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.

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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Khribechy

Answered 2021-08-07
Author has **100** answers

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$(x,y)\in Ri.e.5(x-y)=0$

if R symmetric (x-y) Then$(y,x)\in R$ means 5(y-x)=0

$\Rightarrow y-x=0\Rightarrow y=x$

as$5(x-y)=0\Rightarrow -5(y-x)=0$

$\Rightarrow 5(y-x)=0\Rightarrow (y,x)\in R$

so R is symmetric.

Reflexive:$x\in Z,x-x=0\Rightarrow 5(x-x)=0$

so$(x,x)\in R$

Hence R is reflexive.

Transitive: Let$(x,y)\in R{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}(y-z)\in R$ so

That$5(x-y)=0{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}5(y-z)=0$

add both$5(x-y)+5(y-z)=0$

$\Rightarrow 5(x-y+y-z)=0$

$5(x-z)=0$

$i.e.(x,z)\in R$

Hence R is transitive.

So, that R is an equivalence selation.

Symmetric let

if R symmetric (x-y) Then

as

so R is symmetric.

Reflexive:

so

Hence R is reflexive.

Transitive: Let

That

add both

Hence R is transitive.

So, that R is an equivalence selation.

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