Solution to For real number x and y, we write xRy⇔x-y+√2 is an irrational number. Is the … - Sikademy
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Archangel Macsika

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

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real number is the union of rational and irrational number

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




relation (R) XRY\iffX-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

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