Which one of these two is an equivalence relation

Question

I'm having an issue with the following exercise:

Given $\alpha$ and $\beta$ two binary relationships defined in $Z$ such that: $$\forall m,n \in Z, n\ \alpha\ m \iff n = m\ \ \vee\ \ rest(n,7)\ +\ rest(m,7) = 7$$ $$\forall m,n \in Z, n\ \beta\ m \iff n \equiv_7 m\ \ \vee\ \ rest(n,7)\ +\ rest(m,7) = 7$$ One of them is an equivalence relation. Which one?

So, first thing I thought upon seeing this exercise is that I have to verify the definition of equivalence relation:

1. Reflexivity
2. Symmetry
3. Transivity

I am able to prove for both that they are reflexive. I am having troubles proving that they are both symmetric and transitive. Because the OR is confusing me in the demonstration.

Could you please guide me to solve this exercise? Thank you.

Answer 1

Clearly the first is not transitive:
$1+6=7$ and $\operatorname{rest}(6,7)+\operatorname{rest}(8,7)=7$ but neither $1=8$ nor $\operatorname{rest}(1,7)+\operatorname{rest}(8,7)=7$ is true.
Now consider the second relation:
in fact we can write $$n\ \beta\ m\iff n^2\equiv m^2\pmod7.$$ Then can you start from this definition and prove that it is both symmetric and reflexive? Can you show that this is equivalent with the original definition?

Hope this helps.

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EDIT:
We know that $a\equiv b\pmod7\iff 7\mid(a-b)$ by definition. Also $n\ \beta\ m\iff n\equiv m\pmod7\ \vee\ n+m\equiv0\pmod7.$ We can then translate the condition to $n\ \beta\ m\iff n-m\equiv 0\pmod7\ \vee\ n+m\equiv0\pmod7.$
Further, we have $n^2\equiv m^2\pmod7\iff 7\mid(n^2-m^2)=(n-m)(n+m).$ Can you finish the proof now?
As to how to show $\beta$ is an equivalence relation, might you tell us where you got stuck? What do you think you should prove which you cannot prove?

Answer 2

Symmetry should be obvious: just switch $m$ and $n$ in the definitions, and see what you get.

As to transitivity, think $n = 1$, $m = 6$, $q = 8$. For one of the two relations (which one?), say $\gamma$, you have $n \gamma m$ and $m \gamma q$ but $n \not\gamma q$.