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$$A = \{foo, bar\}$$

$$R_1 = \{(foo, foo)\}$$

$$R_2 = \{(foo, bar)\}$$

Is $R_1$ transitive on $A$? My gut tells me yes, but I'm not sure if transitivity requires $3$ elements.

What about $R_2$? foo R2 bar ^ foo R2 bar => foo R2 bar?

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4 Answers 4

up vote 2 down vote accepted

Yes, it’s transitive. If it weren’t, you’d be able to find $a,b,c\in A$, not necessarily distinct, such that $\langle a,b\rangle\in A$, $\langle b,c\rangle\in A$, and $\langle a,c\rangle\notin A$, and you can’t: the only way to get $\langle a,b\rangle\in A$ and $\langle b,c\rangle\in A$ is to set $a=b=c=\text{foo}$, in which case $\langle a,c\rangle$ is in $A$.

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Transitivity requires three elements, but they need not be distinct. This is one of the many reasons why mathematicians are considered strange by laypeople. We say three, but we really mean three, two or one.

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The relation $R$ is transitive if whenever $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$.

Since $R$ is a singleton there is only one pair so $\newcommand{\foo}{\mathbf{foo}}(\foo,\foo)\in R$ as well $(\foo,\foo)\in R$ and indeed $(\foo,\foo)\in R$.

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Yes it is transitive.

Proof :

defination of Transitivity :if ( aRb and bRC )then { aRc}

here (foo, foo) is aRb..where a=foo and b=foo .

Since there is nothing to represent bRc, the condition ( aRb and bRC ) is false

defination of if...then

if..then

By the defination of if...then , we can say when p is false the answer is true as can be seen from the image.

THUS , The Given Relation is Transitive :)

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