Answer
a) The union of two relations is the union of these sets. Thus R1 ∪ R3 holds between two real numbers if R1
holds or R3 holds (or both, it goes without saying). Here this means that the first number is greater than the
second or vice versa—in other words, that the two numbers are not equal. This is just relation R6 .
b) For (a, b) to be in R3 ∪ R6 , we must have a > b or a = b. Since this happens precisely when a ≥ b, we
see that the answer is R2 .
c) The intersection of two relations is the intersection of these sets. Thus R2 ∩ R4 holds between two real
numbers if R2 holds and R4 holds as well. Thus for (a, b) to be in R2 ∩ R4 , we must have a ≥ b and a ≤ b.
Since this happens precisely when a = b, we see that the answer is R5 .
d) For (a, b) to be in R3 ∩ R5 , we must have a < b and a = b. It is impossible for a < b and a = b to hold
at the same time, so the answer is Ø, i.e., the relation that never holds.
e) Recall that R1 − R2 = R1 ∩ R2 . But R2 = R3 , so we are asked for R1 ∩ R3 . It is impossible for a > b
and a < b to hold at the same time, so the answer is Ø, i.e., the relation that never holds.
f) Reasoning as in part (f), we want R2 ∩ R1 = R2 ∩ R4 , which is R5 (this was part (c)).
g) Recall that R1 ⊕ R3 = (R1 ∩ R3) ∪ (R3 ∩ R1). We see that R1 ∩ R3 = R1 ∩ R2 = R1 , and R3 ∩ R1 =
R3 ∩ R4 = R3 . Thus our answer is R1 ∪ R3 = R6 (as in part (a)).
h) Recall that R2 ⊕ R4 = (R2 ∩ R4) ∪ (R4 ∩ R2). We see that R2 ∩ R4 = R2 ∩ R1 = R1 , and R4 ∩ R2 =
R4 ∩ R3 = R3 . Thus our answer is R1 ∪ R3 = R6 (as in part (a)).
Work Step by Step
a) The union of two relations is the union of these sets. Thus R1 ∪ R3 holds between two real numbers if R1
holds or R3 holds (or both, it goes without saying). Here this means that the first number is greater than the
second or vice versa—in other words, that the two numbers are not equal. This is just relation R6 .
b) For (a, b) to be in R3 ∪ R6 , we must have a > b or a = b. Since this happens precisely when a ≥ b, we
see that the answer is R2 .
c) The intersection of two relations is the intersection of these sets. Thus R2 ∩ R4 holds between two real
numbers if R2 holds and R4 holds as well. Thus for (a, b) to be in R2 ∩ R4 , we must have a ≥ b and a ≤ b.
Since this happens precisely when a = b, we see that the answer is R5 .
d) For (a, b) to be in R3 ∩ R5 , we must have a < b and a = b. It is impossible for a < b and a = b to hold
at the same time, so the answer is Ø, i.e., the relation that never holds.
e) Recall that R1 − R2 = R1 ∩ R2 . But R2 = R3 , so we are asked for R1 ∩ R3 . It is impossible for a > b
and a < b to hold at the same time, so the answer is Ø, i.e., the relation that never holds.
f) Reasoning as in part (f), we want R2 ∩ R1 = R2 ∩ R4 , which is R5 (this was part (c)).
g) Recall that R1 ⊕ R3 = (R1 ∩ R3) ∪ (R3 ∩ R1). We see that R1 ∩ R3 = R1 ∩ R2 = R1 , and R3 ∩ R1 =
R3 ∩ R4 = R3 . Thus our answer is R1 ∪ R3 = R6 (as in part (a)).
h) Recall that R2 ⊕ R4 = (R2 ∩ R4) ∪ (R4 ∩ R2). We see that R2 ∩ R4 = R2 ∩ R1 = R1 , and R4 ∩ R2 =
R4 ∩ R3 = R3 . Thus our answer is R1 ∪ R3 = R6 (as in part (a)).
