Is $R = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x - y \in \mathbb{N}\}$ an equivalence relation? Note: I am also trying to answer my own question, but I am not sure if it is correct, please correct it, if it's wrong. Thanks :)
I have an exercise where I have to say if a relation $R$ is or not an equivalence relation, and this are the 2 examples:


*

*$R = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x - y \in \mathbb{N}\}$

*$R = \{ (x, y) \in \mathbb{R} \times \mathbb{R} \mid x - y \in \mathbb{Q}\}$
My answer is no for first case, because $R$ is not symmetric, but reflexive. Suppose $x = 1.313$ and $y = 0.313$, then it's not symmetric, since $x - y \in \mathbb{N}$, but $y - x \not \in \mathbb{N}$.
For second case, my answer is yes, since it's symmetric and transitive, and of course also reflexive. The following is my proof:


*

*Reflexive:
Suppose $x = 3.14$, then $x - x = 0 \in \mathbb{Q}$

*Symmetric:
Suppose $x = 1.5$ and $y = 2.0$.
Then $x - y = -0.5 = -\frac{1}{2} \in \mathbb{Q}$, and $y - x = 0.5 = \frac{1}{2} \in \mathbb{Q}$
So $R$ is symmetric.

*Transitive:
Suppose $x = 1.5$, $y = 2.0$ and $z = 3.5$.
We have that:
$x - y = -0.5 \in \mathbb{Q}$
$y - z = -1.5 \in \mathbb{Q}$
$x - z = -2.0 \in \mathbb{Q}$
So $R$ is transitive.
Since $R$ is reflexive, symmetric and transitive, we can say that $R$ is an equivalence relation (for the second case).
 A: You seem to have some misunderstanding on how proofs work. To show that a relation is reflexive, you must show that $(x, x) \in R$ for ALL $x$, not just one. To show it is not reflexive, you simply demonstrate some $x$ for which it is not reflexive. Your second proof is invalid because it does not show that it always true, it only gives an example. Consider the following relation defined on the numbers $0$ and $1$:
$$ D = \{(0, 0), (0, 1) \} $$
Is $D$ reflexive? No. Even though $(0,0) \in D$, we have $(1, 1) \not\in D$. One example does not prove a statement (although a counterexample will disprove a statement).
Here is another statement: $x \ge 0$ for all $x \in \mathbb R$. You cannot prove this statement by exhibiting one example or even infinitely many, since it is false for any negative $x$.
So let's try and prove your statement correctly.


*

*Reflexive. Since $x-x=0\in \mathbb Q$, we know $xRx$ so $R$ is reflexive.

*Symmetric. We have to prove that if $x R y$, then $y R x$. This means that if $x - y \in \mathbb Q$, then $y-x \in \mathbb Q$. Let $x - y = \frac{r}{s}$, where $r$ and $s$ are integers (we can do this since $x-y\in \mathbb Q$, so it is the quotient of integers). Now $y-x = \frac{-r}{s}$, which is the quotient of integers, and so $y-x\in\mathbb Q$. Therefore the relation is symmetric. (Basically, you want to show that if $q\in\mathbb Q$ then $-q \in\mathbb Q$.)

*Transitive. We need to show if $x - y \in \mathbb Q$ and $y - z \in \mathbb Q$, then $x - z \in \mathbb Q$. Note that the rationals are closed under addition, so $(x-y) + (y-z) = x - z \in \mathbb Q$. Therefore the relationship is transitive.

