Proving Equivalence Relation in $\mathbb{N}\times\mathbb{N}$

Question

I just need to see if this is correct

Define a relation R on $\mathbb{N} \times \mathbb{N}$ by $<a,b>$ R $<c,d> $ iff $a+d=c+b$. Prove this is an equivalence relation.

I am having an issue with the reflexive case. I wrote let $<a,b>\in\mathbb{N}\times\mathbb{N}$. Then $a+b=b+a$ by commutativity, thus $<a,b>$ R $<a,b>$, but should I instead be using $<a,a>$.

Then for symmetric, I have that $<a,b>,<c,d>\in\mathbb{N}\times\mathbb{N}$ such that $<a,b>$ R $<c,d>$, $a+d=c+b$. Then,$a+d=b+c$= $d+a=b+c$, which we can just rewrite as $d+a=c+b$. Then, this shows $<c,d>$ R $<a,b>$

For transitivity, I have let $<a,b>,<c,d>,<x,y>\in\mathbb{N}\times\mathbb{N}$ such that $<a,b>$ R $<c,d>$ and $<c,d>$ R $<x,y>$. So, we have that $a+d=b+c$ and $c+y=d+x$. Then, we can do $(a+d)+(c+y)=(b+c)+(d+x)$. We can cancel out the $c$ and $d$ from both sides by subtracting and are left with $a+y=b+x$. Then, $<a,b>$ R $<x,y>$.

Answer 1

You've correctly shown that $R$ is an equivalence relation on $\mathbb{N}\times\mathbb{N}$.

To answer your question regarding reflexivity: No, you should not use $\langle a,a \rangle$. A relation $R$ on a set $X$ is reflexive if, for any $x\in X$, we have $x R x$. In your case, $X=\mathbb{N}\times\mathbb{N}$, so you need to show that $x R x$ for an arbitrary element $x\in\mathbb{N}\times\mathbb{N}$. This element should be of the form $\langle a,b\rangle$, not $\langle a,a\rangle$. (Which is just as you did :))

Answer 2

Yes, it is correct. Concerning reflexivity, what you should (and did) prove was that$$(a,b)\operatorname{R}(c,d)\iff(c,d)\operatorname{R}(a,b).$$

Answer 3

For another approach, note that $(a,b) R (c,d) $ iff $f(a,b)=f(c,d)$, where $f: \mathbb{N}\times\mathbb{N} \to \mathbb{Z}$ is given by $f(x,y)=x-y$. Since equality is an equivalence relation, so is $R$.