Question

For real number x and y, we write xRy⇔x-y+√2 is an irrational number. Is the relation
(a) Equivalence
(b) Partial order

Answer 1

real number is the union of rational and irrational number

set of real number(A)= {$1,2$ , $\sqrt{2}$ )

AXA={

$${(1,1),(1,2)(1,\sqrt{2}),(2,1)(2,2)(2,\sqrt{2}),(\sqrt{2},1)(\sqrt{2},2)(\sqrt{2},\sqrt{2})}$$

}

relation (R) XRY$\iff$$X-Y+\sqrt{2}$ is an irrational

(1,1)=$1-1+\sqrt{2}$ =$\sqrt{2}$ is irrational number

R= {${(1,1),(1,2),(2,1)(2,2),(\sqrt{2},1)(\sqrt{2},2)(\sqrt{2},\sqrt{2})}$ }

R is the set of irrational number

relation R IS Equivalence relation because relation R FOLLOWS the 3 property

1)reflexive: ${(1,1),(2,2),(\sqrt{2},\sqrt{2})}$ xRX

2)symmetric:(1,2)(2,1) XRY and YRX THEN X IS NOT equal to Y

3)transitive : xRy and yRZ then xRz

(1,2),(2,1)(1,1)follows transitivity property

partial order: relation R IS not partial relation because relation R not FOLLOWS the one property (antisymmertic )out of 3 property

1)reflexive ${(1,1),(2,2),(\sqrt{2},\sqrt{2})}$ i.e xRX

2) antisymmetri property shows xRY and YRx then x=y

but relation R IS not antisymmetric because the order of x and y change

(1,2)(2,1) x is not equal to y

3)transitive :xRy and yRZ then xRz

relation (1,2),(2,1)(1,1) follows transitivity property