The less-than relation, $<,$ on reals is
Relation $<$ is :
Relation is not POSET because it is irreflexive.
Check AntiSymmetry: $aRb \neq bRa$ unless $a=b.$
A relation may be 'not Asymmetric and not reflexive' but still Antisymmetric.
Not Asymmetric and Irreflexive = Antisymmetric
Definition:A binary relation is a POSET if it is reflexive,antisymmetric and transitive.
Here, "<" is not a POSET because (1,1) is not possible(we can't say 1<1).
By definition of a POSET, we have nothing to do with the asymmetry.
Hence, (E) is the answer.
This relation is anti-symmetric.
Kindly see this link: https://math.stackexchange.com/questions/255683/antisymmetric-relations
Since "<" relation neither reflexive nor patialy ordering on set of real number.
But it is anti Symmetric relation.Therefor Option E will be appropriate option for it.
definitely the less than relation won't be reflexive bcz a<a is always false for every a belongs to real number.
now for antisymmetric (a<b & b<a) implies a=b , as we know if a<b then b<a can never be true at once therefore (a<b & b<a) results false. but false can implies anything therefore it is antisymmetric.
for a relation to be POSET: it must be reflexive, antisymmetric and transitive
here it is not reflexive but antisymmetric in nature therefore only E option is correct.