in Set Theory & Algebra recategorized by
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18 votes
18 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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2 Comments

To be a POSET ...  It hav to be reflexive,Antisyammetric and transitive .... So option E .. correct me if i am wrong ....
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"reals" means set of Real numbers.

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3 Answers

26 votes
26 votes
Best answer

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

4 Comments

$(a<b)$ will be Anti-symmetric if

$((a < b) \wedge (b < a)) \implies (a=b)$ is $True$.

Note the implication, here, $F \implies <anything>$ is automagically $True$, and hence the relation is Anti symmetric.
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Antisymmetric and Irreflexive = Asymmetric

Can anybody explain this??

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Asymmetric don't allow flipping as well as same elements.

so here antisymmetric don't allows flipping and irreflexive don't allow same elements thus combining both we get asymmetric.

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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.

3 Comments

Can you explain how it is anti symmeteric?
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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
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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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