 # Is R an equivalence relation? If so, prove it discrete math and if not, explain why it is not. Let R be a relation on Z defined by (x.y) \in R if and only if 5(x-y)=0. Formally state what it means for R to be a symmetric relation. Haven 2021-08-06 Answered
Let R be a relation on Z defined by (x.y) $\in$ R if and only if 5(x-y)=0. Formally state what it means for R to be a symmetric relation. Is R an equivalence relation? If so, prove it discrete math and if not, explain why it is not.
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Let R be a relation on Z defined by x, y $\in$ Z. Then (x, y) $\in$ R if 5(x-y)=0
Symmetric let $\left(x,y\right)\in Ri.e.5\left(x-y\right)=0$
if R symmetric (x-y) Then $\left(y,x\right)\in R$ means 5(y-x)=0
$⇒y-x=0⇒y=x$
as $5\left(x-y\right)=0⇒-5\left(y-x\right)=0$
$⇒5\left(y-x\right)=0⇒\left(y,x\right)\in R$
so R is symmetric.
Reflexive: $x\in Z,x-x=0⇒5\left(x-x\right)=0$
so $\left(x,x\right)\in R$
Hence R is reflexive.
Transitive: Let $\left(x,y\right)\in R\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(y-z\right)\in R$ so
That $5\left(x-y\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}5\left(y-z\right)=0$
add both $5\left(x-y\right)+5\left(y-z\right)=0$
$⇒5\left(x-y+y-z\right)=0$
$5\left(x-z\right)=0$
$i.e.\left(x,z\right)\in R$
Hence R is transitive.
So, that R is an equivalence selation.