Question: Let $A$ be $\{a,b,c\}$. Let the relation $R$ be $A \times A$ - $\{(a,a), (a,b), (b,b), (c,a)\}$ Which of the following statements about $R$ is true?
a. $R$ is reflexive, is symmetric, and is transitive.
b. $R$ is reflexive, is symmetric, and is not transitive.
c. $R$ is not reflexive, is not symmetric, and is not transitive.
d. $R$ is not reflexive, is symmetric, and is not transitive.
e. $R$ is reflexive, is not symmetric, and is transitive.
I understand why the relation is neither reflexive nor symmetric but as for whether it's not transitive I kind of understand why its not but cannot explain it.
For a relation to be transitive $(a R b) ∧ (b R z) → (a R z)$ but in the context of the question this becomes $(a R b) ∧ (b R b) → (a R b)$ however since I since I already used the ordered pair $(a R b)$ in the premise I understand I can't use it again so that being said how do I explain why it is not transitive?
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CalciumTablet
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2 Answers
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There is no problem with the one you are discussing (note that the elements of $A\times A-\{(a,a),(a,b),(b,b),(c,a)\}$ are all the elements of $A\times A$ without those specified after the $"-"$, so you do not even have $(a,b)$ or $(b,b)$ in $\mathcal{R}$).
Rather note that you have $(c,b)$ and $(b,c)$ in $\mathcal{R}$ but not $(c,c)$.
Balloon
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In a similar question the relation is A x A - {(b,b),(c,b).} So how would I go explaining why this relation is transitive? I understand what you are showing I should have rephrased the question so it specifies just what I was showing. – CalciumTablet Jan 02 '18 at 14:34
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In this case it is also not transitive: since in $A\times A-{(b,b),(c,b)}$ you have $(a,b)$ and $(b,c)$ but not $(c,b)$, the transitivity cannot be guaranteed. – Balloon Jan 02 '18 at 14:38
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Note that $(c,b) \in R, (b,b)\in R $ and $(c, b) $ is also in R. Finished. – Jan 02 '18 at 14:38
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@baloown What do you mean the transitivity cannot be guaranteed? – CalciumTablet Jan 02 '18 at 14:53
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I mean the relation is not transitive. – Balloon Jan 02 '18 at 14:53
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@Baloown Oh so would the explanation be more in the sense that you can never have two of the same ordered pairs in a relation and so transitivity can never be satisfied? – CalciumTablet Jan 02 '18 at 14:59
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In my first comment there is an error, we should read "you have $(b,a)$ and $(a,b)$ but not $(b,b)$". No, relations are sets and in sets the repetition of elements doesn't count: ${a,a}={a}$. Here the this relation is not transitive since you have $b\mathcal{R}a$ and $a\mathcal{R}b$, but no $b\mathcal{R}b$. – Balloon Jan 02 '18 at 15:09
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$(c,a) \in R$, $(a.b) \in R$ but $(c,b)$ not in $R$