The question asks to prove that the following relation is an equivalence relation on the set of integers and find the equivalence class of the number 5.
$ (x,y) ∈ R$ if and only if $x + 2y$ is divisible by 3 .
I was able to prove that the relation is reflexive but I'm not quite sure how to prove that the relation is symmetric and transitive, as well as [5].
On
if $(x,y)$ then $x+2y = 3k$
$x = 3k-2y$
If the relation is symmetric then $(y,x)$ and $y +2(3k+2y) = 9k-3y$ must also be divisible by 3. And it is!
If the relation is transative.
$(x,y)$ and $(y,z)$
$x = 3j - 2y\\ y = 3k - 2z$
imply $(x,z)$
$x = 3j - 6k + 4z\\ x+2z = 3j - 6k + 6z$
And the RHS is indeed divisible by 3.
This has all been a bit messy.
It might be cleaner to say $(x,y)\implies x \equiv y \pmod 3$
Prove to yourself that these are the same relations.
The equivalence class that includes 5.
$\{\cdots,-1,2,5,8,\cdots \}$
To prove symmetry, you need to prove: if $(x,y) \in R$, then $(y,x) \in R$. That is, if $x+2y$ is divisible by $3$, then $y+2x$ is divisible by $3$. Hint: What is $(x+2y)+(y+2x)$?
To prove transitivity, you need to prove: if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$. That is, if $x+2y$ is divisible by $3$ and $y+2z$ is divisible by $3$, then $x+2z$ is divisible by $3$. Hint: Add $(x+2y)+(y+2z)$.
To find the equivalence class of $5$, you need to find the set of all numbers $x$ such that $(x,5) \in R$. What does it mean to have $(x,5) \in R$? Can you characterize all such $x$ that satisfy this?