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19 votes
19 votes

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
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3 Answers

Best answer
28 votes
28 votes

Relation $<$ is :

  • Irreflexive and hence not reflexive
  • Asymmetric and hence antisymmetric and also not symmetric

The relation is not POSET because it is irreflexive.

Condition for Antisymmetric: $\forall a,b \in \mathbb{R}, aRb \neq bRa$ unless $a=b.$

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
  • not asymmetric because $(1,1)$ is present
  • but is anti-symmetric. 

Antisymmetric and Irreflexive $=$ Asymmetric

Correct Option E.

edited by
5 votes
5 votes

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.

4 votes
4 votes

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.

Answer:

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