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My book says the answer is B. But how. I understand that the relation is reflexive as (a,a) belongs to R for all a belonging to the given set. Furthermore I understand that the relation is not reflexive as (a,b) belongs to R does not imply that (b,a) belongs to R for all a,b belonging to the given set. What I don't understand is how is the relation transitive ? I mean they have given (1,3) and (3,2) and (1,2) in the relation which makes it transitive for one case. But doesn't transitivity mean that (a,b) belongs to R and (b,c) belongs to R implies that (a,c) belongs to R for all a,b,c belonging to the given set. Isn't the for all violated here? In transitivity? So shouldn't it be not transitive ? Or am I missing something.

EDIT

My book's statement:

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2 Answers 2

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For it not to be transitive, you need an a,b,c such that (a,b) and (b,c) are in R, but (a,c) is not (a,b,c need not be different). Are seeing any instance of this? Let's see. Here are all pairs of pairs (a,b) and (b,c) in R:

(1,1) and (1,1)

(1,1) and (1,2)

(1,1) and (1,3)

(1,2) and (2,2)

(2,2) and (2,2)

(3,2) and (2,2)

(1,3) and (3,2)

(3,3) and (3,2)

(1,3) and (3,3)

(3,3) and (3,3)

Now, let's see if the (a,c) pair is in R in all those cases as well:

(1,1) and (1,1) => (1,1) Yes!

(1,1) and (1,2) => (1,2) Yes!

(1,1) and (1,3) => (1,3) Yes!

(1,2) and (2,2) => (1,2) Yes!

(2,2) and (2,2) => (2,2) Yes!

(3,2) and (2,2) => (3,2) Yes!

(1,3) and (3,2) => (1,2) Yes!

(3,3) and (3,2) => (3,2) Yes!

(1,3) and (3,3) => (1,3) Yes!

(3,3) and (3,3) => (3,3) Yes!

Yes, they all are ... so it is transitive.

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  • $\begingroup$ But shouldn't it be transitive for the other elements of the set as well? There is a for all isn't there? I can see that it's not not transitive. But it isn't transitive for all.. $\endgroup$ Mar 19, 2017 at 16:50
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    $\begingroup$ Transitive says "if (a,b) and (b,c) in R, then (a,c) in R" ... if the if is not satisfied, then it's ok! So, for example, we have (1,3) in R but we don't have (3,4) in R ... So, we don't need (1,4) in R either. $\endgroup$
    – Bram28
    Mar 19, 2017 at 16:52
  • $\begingroup$ @KunalPawar So which pair(s) do you think is/are missing? And by the way, if it's not not transitive, then it is transitive. $\endgroup$
    – Bram28
    Mar 19, 2017 at 16:54
  • $\begingroup$ I think I'm confused on the for all statement :| $\endgroup$ Mar 19, 2017 at 16:54
  • $\begingroup$ The for all has led me to believe that every element in the set on which the relation is defined should have such a pair which shows transitivity. I'm still confused. I'll edit the question to show the statement my book has. $\endgroup$ Mar 19, 2017 at 16:56
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Your book defines transitive to be, if (a implies b) and (b implies c) then (a implies c).

It seems easy to be confused from the fact that there is no such case in the relation where the above statement occurs. In this instance the proof of transitivity is known as the vacuous proof. Also known as a vacuous truth. This occurs when we have a statement If P-->then Q, when P is known to be false.

Take for example the statement,

----"All elephants you own have rainbow stripes",

Explanation: When it is known you own no elephants. This statement is still true.

A more practical example is given by the wikipedia article on "Vacuous truth", which states,

----For example, the statement "all cell phones in the room are turned off" will be true whenever there are no cell phones in the room. In this case, the statement "all cell phones in the room are turned on" would also be vacuously true, as would the conjunction of the two: "all cell phones in the room are turned on and turned off"."

In the case of the given relation. It is known we have no (a1,a2,) in R such that (a2,a1) is in R. So the transitivity is true.

Hope that helps(:

--Link to wikipedia article on vacuous truth https://en.wikipedia.org/wiki/Vacuous_truth

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