Question

Let R1 and R2 be two relation on real number such that R1 = {(x, y)|x < y} and R2 =

{(x, y)|x > y}, then find R1 ∪ R2,R1 ∩ R2,R1 − R2, R2 − R1, and R1

LR2.

Answer 1

Solution

Given that

${R_1} = \left\{ {\left( {x,y} \right)|\,x < y} \right\}$

It means that there are ordered pairs like (1, 2), (2, 3), (1, 3), (3, 4), ….

And

${R_2} = \left\{ {\left( {x,y} \right)|\,x > y} \right\}$

It means that there are ordered pairs like (2, 1), (3, 2), (4, 3), (5, 1), …

Solution (a)

${R_1} \cup {R_2} = \left\{ {\left( {x,y} \right)|\,x \ne y} \right\}$

It will contain both the sets and therefore all the elements of both the sets ${R_1}$ and ${R_2}$ and its elements will be of the form $(x, y)$ such that either $x > y$ or $y > x$ , or we can say $x \ne y$

Solution (b)

{R_1} \cap {R_2} = \phi \

An empty set. This is because in set ${R_1}$the elements are of the form $(x, y)$ such that $y > x$

and in the set ${R_2}$the elements are of the form $(x, y)$ such that $x > y$, so there is nothing common in between both the sets.

Solution (c)

{R_1} - {R_2} = {R_1}\

We know in set ${R_1}$the elements are of the form $(x, y)$ such that $y > x$ and in the set ${R_2}$the elements are of the form $(x, y)$ such that $x > y$, so there is nothing common in between both the sets. So when we subtract the second set from the 1st one, nothing will be subtracted and the 1st set will apear as it is.

Solution (d)

${R_2} - {R_1} = {R_2}$

We know in set ${R_1}$the elements are of the form $(x, y)$ such that $y > x$ and in the set ${R_2}$the elements are of the form $(x, y)$ such that $x > y$, so there is nothing common in between both the sets. So when we subtract the first set from the second one, nothing will be subtracted and the second set will apear as it is.