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$I.$ is not reflexive

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i did very silly mistake while checking for 2 i checked reflexive and symmetric but for poset it should be antisymmetric which is true here :(

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Happens :|
for quick revision

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have you givem CBT ? @Mk Utkarsh

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No

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@Mk Utkarsh

I ask you very silly question but i have to clear my doubt :p

{(1,1),(2,2) , (3,3)}  ---> this is POSET

(3,3) R (1,1) --- > but his is not even hold the 2nd condition right ??

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what's the approach that's we have to follow ?

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but (1,1)R(3,3) will hold

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or you can interchange postions of (1,1) and (3,3) but still it will be anti-symmetric

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but it also accept

(1,3) R (3,1) nah ?? which is not valid here ?

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why not valid? (am i missing something :|)

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(1,3) R (3,1) is not closed under anti-symmetric matrix right ???

but here The statement II accept that relation too which is not correct right ?

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anti-symmetric*

(1,3) R (3,1) is closed because 1 < 3

but (3,1) R (1,3) is not in relation as it is a anti-symmetric

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Now i got it completely thanks

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Magma you made me question myself :p

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xD

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@jatin khachane 1 the this statement must be false too

Round robin sheduling algorithm always five better performance compared to FCFS sheduling algorithm