# Thread: Reflexive and symmetric but not an equivalence relation

1. ## Reflexive and symmetric but not an equivalence relation

Define a binary relation R on {n Z: ≥ 2} such that aRb if and only if a and b havea common factor not equal to 1.
Show R is reflexive and symmetric but not an equivalencerelation.

hmm so n Z: ≥ 2
and a and b does not have a common factor of 1.

so if we make a all even squares ≥ 2 and b, all squares powers..from 1 upwards
{1,2,3}

we use a subset we could have would be all Z ≥ 2 a ={2,4,8} b ={1,2,3}

Is this on the right track
any help would be appreciated thanks

2. ## Re: Reflexive and symmetric but not an equivalence relation

Originally Posted by bee77
so if we make a all even squares ≥ 2 and b, all squares powers..from 1 upwards
{1,2,3}

we use a subset we could have would be all Z ≥ 2 a ={2,4,8} b ={1,2,3}

Is this on the right track
any help would be appreciated thanks

no.

$aRb \Rightarrow a = u v,~b = u w,~u \neq 1$

First show it's reflexive and symmetric.

Reflexive means $aRa$

does $aRa$ ?

well $a$ is a common factor and by assumption $a\geq 2$

so yes $aRa$ and $R$ is reflexive

Can you show that $R$ is symmetric, i.e. that $a R b \Rightarrow b R a$ ?

Then you have to show that despite the above $R$ is not an equivalence relation.

What further property must an equivalence relation have?

Show that $R$ does not possess this further property and thus is not an equivalence relation.