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Suppose $A = \{ a , b , c \} $ and relation $R = \{ (a,b) , (a,c) \} $ is defined . Does the relation $R$ have transitive property ? Why ?

Note : I think it doesn't but my teacher told me that it does and we can prove it by logic easily . After that he gave me some explanation but I didn't grasp it .

S.H.W
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  • Transitivity holds IF $(a,b)\land(b,c)\rightarrow (a,c)$ in the given relation there is no $(b,c)$ you can't say $R$ has not transitivity, but I don't understand the reason why your teacher says that $R$ has transitivity. – Raffaele Jul 09 '17 at 17:59
  • It is indeed transitive. Do you remember what the transitive property says? It says if we can find some elements $x,y,z$ such that $(x,y)$ and $(y,z)$ are both in the relation then so too must $(x,z)$ be in the relation. Not being able to find any $x,y,z$ at all where $(x,y)$ and $(y,z)$ are in the relation does not stop it from being transitive, rather it makes it transitive vacuously. – JMoravitz Jul 09 '17 at 17:59
  • Did your teacher talk about vacuous truth or something like the principle of explosion? – autodavid Jul 09 '17 at 18:01
  • If you think of relations as (directed multi-)graphs with arrows leading from one element to another if they are related and loops if an element is related to itself, transitivity is the property that "if you can get from point $a$ to point $b$ in potentially multiple steps by traveling along the arrows in their respective directions, then you must also be able to get from point $a$ to point $b$ in a single step." (similarly, reflexivity is there is a loop on every vertex, symmetry is all arrows are doublesided, and antisymmetry is all arrows are singlesided) – JMoravitz Jul 09 '17 at 18:02
  • @autodavid I think he used the truth table . – S.H.W Jul 09 '17 at 18:05
  • @JMoravitz Can you show an example ? – S.H.W Jul 09 '17 at 18:06
  • @S.H.W. https://math.stackexchange.com/questions/1475354/can-a-relation-be-both-symmetric-and-antisymmetric-or-neither/1475381#1475381 In my example in my answer in the linked question, the relation is $R={(1,1),(1,2),(2,3),(3,2)}$ on the set ${1,2,3}$ and is not transitive because $(2,3)$ and $(3,2)$ are in the relation but $(2,2)$ is not. Similarly $(3,3)$ is not. I.e., you could travel from $2$ to $2$ in my example, but it would take more than one step. Had it been transitive, you should have been able to make such a trip in exactly one step. – JMoravitz Jul 09 '17 at 18:08
  • The logical statement $(P \rightarrow Q)$ holds whenever $P$ is false (maybe a bit weird). This could be seen by truth table. Now take $P:=(a,b)\in R \wedge(b,c)\in R$, which never happens, and take $Q :=(a,c)\in R$. Thus $P\rightarrow Q$ holds, ie, the def of transitivity holds. – autodavid Jul 09 '17 at 18:18

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