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

R1$\cup$ R2={(x,y)|x>y or x<y}={(x,y)|x!=y}

R1$\cap$ R2={(x,y)|x>y and x<y}=$\varnothing$

R1-R2=R1-R1$\cap$ R2=R1 -{(x,y)|x>y and x<y}=R1-$\varnothing$=R1

R2-R1=R2-R1$\cap$ R2=R2 -{(x,y)|x>y and x<y}=R2-$\varnothing$=R2

R1+LR2=R1$\cup$ LR2-R1$\cap$ LR2={(х,у)$|\forall x \in\R,\forall y \in \R$}-∅={(x,y)$|\forall x \in\R,\forall y \in \R$ }