# Math Help - Relation and Their properties - Composite

1. ## Relation and Their properties - Composite

Can someon show me how to do one of these problems step by step so I have an idea of how to work these problems. I do not know what to do.
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 "equal to" relation

Find:

a.) R1 o R1
b.) R1 o R2
c.) R1 o R3
d.) R1 o R4
e.) R1 o R5
f.) R1 o R6
g.) R2 0 R3
h.) R3 o R3

2. Let’s do part b). Start with an element is the composition.
$(a,b) \in R_1 \circ R_2 \Rightarrow \quad \left( {\exists c} \right)\left[ {(a,c) \in R_2 \wedge (c,b) \in R_1 } \right]$.

Using the definitions of the two relations we get:
$(a,c) \in R_2 \Rightarrow a \ge c~ ~\&~ ~(c,b) \in R_1 \Rightarrow c > b$.

Putting those together we see that $a > b$ or $(a,b) \in R_1$.

Thus the answer is $R_1$.

3. I see Plato. Thanks.

Ima try to work on the others.

4. Let me see if I got this right for part c.

R1 ○ R3 = {(a, b) / (c, b) element R1 and (a, c) element R3}
= {(a, b) / c > b and a < c}
= {(a, b) / a < b}
=
R3

Is this correct?

5. No that is not correct. But I will say that this one is very tricky.
$\begin{gathered}
(x,x + 1) \in R_3 \,\& \,\left( {x + 1,x} \right) \in R_1 \Rightarrow \quad \left( {x,x} \right) \in R_1 \circ R_3 \hfill \\
x < y \Rightarrow \quad \left( {x,y + 1} \right) \in R_3 \,\& \,\left( {y + 1,y} \right) \in R_1 \Rightarrow \quad \left( {x,y} \right) \in R_1 \circ R_3 \hfill \\\end{gathered}$

$x > y \Rightarrow \quad \left( {x,x + 1} \right) \in R_3 \,\& \,\left( {x + 1,y} \right) \in R_1 \Rightarrow \quad \left( {x,y} \right) \in R_1 \circ R_3
$

Does this show that every pair it there?

6. Originally Posted by Plato
No that is not correct. But I will say that this one is very tricky.
$\begin{gathered}
(x,x + 1) \in R_3 \,\& \,\left( {x + 1,x} \right) \in R_1 \Rightarrow \quad \left( {x,x} \right) \in R_1 \circ R_3 \hfill \\
x < y \Rightarrow \quad \left( {x,y + 1} \right) \in R_3 \,\& \,\left( {y + 1,y} \right) \in R_1 \Rightarrow \quad \left( {x,y} \right) \in R_1 \circ R_3 \hfill \\\end{gathered}$

$x > y \Rightarrow \quad \left( {x,x + 1} \right) \in R_3 \,\& \,\left( {x + 1,y} \right) \in R_1 \Rightarrow \quad \left( {x,y} \right) \in R_1 \circ R_3
$

Does this show that every pair it there?
You just confused me here, because I thought it was similar step that you showed me before with the Unions, Intersection, etc when we had R1 U R2 for an example.

I am lost now

7. Composition of relations has very little in common with union, intersection, etc of relations.
Those are just ordinary set operations: membership of pairs.

Whereas, composition of relations (leads to composition of functions) has to do with the existence of two particular pairs implying the existence of a third pair.
$\left( {x,y} \right) \in T \circ S\quad \Leftrightarrow \quad \left( {\exists z} \right)\left[ {\left( {x,z} \right) \in S \wedge \left( {z,y} \right) \in T} \right]$.

Notice that the definition is ‘if and only if’.
Also we work from right to left in the composition, start with the $S$ and then the $T$.