equivalence relation $3y+x=5t$ let $R=\{(x,y)\in\mathbb{Z}^2\ |\ \exists t\in \mathbb{Z},\ 3y+x=5t\}$ be a relation. prove/disprove that it equivalence relation.
So I first tried to disprove that its reflexivity: let $x\in\mathbb{Z}$ so $3x+x=4x$. there are not $t\in\mathbb{Z}$ so $x=\frac{5}{4}t$ because $x\in\mathbb{Z}$. So $R$ isn't equivalence relation.
Is it correct?
 A: It is enough to show that the given relation $R=\{(x,y) \in \mathbb{Z}^2 | \exists t \in \mathbb{Z}, 3y+x=5t \}$ is not reflexive to disprove that it's an equivalence relation. One easy counter-example for reflexivity that does not require any knowledge of elementary number theory is to take $x=1$.
Then $1=\frac{5}{4}t$. It is obvious that $t \neq 0$. Therefore, $|t| \geq 1$ and we will have:
$$1=|\frac{5}{4}t|=\frac{5}{4}|t| \geq \frac{5}{4}$$
But $1 \geq \frac{5}{4}$ is absurd. Therefore, no $t \in \mathbb{Z}$ exists for $(1,1) \in \mathbb{Z}^2$ and $R$ is not reflexive.
ADDENDUM:
If we instead work with $\bar{R}=\{(x,y) \in \mathbb{Z}^2 | \exists t \in \mathbb{Z}, 9y+x=5t \}$ then it is indeed an equivalence relation because:


*

*Reflexivity holds because for any $x \in \mathbb{Z}$, we can take $t=2x$ and it works.

*Symmetry holds too! Suppose that $(x,y)$ works and we have $9y+x=5t_0$ for some $t_0\in\mathbb{Z}$. Then:
$$9x+y=(10y+10x)-(9y+x)=5(2y+2x-t_0)$$
Now, take $t=2x+2y-t_0\in \mathbb{Z}$ and you will see that symmetry works.


*Transitivity works too because of the following equality:


$$9z+x=(9z+y)+(9y+x)-10y=5(t_0+t_1-2y)$$
where $9z+y=5t_0$ and $9y+x=5t_1$ by assumption.
This proves that in this case, $\bar{R}$ will indeed be an equivalence relation.
