36 questions · timed · auto-graded
R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)}
Reflexive and transitive but not symmetric.
$\therefore$
R can be written as ordered pair as R = {(2, 4), (3, 3), (4, 4)}As,
$(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$ So, R is a reflexive relation. Also, $(\text{a, b})\in\text{R}$ and $(\text{b, a})\in\text{R}$ So, R is a symmetric as well And, $(0,1)\in\text{R}$ but $(1,2)\notin\text{R}$ and $(2,3)\notin\text{R}$ So, R is not a transitive relation.xy is square of an integer, $\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), \$8, 2), (16, 1), (1, 16), .....\big\}$
$$This is reflexive as (1, 1), (2, 2), .... are present.
This is also symmetric because if aRb ⇒ bRa, for all $\text{a, b}\in\text{N.}$
This is transitive also because if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{N.}$
$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3),\\ (4, 6), (6, 4), (5, 5)\big\}$
$$This is not reflexive as (1, 1), (2, 2), .... are absent.
This only follows the condition of symmetric set as $(1,9)\in\text{R}$ also $(9,1)\in\text{R}$ Similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9), (9, 1)} $\in\text{R}$ but (1, 1) is absent.
The equivalence classes can be taken as [0], [1].
Note that, $\text{for}\ 0\leq\text{i}\leq1,$ [i] = {2n + i: $\text{n}\in\text{Z}$}
So equivalence class [0] = {2n: $\text{n}\in\text{Z}$}
It is clear that all the elements of equivalence class [0] are even.
Hence, equivalence class $[0]=\{0,\pm2,\pm4,\pm6\ ...\}$
$\text{x}>\text{y, x, y}\in\text{N}$
$(\text{x, y})\in\big\{(2, 1), (3, 1),..., (3, 2), (4, 2),....\big\}$
$$This is not reflexive as (1, 1), (2, 2), .... are absent.
This is not symmetric as (2, 1) is present but (1, 2) is absent.
This is transitive as
$(3,2)\in\text{R}$ and $(2,1)\in\text{R}$ also $(3,1)\in\text{R},$ similarly this property satisfies all cases.$\text{x} + 4\text{y} = 10, \ \text{x, y}\in\text{N}$, $$
$(\text{x, y})\in\big\{(6, 1), (2, 2)\big\}$
$$This is not reflexive as (1, 1), (2, 2), .... are absent.
This is also symmetric because
$(6,1)\in\text{R}$ but (1, 6) is absent.This is not transitive as there are only two elements in the set having no element common.