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prove that if $n>0$ is a positive integer then relation $\equiv_n$ on integers defined by $a\equiv_n b$ if $n\mid (a − b)$ is an equivalence relation. what if $n=2$?

so i know we have to proof reflexivity, symmetry and transitivity.

$\tag 1 a\sim a ?$ $\tag 2 a\sim b \text{ then } b\sim a$ $\tag 3 a\sim b, b\sim c \text{ then } a=c$

but im kind of confused of how to prove it

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What have you tried? This is a very straight forward verification of the definitions. – EuYu Oct 30 '12 at 0:24
Dear Jack F, Welcome to math.SE. since you are a new user, we wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Further, it would be better if you could typeset your problem so that it is easy for people to read. Kindly look here… for more details on typesetting. – user17762 Oct 30 '12 at 0:26

1 Answer 1

up vote 2 down vote accepted

You are correct that we need to check 1) reflexivity, 2) symmetry, and 3) transitivity. (except note your transitivity should have a (~) not $=$ in the last part)

So lets check these:

1) Fix any $a$. Then $(a - a) = 0$, so $n \mid (a - a)$ and hence $ a \equiv_n a$.

2) Suppose $a \equiv_n b$. Then $n \mid (a-b)$. So we have some $k$ so that $nk = (a-b)$, but then $n(-k) = (b-a)$ and hence $n \mid (b-a)$ so we have $b \equiv_n a$.

3) Suppose $a \equiv_n b$ and $b \equiv_n c$. Then $n \mid (a-b)$ and $n\mid (b-c)$ so we have some $k$ so that $nk = (a-b)$ and $j$ so that $nj = (b-c)$ Then $n(j + k) = nk + nj = (a-b) + (b-c) = a - c$. So $n \mid (a-c)$ and $a \equiv_n c$.

So $\equiv_n$ is an equivalence relation.

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\mid gives better spacing than | for divides. Since this is homework, it would be better to give hints or to answer part of the question with hints for the rest. – Brian M. Scott Oct 30 '12 at 0:51
@BrianM.Scott Thanks for \mid didn't know about that. And okay, I'll be sure to look for homework in the tags from now on. – Deven Ware Oct 30 '12 at 1:03
@Deven Ware how does the equivalence class is different? is it like a whole set that is compared or what? – Jack F Oct 30 '12 at 1:03
Not to worry: I sometimes miss it myself, but since I was commenting anyway on the \mid, I thought that I’d mention it. – Brian M. Scott Oct 30 '12 at 1:05
@DevenWare what would the equivalence class be when n=3? I am having trouble figuring this out – Jack F Oct 30 '12 at 16:29

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