$R_1 = \{(a,b)$ such that $a \leq b \}$
$R_2 = \{(a,b)$ such that $a>b \}$
$R_3 = \{(a,b)$ such that $a=b$ or $a=-b \}$
$R_4 = \{(a,b)$ such that $a=b \}$
$R_5 = \{(a,b)$ such that $a=1+b \}$
$R_6 = \{(a,b)$ such that $a+b <=3 \}$
Those are the relations,
and these are the pairs that are given:
(1,1),(1,2),(2,1),(1,-1), and (2,2)
May someone please explains why $R_3$ and $R_4$ symmetric ?
if we look at the pairs , only (1,2) and (2,1) if they're in the relation then the relation is symmetric , but if we apply those to $R_3$, $R_4$ it won't be there since $1 \neq 2$ and $2 \neq 1$
In order for the relation to be symmetric, take a point $(a, b) \in R$. Is the point $(b, a) \in R$ as well? If so, the relation is symmetric. Note that I am taking an arbitrary point in $R$, so I apply the universal quantifier.
The relation $R$ is symmetric if $\forall{(a, b)} \in R$, $(b, a) \in R$.
So for $R_{1}$, take $(1, 2)$. That satisfies the condition that $1 \leq 2$, right? Now is $(2, 1) \in R_{1}$? The answer is no, since $2 \not \leq 1$. Does that make sense? You apply the same logic for $R_{2}$, which is the complement of $R_{1}$.
Now look at $R_{3}$. Take $(a, b) \in R_{3}$. So we have $a = -b$. Isn't it true then that $b = -a$? So is $(b, a) \in R_{3}$? The answer is yes.
$R_{4}$ is the standard equivalence relation- the equality operator. So symmetric is pretty obvious.
Are you able to figure out why $R_{5}$ and $R_{6}$ are not symmetric?
Hope this helps some!