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I need to determine if the following relation is reflexive, symmetric, antisymmetric, and/or transitive. I have been reading a lot of similar posts about these topics on here, but I am still stumped.

The relation is $(x,y)\in R$ iff $3$ divides $x + 2y$.

I'm stuck on the reflexive proof right now. I know that a relation is reflexive iff $(a,a)\in R$ for all $a\in A$ where $R$ is a binary relation on the set $A$. So knowing that I do this:

$$\frac{3}{x + 2x} = \frac{1}{x}.$$

I have no idea what that means, if I'm doing it right, or how to proceed. Examples of a similar nature on this site take the equation, replace the $y$ with $x$, and simply to get something like $x-x=0$. That is obviously not the case here. Could somebody please help? Thanks.

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  • $\begingroup$ Presumably x and y are integers (you should edit your question to state this). You also have the "divides" concept reversed. 3 divides 6; 6 does not divide 3. "3 divides x+2y" means (x+2y)/3 is an integer, not 3/(x+2y). $\endgroup$ – Ted Sep 14 '15 at 3:03
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I assume that you mean for $R$ to be defined over the integers.

Indeed, the relation is reflexive. Let $x$ be any integer. Then we have $$ x + 2x = 3x $$ Since $3x$ is divisible by $3$ for any integer $x$ (or as I would write, $3 \mid 3x$ for any $x$), we may conclude that $(x,x) \in R$ for any integer $x$, which is to say that $R$ is reflexive.


It is also useful to note that since $3y$ is a multiple of $3$, we will have $$ (x,y) \in R \iff\\ 3 \mid (x + 2y) \iff\\ 3 \mid (x + 2y - 3y) \iff\\ 3 \mid (x - y) $$ You will probably find this equivalent definition of the relation easier to work with.

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