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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.

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    $\begingroup$ It suffices to show that at leas tone of the three conditions i snot met. $\endgroup$ – Hagen von Eitzen Jul 8 '16 at 17:20
  • $\begingroup$ Oh so all I have to do is show that is not reflexive. $\endgroup$ – Jon Jul 8 '16 at 17:21
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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 $3$.

This equivalence also fails the transitive property, but you need only show one property is not satisfied to show it is not an equivalence. However, you may find it instructive to find a counterexample to the transitive property.

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  • $\begingroup$ So in otherwords $3 \not | \; x $ Or $2x \not = 3a$. $\endgroup$ – Jon Jul 8 '16 at 17:54
  • $\begingroup$ As long as the conditions are satisfied that is good enough for me. $\endgroup$ – Jon Jul 8 '16 at 17:56
  • $\begingroup$ Yep, $3$ need not necessarily divide $x$, so reflexitivity (...that was a mouthful) fails. :-) $\endgroup$ – Zain Patel Jul 8 '16 at 17:59
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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.

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