Test the binary relation on the set for reflexivity, symmetry, antisymmetry, and transitivity.

Question

$S = \{0,1,2,3,4,5\}$

$xRy, x+y = 5$

I'm not entirely sure on how to test this for reflexivity, symmetry, antisymmetry, and transitivity, though I understand the rules for each.

I guess I'm confused because I'm not sure what to use for my $x$ and $y$ values since there are no ordered pairs given.

Can you all provide a hint to get me started?

Answer 1

$\langle x, y\rangle$ is an ordered pair that is a member of $R$ if, and only if, $x+y=5$. For example, the ordered pair $\langle 1 , 4\rangle$ is a member of $R$, but the ordered pair $\langle 1, 3\rangle$ is not.

Does that clear it up for you?

Answer 2

$R$ is not reflexive because $xRx \iff x + x = 5 \iff x = 2.5$, but $2.5 \notin S$. Further it is not transitive because $xRy$, and $yRz$ implies: $x+y=5,y+z=5 \to x = z$, and $xRz \to x+z = 5$. So $x = z = 2.5$, again $2.5\notin S$ so $R$ is not transitive.

Answer 3

For reflexivity you must verify if all element in $S$ relates to itself, this is, if $xRx$ for all $x\in S$. A quick view shows that this is not true. For $3$, $3$ doesn't relate to itself because $3+3\neq 5$.

For symmetry, you must show that if $xRy$, then $yRx$. Suppose that $xRy$ for $x,y\in S$. This means that $x+y=5$. Because the sum is (supposedly) commutative, this implies that $y+x=5$, and this implies that $yRx$.

Note that for reflexivity it is sufficient to give a counterexample (because the statement its not true for all elements of $S$) but for symmetry you need a proof (after convincing yourself that this is true).

I hope this helps when verifying transitivity and antisymmetry.