This exercise is given in my textbook and I am trying to solve it.
Determine whether they are symmetric, antisymmetric or reflexive.
$R_1=\{(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}$
$R_2=\{(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)\}$
$R_3=\{(2,4), (4,2)\}$
$R_4=\{(1,2), (2,3), (3,4)\}$
$R_5=\{(1,1), (2,2), (3,3), (4,4)\}$
$R_6=\{(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)\}$
My answers:
1- $R_1$ is symmetric.
2-$R_2$ is reflexive, symmetric.
3-$R_3$ is symmetric.
4-$R_4$ is antisymmetric.
5-$R_5$ is reflexive, antisymmetric.
6-$R_6$ is symmetric,
Book's answers:
1-None of the these properties.
2-$R_2$ is reflexive and symmetric.
3-$R_3$ is symmetric.
4-$R_4$ is antisymmetric.
5-$R_5$ is reflexive, symmetric and antisymmetric.
6-None of these properties.
You can see that some of my answers don't match the answers given in book. Is that probably a misprint or I am wrong somewhere?
For a relation $R$ to be symmetric, we have to have for all elements in $R$ that if $(x,y) \in R$, then also $(y,x) \in R$. You have found some elements in $R_1$ such that both $(x,y) \in R_1$ and $(y,x) \in R_1$, but for example $(2,4) \in R_1$ but $(4,2) \notin R_1$, hence it it not symmetric because it doesn't satisfy the criterion for every element.
There is also an element in $R_6$ that makes it non-symmetric, can you find it?
As for $R_5$, since every element is of the form $(x,x)$, it also (obviously) holds that $(x,x) \in R_6$, so it is symmetric.