# For the relation R = {(x,y) : x + 2y <= 3}, defined by A = {0,1,2,3}, determine if it is reflexive, symmetric, antisymmetric and transitive.

For the relation $R=\left(x,y\right):x+2y\le 3$, defined by $A=\left\{0,1,2,3\right\}$, determine if it is reflexive, symmetric, antisymmetric and transitive.
The 2y is throwing me a bit here. To determine if it is reflexive, I have done the following: $x+2x\le 3$, which is a no with counter example (1,2), as that would give me (1,4) and be greater than 3. Is every second number doubled? Ie. (x,2x) or (x,2y)?
I have concluded this relation is not symmetric, as it does not imply $y+2x\le 3$, on the basis that is , this would result in for $y+2x$, which is greater than 3. I have no confidence in this answer however.
I'm floundering on this one. Every time it was touched on during the lecture, it was simple examples, like $x+y$ is even, or if there was an equality, there wasn't a defined set for it.
For determining transitive, what would I use as z?
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Reese King
Step 1
It might be more helpful to think of the relation like this. Don't forget that, here, R is a subset of $A×A.$ Then
$R=\left\{\left(x,y\right)\in A×A|x+2y\le 3\right\}$
With that established, the (perhaps) cleaner way of thinking of this:
$\text{x and y are related by R (i.e. xRy)}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(x,y\right)\in R\subseteq A×A\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x+2y\le 3$
In testing reflexivity, we seek $\left(x,x\right)\in R$ for all $x\in A$. By the above, that would require $x+2x=3x\le 3$. As you have seen, there is a counterexample.
Step 2
For symmetry, whenever $\left(x,y\right)\in R$, we want $\left(y,x\right)\in R$. That would mean that, whenever $x+2y\le 3$, that $y+2x\le 3$. Again, you have a counterexample to this.
For transitivity, you want whenever $\left(x,y\right),\left(y,z\right)\in R$ that $\left(x,z\right)\in R$. This would mean that
$\begin{array}{c}x+2y\le 3\\ y+2z\le 3\end{array}\right\}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x+2z\le 3$
I'll tell you right away that there is a counterexample to this as well.
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Tamara Bryan
Step 1
I think you got the first two.
Let's try transitive: if $\left(x,y\right)\in \mathcal{R}$ and $\left(y,z\right)\in \mathcal{R}$, then $x+2y\le 3$ and $y+2z\le 3$. We need to check if $x+2z\le 3$. Well, after trying a few things, I think I have a counter example. Namely $\left(3,0\right)\in \mathcal{R}$ and $\left(0,1\right)\in \mathcal{R}$. Clearly though, $\left(3,1\right)\notin \mathcal{R}$.
Step 2
Thus R isn't transitive.
Finally for antisymmetry, note that $\left(0,1\right)\in \mathcal{R}$. And $\left(1,0\right)\in \mathcal{R}$. But $0\ne 1$.