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The less-than relation, $<,$ on reals is

1. a partial ordering since it is asymmetric and reflexive
2. a partial ordering since it is antisymmetric and reflexive
3. not a partial ordering because it is not asymmetric and not reflexive
4. not a partial ordering because it is not antisymmetric and reflexive
5. none of the above
edited | 852 views
0
To be a POSET ...  It hav to be reflexive,Antisyammetric and transitive .... So option E .. correct me if i am wrong ....

Relation $<$ is :

1. not reflexive
2. Irreflexive
3. not symmetric
4. Asymmetric
5. Anti symmetric

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.
as $\{(1,1),(1,2)\}$

Not Asymmetric and Irreflexive = Antisymmetric
Option E.

selected
+2
given relation is not Poset and reason it it is not reflexive..
it has nothing to do with asymmetry.
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Sir, please let me know why option 'C' cannot be the answer. I'm confused :(

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.

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Can you explain how it is anti symmeteric?
+1
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.
+1 vote

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.