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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 | 1.3k views
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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.

by Veteran (60.5k points)
selected
+4
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 :(
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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.

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The given relation is not even antisymmetric a<b and b<a cant be true simultaneously even  if a=b. Why u considering it to be antisymmetric
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Since < is not antisymmetric and not reflexive then why is option D not the answer.

Oh got it option D means not antisymmetric and "reflexive".....

So yes answer has to be E
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This relation is anti-symmetric.

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.

by Boss (10k points)
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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.
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The given relation is not even antisymmetric a<b and b<a cant be true simultaneously even  if a=b. Why u considering it to be antisymmetric

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.

by Active (3.6k points)