Proving that S is not an equivalence relation.

Consider the relations R and S on $\Bbb N$ defined by $x\; R\; y$ iff

$2 \;$divides $x + y$ and $x \;S \;y$ iff $3$ divides $x + y.$

$\text{QN: Prove that S is not an equivalence relation.}$

For this question I know one must prove that the properties reflexive, symmetric, and transitive are not an equivalence relation. Thus they are false.

$(i)$ To prove that it is not reflexive one must give a counter example. So let $x =1$ and $y =3$. Then $4 \not = 3.$ This is so far what I have I just do not know how one can give a counter example for transitive and symmetric. Any hints would be appreciated.

• It suffices to show that at leas tone of the three conditions i snot met. – Hagen von Eitzen Jul 8 '16 at 17:20
• Oh so all I have to do is show that is not reflexive. – Jon Jul 8 '16 at 17:21

Symmetry follows from commutative of addition. However, it is easy to note that $xRx$ iff $3 | (x+x) \iff 3|2x \iff 3|x$. So the reflexive property fails for any $x$ not divisible by $3$.
• So in otherwords $3 \not | \; x$ Or $2x \not = 3a$. – Jon Jul 8 '16 at 17:54
• Yep, $3$ need not necessarily divide $x$, so reflexitivity (...that was a mouthful) fails. :-) – Zain Patel Jul 8 '16 at 17:59
The given relation is symmetric since for all $x$ and $y$, $(x+y) = (y+x)$. It isn't transitive since $1R2$ and $2R1$ but 1 is not related to 1.