Answer to Question #117146 in Discrete Mathematics for Priya

Question #117146
For real number x and y, we write xRy⇔x-y+√2 is an irrational number. Is the relation
(a) Equivalence
(b) Partial order
1
Expert's answer
2020-06-11T16:49:25-0400

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



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS