I have the relation $R$ on $ℚ$, and $R$ is set to be the relation $R$ = {$(a,b): a - b ∈ ℤ$}. I am supposed to prove that the relation $R$ is an equivalence relation.
I know I'm supposed to prove that $R$ is reflexive, symmetric, and transitive in order to prove that $R$ is an equivalence relation, so first I attempted to prove that $R$ is reflexive:
Reflexive Proof:
Let $x ∈ ℚ$ be arbitrary
Since $x - x = 0$, $xRx$
Hence, $(x,x) ∈ R$
After attempting to prove $R$ is reflexive, I then tried to prove that it is symmetric, but I got stuck with the following:
Symmetric Proof:
Let $x,y ∈ ℚ$ be arbitrary, assume $(x,y) ∈ R$
$xRy$
... (This is the point where I get stuck)
$yRx$
$(y,x) ∈ R$
I have yet to attempt to prove that $R$ is transitive.
I am unsure if my proof that $R$ is reflexive is correct and I do not know how complete the symmetric proof.
(Apologies if my formatting is strange. This is the first time I ask a question on this site.)
(1) $R$ is reflexive.
Let $x∈Q$ be arbitrary
Since $x−x=0$, $xRx$
Hence, $(x,x)∈R$.
(2) $R$ is symmetric.
Let $x, y \in Q$ be arbitrary.
Suppose that $xRy$ or $(x, y) \in R$.
This shows that $$ x - y = m $$ where $m \in \mathbf{Z}$.
Hence, $$ y - x = -m $$ and $-m$ is also an integer.
This shows that $yRx$ or $(y, x) \in R$.
Hence, $R$ is symmetric.
(3) $R$ is transitive.
Let $x, y, z \in Q$ be arbitrary.
Suppose that $xRy$ and $yRz$.
That is, $(x, y) \in R$ and $(y, z) \in R$.
This shows that $$ x - y = m, \ \ \ y - z = n $$ where $m, n \in \mathbf{Z}$.
Adding the two equations, we get $$ x - z = m + n $$ and note that $m+n$ is also an integer.
This shows that $xRz$ or $(x, z) \in R$.
Hence, $R$ is also transitive.
Combining the three cases, we conclude that $R$ is an equivalence relation.