Given R = Set of all real numbers, define the following relations:
R1 = {(a, b) ∈ R^2 | a > b}, the “greater than” relation,
R2 = {(a, b) ∈ R^2 | a ≥ b}, the “greater than or equal to” relation, R3 = {(a, b) ∈ R^2 | a < b}, the “less than” relation,
R4 = {(a, b) ∈ R^2 | a ≤ b}, the “less than or equal to” relation,
R5 = {(a, b) ∈ R^2 | a = b}, the “equal to” relation,
R6 = {(a, b) ∈ R^2 | a != b}, the “unequal to” relation.
Find:
a) R1 ◦ R1 - My answer: R1
b) R1 ◦ R2. - My answer: R1
c) R1 ◦ R3. - My answer: R^2 because if a <= b and b > c, then both a and c might be arbitrary (doubt here)
d) R1 ◦ R4. - My answer: R^2 because if a < b and b > c, then both a and c might be arbitrary (doubt here)
e) R1 ◦ R5. - My answer: R1 because if a = b and b > c, then a > c which is R1.
f ) R1 ◦ R6. - My answer: R^2 because if a != b and b > c, then both a and c might be arbitrary (doubt here)
g) R2 ◦ R3. - My answer: R^2 because if a < b and b >= c, then both a and c might be arbitrary (doubt here)
h) R3 ◦ R3. - My answer: R3 because if a < b and b < c, then then a < c
Please correct me if I went wrong.
Clarifying the doubts:
$R_1 \circ R_3 = \mathbb R^2$ $$\text{Yes, you are right.}$$
$R_1 \circ R_4 = \mathbb R^2$ $$\text{Yes, you are right.}$$
$R_1 \circ R_6 = \mathbb R^2$ $$\text{Yes, you are right.}$$
$R_2 \circ R_3= \mathbb R^2$ $$\text{Yes, you are right.}$$
And the others are also right. Well done!