For asymmetric we have a stronger requirement excluding the unless part from the antisymmetric requirement. i.e., $\forall a,b \in \mathbb{R}, aRb \neq bRa$

A relation may be 'not Asymmetric and not reflexive' but still Antisymmetric. Example: $\{(1,1),(1,2)\}$ over the set $\{1,2\}$ which is

not reflexive because $(2,2)$ is not present

not irreflexive because $(1,1)$ is present

not symmetric because $(1,2)$ is present and $(2,1)$ is not

A binary relation $R$ is anti symmetric if for two elements $x$ and $y$, $xRy$ is true but $yRx$ is not. The binary relation < is anti symmetric because for $x$ and $y$, if $x<y$ is true, then certainly $y<x$ is not.

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.