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I think I'm missing something about equivalence classes here. My answer is very different from the books answer and I'm confused as to what is happening.

The task:

On the set of nonnegative integers, we can define a relation x (triple bar) y if and only if x mod 3 = y mod 3.

Use this this equivalence to partition the set {2, 4, 5, 6,9,22,24,25,31,37} into equivalence classes.

My answer: [0] = {6,9,24} [1] = {4,22,25,31,37} [2] = {2,5}

The books answer: [0] = {6,9,24} [1] = {1,4,25,31,37} [2] = {2,5,23}

My question: Where did 1 and 23 come from and where did 22 go? Aren't the union of all partitions supposed to equal the given set?

Edit: Fixed a typo.

Eric W
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    Neither $1$ nor $23$ were in the original set. I would assume these are just typos/mistakes unless I'm overlooking something critical. EDIT: Weirdly $22$ is also present nowhere in the book's answer... I'm also anecdotally not a fan of the way the question is posed but that might be me. – PrincessEev Sep 05 '19 at 03:13
  • Or to rephrase my comment: let's say you partition a set $A$ into equivalence classes $B_1, \cdots, B_n$. Then $B_i \cap B_j = \emptyset$ for all $i \ne j$, and $A = \cup_{i=1}^n B_i$. This means that each element of the original set should appear in exactly one of the classes - no more or less (no elements go missing or suddenly appear). Thus why I feel this is a mistake on the book's part. – PrincessEev Sep 05 '19 at 03:16
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    That was my understanding as well Eevee Trainer. I think you're right about it being a typo of some sort. Thank you for your replies! – Eric W Sep 05 '19 at 03:21

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