The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+11 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
asked in Set Theory & Algebra by Veteran (59.6k points)
edited by | 925 views
To be a POSET ...  It hav to be reflexive,Antisyammetric and transitive .... So option E .. correct me if i am wrong ....

3 Answers

+15 votes
Best answer

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.

answered by Veteran (55.5k points)
selected by
given relation is not Poset and reason it it is not reflexive..
it has nothing to do with asymmetry.
Sir, please let me know why option 'C' cannot be the answer. I'm confused :(

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.

Hence, (E) is the answer.

+4 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.

answered by Loyal (9.1k points)
Can you explain how it is anti symmeteric?
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.
+2 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.

answered by Active (3.5k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

42,658 questions
48,639 answers
63,953 users