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+7 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 (67.5k points) | 507 views

3 Answers

+10 votes
Best answer
relation less than is :
a. not Reflexive
b. Irreflexivew
b. not symmetric
c. Asymmetric
d. Anti symmetric

relation is not POSET because it is irreflexive.
check AntiSymmetry..
aRb != bRa unless a=b.

A relation may be 'not Asymmetric and not reflexive' bt still Antisymmetric.
as {(1,1) (1,2)}

not Asymmetric and Irreflexive = Antisymmetric
Option E
answered by Veteran (48.3k points)
selected by
given relation is not Poset and reason it it is not reflexive..
it has nothing to do with asymmetry.
+2 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 Boss (6.1k 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.
+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.

answered by Active (1.9k points)

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