shows up a lot if you go on with this type of math so it's good to have a solid understanding. $\endgroup$ – Chris Brooks Mar 16 '17 at 18:33

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