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The following relation is defined on $\mathbb{N}^{+}$

True or false? This relation is an equivalence relation: $xRy \Leftrightarrow x \cdot y$ is even

First, relation is equivalence relation if it have property: reflexive, symmetric, transitive.

$xRy \Leftrightarrow x \cdot y$ is even $\text{ }\text{ }\text{ }\text{ }$ This says that when product is even, you have relation that satisfy all these property i.e. is equivalence relation, right?

Is allow I negate the relation and find counterexample?

Negation say: If product of $x,y$ is uneven, you don't get equivalence relation.

Let me make example, $x=y=1:$ $1 R 1 \Leftrightarrow 1$ is uneven

But $(1,1)$ is equivalence relation. Statement is false.

Is all good work or I do wrong?

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    $\begingroup$ Yes, one conterexample to reflexivity such as $1\not R1$ is enough to show that $R$ is not an equivalence relation. Additionally, $R$ fails to be transitive, for example $1R2$ and $2R3$, but $1\not R3$. $\endgroup$ – Hagen von Eitzen Nov 13 '17 at 10:55
  • $\begingroup$ Pls no give me negative point!! $\endgroup$ – roblind Nov 13 '17 at 10:56
  • $\begingroup$ Exercise. Shiw the relation is symmetric and not transtive. $\endgroup$ – William Elliot Nov 13 '17 at 11:44
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$R$ is a relation. A relation can either be True or False for a pair. So when we look at $1R1$ this state that $1*1$ is even, which is clearly false, just like you have stated.

As you say, an equivalence relation has to be reflexive i.e. for each positive integer x, $xRx$ should hold. But you have shown that $1R1$ does not hold, and thus $R$ is not reflexive and thus $R$ is not an equivalence relation.

Your proof is correct and good up to the English. It would also be nice to see that you recognize that you are contradicting the reflexivity.

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