Kenneth Rosen Edition 6th Exercise 7.1 Question 35 (Page No. 473)

Question

Let $R_1,R_4,R_6$ be relations on the set of real numbers to the set of real numbers

$R_1=\{(a,b) \in R^2 \, | \, a>b\}$

$R_4=\{(a,b) \in R^2\, | \, a \leq b\}$

$R_6=\{(a,b) \in R^2 \, | \, a \neq b\}$

Find

(d) $R_4 o R_1$

(g)$R_4 o R_6$

Answer 1

d : $R_4\circ R_1$ would have Those pairs $(a,b)$ such that For some $c$, $(a,c) \in R_1$ and $(c,b) \in R_4$. So, Apply this for Question d.  

$(a,c) \in R_1$ Iff $a > c$, and $(c,b) \in R_4$ iff $c \leq b$, For some $c$. So, $(a,b)$ will belong to $R_4\circ R_1$ iff $a > c \,\,and\,\, b \geq c$ For some $c$. So, All the pairs of $R^2$ will satisfy this.

Hence, $R_4\circ R_1$ $=$ $R^2$.

g : Similar logic for  $R_4\circ R_6$,  $R_4\circ R_1$ would have Those pairs $(a,b)$ such that For some $c$, $(a,c) \in R_6$ and $(c,b) \in R_4$. So, Apply this for Question g.

$(a,c) \in R_6$ Iff $a \neq c$, and $(c,b) \in R_4$ iff $c \leq b$, For some $c$. So, $(a,b)$ will belong to $R_4\circ R_6$ iff $a \neq c \,\,and\,\, b \geq c$ For some $c$. So, All the pairs of $R^2$ will satisfy this.

Hence, $R_4\circ R_6$ $=$ $R^2$.

Comments

I got to understand your reasoning deepak but I have a doubt

I want to take a small example and understand this better

Let my domain be D={1,2,3}

Yes, this domain is a subset of $R$.

on top of this domain D I build $R_4$ and $R_1$

$R_4=\{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3) \}$

$R_1=\{(2,1),(3,1),(3,2)\}$

Now I try to build $R_4 \circ R_1=\{(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$ but this ain't equal to $D^2$.

Then how your above reasoning works?

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Then how your above reasoning works?

Because of Set of real numbers. Since it is set of All real numbers, you can always say that there will be some element less/greater than your element. That is why my reasoning works. You can take whatever pair of real numbers, you can always satisfy the condition of both $d$ and $g$ of "$for$ $some$ $c$". But once you make the set Finite, these condition may not satisfy. 

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ohhhh... In that case I should never solve these questions like that.

Thanks for the info :)

Okay and tell me one more thing,

like in above two relations when we do composition,then the new relation so formed will have the conditions as the conjunction of the relations whose composition is taken?

Like for (d) part the composition relation formed has it's condition ANDed over conditions of two relations over which composition was taken right?

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Like for (d) part the composition relation formed has it's condition ANDed over conditions of two relations over which composition was taken right?

Yes. For solving R1oR2 ...Always go using this method :

R1oR2 will have those pairs $(a,c) for which there exists some $b$ such that $$(a,b) \in R2$ , and $(b,c) \in R1$...

Answer 2

$d)$ which holds relation $> ,\leq$ i.e. $> ,=,<$

So, it contains any sets . So, it holds equivalence property

$g)$ which holds only $<$ property , which is not reflexive, not symmetric but transitive. Also it holds irreflexive and asymmetric relation properties