Question

3. (9.1.35) Consider the following relations on ℝ:

R1 = {(a, b) ∈ ℝ2 | a > b}, the “greater than” relation,

R2 = {(a, b) ∈ ℝ2 | a ≥ b}, the “greater than or equal to” relation,

R3 = {(a, b) ∈ ℝ2 | a < b}, the “less than” relation,

R4 = {(a, b) ∈ ℝ2 | a ≤ b}, the “less than or equal to” relation,

R5 = {(a, b) ∈ ℝ2 | a = b}, the “equal to” relation,

R6 = {(a, b) ∈ ℝ2 | a ≠ b}, the “unequal to” relation.

Find the following sets. You don’t have to show your work.

1. R1 ∪ R6

2. R3 ∩ R5

3. R2 ∘ R4

Answer 1

a) R1 U R6 = { (a,b) in R² | a $\neq$ b } = R6

Explanation : for any arbitrary, (a,b) in R², either, a = b or a > b or a < b.

In R1, a > b & in R6, a $\neq$ b, i.e. a > b or a < b, so, combining them, we get, a > b or a < b , i.e. a $\neq$ b

So, R1 U R6 = R6

b) R3 $\cap$ R5 = { } , the empty set.

Explanation : in R3, a < b & in R5, a = b, which are two disjoint cases (as explained in (a) )

So, their intersection is the common part, which is empty.

c) R2 ° R4 = { (a,c) in R² | there exists b in R such that (a,b) belongs to R2 & (b,c) belongs to R4 }

= { (a,c) in R² | there exists b in R such that, a $\geq$ b & b $\leq$ c }

= { (a,c) in R² | there exists b in R such that, a $\geq$ b $\leq$ c }

= R²

Explanation : suppose, any arbitrary (p,q) belongs to R².

Then, either, p = q or p > q or p < q

If, p = q, then there exists p in R such that, (p,p) is in R2 & (p,q) is in R4.

So, in this case, R2 ° R4 = R²

If, p > q, then, there exists b = (p+q)/2 such that,

( p, (p+q)/2 ) is in R2 & ( (p+q)/2 , q ) is in R4.

So, in this case, R2 ° R4 = R²

If, p < q, then, there exists, b = (p+q))2 such that,

( p, (p+q)/2 ) is in R2 & ( (p+q)/2 , q ) is in R4.

So, in this case, R2 ° R4 = R²

So, in all cases exhausted, we have concluded, R2 ° R4 = R²