if r2 is in the set of N*N ( natural numbers) with (X,y) in the subset of r2, if and only if x+y=0
is it reflexive? is it symmetric? is it anti symmetric? is it Transitive?
i said it is reflective because 0+0 =0
but i'm a little stuck beyond that. the only relation in the set would be (0,0) correct?
$\endgroup$1 Answer
$\begingroup$If I understand this correctly, you have a relation $R_2\subseteq\Bbb N\times\Bbb N$ defined by $\langle x,y\rangle\in R_2$ if and only if $x+y=0$. If this is correct, then $\langle 0,0\rangle$ is the only element of $R_2$, but that does not make $R_2$ reflexive: in order for $R_2$ to be reflexive, $\langle n,n\rangle$ would have to be in $R_2$ for every $n\in\Bbb N$, which is clearly not the case. For instance, $\langle 1,1\rangle\notin R_2$, even though $1\in\Bbb N$. Thus, $R_2$ is not reflexive. $R_2$ is both symmetric and transitive, however,
Symmetry: Is it true that if $\langle x,y\rangle\in R_2$, then $\langle y,x\rangle\in R_2$ as well? Yes: if $\langle x,y\rangle\in R_2$, then $x=y=0$, so $\langle y,x\rangle=\langle x,y\rangle=\langle 0,0\rangle\in R_2$.
Transitivity: Is it true that if $\langle x,y\rangle\in R_2$ and $\langle y,z\rangle\in R_2$, then $\langle x,z\rangle\in R_2$? The answer is again yes, and the proof is similar; can you complete it?
