Is the below relation transitive?

Question

Is the below relation transitive? The theorem says if (a,b) and (b,c) belong to a relation, then (a,c) also does.

I am confused because (3,2) and (2,3) is like (a,b) with (b,a) and (3,3) like (a,a), and something is wrong here, or the relation or the theorem

$$R\:=\left\{\left(3,2\right),\:\left(2,3\right),\:\left(3,3\right)\right\}\:$$

You're confusing "symmetric" with "transitive". Your first statement is the transitive property, but when you write "I am confused", it looks like you're testing a transitive property by setting up a test for symmetry. — Adrian Keister, Apr 25 '18 at 17:33
I agree that the above relation is symmetric, but it that transitive as well ? — csandreas1, Apr 25 '18 at 17:37
You have to check every possible choice of $(a,b)$ and $(b,c)$ to make sure that $(a,c)$ is also in the set. — Adrian Keister, Apr 25 '18 at 17:40
i know that dude. My case is that the above relation is somehow confusing. I am not sure if it satisfies that theorem. — csandreas1, Apr 25 '18 at 17:41

Adrian Keister · Accepted Answer · 2018-04-25T17:52:54.143

1

Table for checking: $$ \begin{array}{|c|c|c|c|}\hline (a,b) &(b,c) &(a,c) &(a,c)\in\text{set?} \\ \hline (3,2) &(2,3) &(3,3) &\text{yes} \\ \hline (2,3) &(3,2) &(2,2) &\text{no} \\ \hline (2,3) &(3,3) &(2,3) &\text{yes} \\ \hline (3,3) &(3,2) &(3,2) &\text{yes} \\ \hline (3,3) &(3,3) &(3,3) &\text{yes} \\ \hline \end{array} $$ These are all the possibilities, because the second position of the first ordered pair must equal the first position of the second ordered pair. Since the second line fails to be in the set, the relation is not transitive. I think if you added $(2,2)$ to the relation, it would be transitive.

edited Apr 25 '18 at 17:52

answered Apr 25 '18 at 17:47

Adrian Keister

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so my relation is transitive? – csandreas1 Apr 25 '18 at 17:55
It is not transitive, because, while $(2,3)$ and $(3,2)$ are in the set, $(2,2)$ is not. – Adrian Keister Apr 25 '18 at 17:56

Is the below relation transitive?

1 Answers1