Let $\sim$ be a relation on $\mathbb{Z}^2$ given by $(x_1, x_2) \sim (y_1, y_2)$ exactly when $x_1 + y_2 = y_1 + x_2$

Question

Let $\sim$ be a relation on $\mathbb{Z}^2$ given by $(x_1, x_2) \sim (y_1, y_2)$ exactly when $x_1 + y_2 = y_1 + x_2$. I need to show that $\sim$ is reflexive, symmetric, transitive, and not anti-symmetric. So, an equivalence relation.

I'm not too sure how to go about proving those things. From what I understand, I notice a few things.

$x_1 + y_2 = y_1 + x_2$ is satisfied when $x_1 = x_2 = y_2 = y_1$ which I assume has to do with reflexive.

$x_1 + y_2 = y_1 + x_2$ is satisfied when $x_1 = x_2$ and $y_1 = y_2$ which I assume has to do with symmetric.

Still, I am at a loss on how to write an answer. Specifically, I think it would help if someone could clarify/expand on what is meant by "$(x_1, x_2) \sim (y_1, y_2)$ exactly when $x_1 + y_2 = y_1 + x_2$."

Answer 1

Hint: the given condition $x_1+y_2=y_1+x_2$ is equivalent to $\color{magenta}{x_1-x_2=y_1-y_2}$ (if you move both $x_2$ and $y_2$ to the other side). In this form, it is much more convenient to use, because each side of the equality has components of the same pair of numbers. See if you can apply the definitions of the reflexive, symmetric, and transitive properties now.

Your work is incorrect because you're making too many unjustified assumptions. For example, in what you wrote about both reflexivity and symmetry, you're assuming that $x_1=x_2$. Why?? $(x_1,x_2)\in\mathbb{Z}^2$ can be an arbitrary pair of numbers. Nothing requires its two components to be equal to each other.

Specifically, I think it would help if someone could clarify/expand on what is meant by "$(x_1,x_2)\sim(y_1,y_2)$ exactly when $x_1+y_2=y_1+x_2$."

It means exactly what it says. For any two pairs of integers that we'd like to look at, we denote one of them $(x_1,x_2)$ and the other one $(y_1,y_2)$, we check whether the equality $x_1+y_2=y_1+x_2$ is satisfied, and if it is we say that $(x_1,x_2)\sim(y_1,y_2)$. Here are a couple of examples.

1. Let $(x_1,x_2)=(3,10)$ and $(y_1,y_2)=(-1,6)$; both are elements of $\mathbb{Z}^2$. So here $x_1=3$, $x_2=10$, $y_1=-1$, and $y_2=6$. Then we can calculate that $x_1+y_2=3+6=9$ and $y_1+x_2=(-1)+10=9$. We can see that $x_1+y_2=y_1+x_2$ (i.e. $9=9$) for these two pairs. Since they satisfy this condition, they are in this relation: $(3,9)\sim(-1,6)$.

2. Let $(x_1,x_2)=(5,5)$ and $(y_1,y_2)=(7,8)$; both are elements of $\mathbb{Z}^2$. So here $x_1=5$, $x_2=5$, $y_1=7$, and $y_2=8$. Then we can calculate that $x_1+y_2=5+8=13$ and $y_1+x_2=7+5=12$. We can see that $x_1+y_2\neq y_1+x_2$ (i.e. $13\neq12$) for these two pairs. Since they don't satisfy this condition, they are not in this relation: $(3,9)\not\sim(-1,6)$.

Answer 2

OP here. This is my attempt at writing an answer, please let me know if this is sufficient.

The relation $(x_1,x_2) \sim (y_1,y_2)$ is true when $x_1 - x_2 = y_1 - y_2$

Reflexive: Clearly you can pick a $(y_1, y_2)$ for any $(x_1, x_2)$ such that $x_1 - x_2 = y_1 - y_2$ is equivalent to $x_1 - x_2 = x_1 - x_2$. Therefore the relation is reflexive.

Symmetric: If $x_1 - x_2 = y_1 - y_2$ is true for $(x_1,x_2) \sim (y_1,y_2)$ then it is still true for $(y_1,y_2) \sim (x_1,x_2)$ since $y_1 - y_2 = x_1 - x_2$. Therefore the relation is symmetric.

Transitive: Consider $(x_1,x_2) \sim (y_1,y_2)$ and $(y_1,y_2) \sim (z_1,z_2)$. Then we have $x_1 - x_2 = y_1 - y_2$ and $y_1 - y_2 = z_1 - z_2$. Meaning $x_1 - x_2 = z_1 - z_2$. Thus $(x_1, x_2 \sim z_1,z_2)$. Therefore the relation is transitive.

Anti-symmetric: First assume $\sim$ is anti-symmetric. Then for $(x_1, x_2) \sim (y_1,y_2)$ and $(y_1,y_2) \sim (x_1,x_2)$ we must hold $(x_1, x_2) = (y_1,y_2)$. But for a counter-example $(3,10)$ and $(-1,6)$ is true both ways, yet $(3,10) =/= (-1,6)$. Therefore the relation cannot be anti-symmetric.

Therefore the relation is an equivalence relation.