Determine whether the following relations are reflexive, symmetric and transitive.
$a) A = {2, 3, 4} R = {(2, 2), (3, 3), (4, 4), (2, 3), (3, 4)}$

$b) R = {(x, y) : y = x + 5 & x < 4; x, y \in R}$

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Solution

$a) A = {2, 3, 4} R = {(2, 2), (3, 3), (4, 4), (2, 3), (3, 4)}$

If the relation is reflexive, then $(a, a) \in R \forall a \in {2, 3, 4}$
Since, $(2, 2) \in R, (3, 3) \in R & (4, 4) \in R$
$∴ R$ is reflexive.

For symmetric relation,
If $(a, b) \in R,$ then $(b, a) \in R$
Here, $(2, 3) \in R$ but $(3, 2) \notin R$ and also $(3, 4) \in R$ but $(4, 3) \notin R$
$∴ R$ is not symmetric.

For transitive relation,
If $(a, b) \in R$ and $(b, c) \in R,$ then $(a, c) \in R$
Here, $(2, 3) \in R$ and $(3, 4) \in R$ but $(2, 4) \notin R$
$∴ R$ is not transitive.

$b) R = {(x, y) : y = x + 5 & x < 4; x, y \in R}$
Clearly, we can see that for every value of $x, y$ will be greater than $x$ by $5$ . So, $(a, a) \notin R \forall a \in {R}$
$∴ R$ is not reflexive.

Since, $(x, x + 5) \neq (x + 5, x)$ . So, $(a, b) \in R,$ but $(b, a) \notin R$
$∴ R$ is not symmetric.

And it can not be transitive also because
Say $a_{1} = x a_{2} = x + 5 a_{3} = x + 5 + 5 = x + 10$
$(a_{1}, a_{2}) = (x, x + 5) \in R$
$(a_{2}, a_{3}) = (x + 5, x + 10) \in R$
But $(a_{1}, a_{3}) = (x, x + 10) \notin R$

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