Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a set $\{1,2,3,4\}$, how is the following relation $R$ antisymmetric?

$$R = \{(1, 2), (2, 3), (3, 4)\}$$

Note: Antisymmetric is the idea that if $(a,b)$ is in $R$ and $(b,a)$ is in $R$, then $a = b$. In my textbook it says the above is antisymmetric which isn't the case as whenever $(a,b)$ is in $R$, $(b,a)$ is not.

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

Try this: consider a relation to be antisymmetric, UNLESS there exists a counterexample: unless there exists $(a, b) \in R$ and $(b, a) \in R$, AND $a\ne b$.

Since no such counterexample exists in for your relation, it is trivially true that the relation is antisymmetric.

Another way to put this is as follows: a relation is NOT antisymmetric IF AND ONLY IF there exist $a, b$ such that BOTH $\;(a, b)\in R\;$ AND $\;(b, a) \in R\;$ BUT $\;a\ne b$.

This is true of other properties as well: a property holds for a relation unless there exists a counterexample such that the property fails to hold. Put differently, a property FAILS to hold IF AND ONLY IF a counterexample exists.

share|improve this answer
add comment

You just need to check the cases. You are given a set $A=\{1,2,3,4\}$ and the relation $$\sim\; =\{(1,2),(2,3),(3,4)\}$$

Note that $a\sim b\iff a+1=b$. Thus, it will be never the case that the other pair you're looking for is in $\sim$, and the relation will be antisymmetric because it can't not be antisymmetric, i.e. the truth holds vacuously.

share|improve this answer
    
So it is similar to the idea of implications, if we don't know if the premise is true for sure, the conclusion is always true? –  user1234440 Dec 10 '12 at 19:00
2  
user123440: Not quite. If the premise is never true, then the implication is true, regardless of the truth of the conclusion. For example, consider the implication, "If 2+2=5, then you will pass the course." Since 2+2 is not 5, the statement is a true statement, regardless of whether or not you pass the course. –  Doug Chatham Dec 10 '12 at 19:10
add comment

$R$ is antisymmetric iff whenever both $(a,b)$ and $(b,a)$ are in $R$ then $a=b$.

In your example, there is no pair $(a,b) \in R$ that also has $(b,a) \in R$, so the statement is vacuously true.

Another (equivalent) way of looking at it is that $R$ is not antisymmetric iff there are elements $a,b$ with $a\neq b$ and both $(a,b),(b,a) \in R$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.