$R_1$ = {(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)}
Is this not symmetric ONLY due to the ordered pair (2,4) not having symmetry with (3,4)? I can't seem to find a lot of information on how to deal with a relation being not a "property" and I'm not sure if my logic is right in how to deal with them (not symmetric, not transitive, not reflexive). Should I be looking for only one missing element?
Symmetry requires that if $\langle a,b\rangle\in R$, then $\langle b,a\rangle\in R$ as well. In this case there are two violations of symmetry: $\langle 2,4\rangle$ is in $R$ while $\langle 4,2\rangle$ is not, and $\langle 3,4\rangle$ is in $R$ while $\langle 4,3\rangle$ is not. These two violations are unrelated. It’s not that $\langle 2,4\rangle$ doesn’t ‘have symmetry with’ $\langle 3,4\rangle$: $R$ would still not be symmetric even if we removed $\langle 3,4\rangle$ from it, because it would still contain $\langle 2,4\rangle$ but not $\langle 4,2\rangle$.
To show that a relation does not have one of these properties you need only find one failure of the property. Here, for instance, the moment you find that $\langle 2,4\rangle\in R$ and $\langle 4,2\rangle\notin R$, you’ve shown that $R$ is not symmetric. You also know that $R$ is not reflexive: the ordered pair $\langle 3,4\rangle$ shows that $4$ is in the field of the relation, but the pair $\langle 4,4\rangle$ is not in $R$.