Set $A = \{0,7,1\}$
1. So for a relation that is reflexive and transitive but neither an equivalence relation nor partial order...Can a relation be both partial order and equivalence?
Attempt: $R_1=\{(0,0),(7,7),(1,1),(0,1),(1,0),(0,7)\}$
Is $R_1$ Reflexive? Yes because $(0,0),(7,7),(1,1)$ exist in the relation.
Is $R_1$ Symmetric? No because every edge of the relation does not have either a two way street or a loop (if you drew $R_1$ as a diagraph)
Is $R_1$ Antisymmetric? No because there exists a two way street between two distinct vertices.
Now the part where I doubt myself...
Is $R_1$ Transitive? Yes? I don't think $(0,7)$ would break transitivity. If $R_2=\{(0,7)\}$ is transitive then the addition of $(0,7)$ to an already transitive relation wouldn't make it not transitive? Could someone please help clarify?
2. Relation on $A$ that is not reflexive, not transitive, not antisymmetric, but is symmetric.
I got $R_2=\{(0,7),(7,0),(7,7)\}$, I believe that is correct since it misses $(0,0)$ and $(1,1)$ and therefore is not reflexive. It is not antisymmetric since the diagraph has no one way streets. It is symmetric because there exists a two way street between each distinct vertices.
Could some please let me know if I am correct?