I am having some trouble understanding how to prove that a given binary relation is an equivalence relation. I understand that to prove that a relation is an equivalence relation, you must prove reflexivity, symmetry, and transitivity. However, I'm not too sure on how to apply them to my given relation.
The relation is: (a,b)R(c,d) ↔ a≡c(mod2) and b≡d(mod3)
Thanks for your help!
On
There is a general principle: If $f:\>X\to Y$ is a function between nonempty sets then $$x\sim x'\quad:\Leftrightarrow\quad f(x)=f(x')$$ is an equivalence relation on $X$. In your case $$X:={\mathbb Z}\times{\mathbb Z},\qquad Y:={\mathbb Z}_2\times{\mathbb Z}_3\ ,$$ and $$f:\quad (a,b)\mapsto\bigl(a\ \> {\rm mod}(2),\ \ b\ \> {\rm mod}(3)\bigr)\ .$$
For your relation reflexivity means that for any two integers $a$ and $b$ there holds $(a,b)R(a,b)$ but this is true since $a\equiv a \mod 2$ and $b\equiv b \mod 3$ is always true.
Symmetry means that if $(a,b)R(c,d)$ for some integers $a$, $b$, $c$ and $d$, than also $(c,d)R(a,b)$. But this is true since from $a \equiv c \mod 2$ it follows that $c \equiv a \mod 2$ and the same symmetry holds for $b$ and $d$.
At last transitivity means that from $(a,b)R(c,d)$ and $(c,d)R(e,f)$ for some integrers $a$, $b$, $c$, $d$, $e$ and $f$ it follows that $(a,b)R(e,f)$. But this is also true since $\equiv$ is transitive (from $a \equiv c$ and $c\equiv e$ it follows that $a\equiv e$).