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For it to be transitive (1,2) would need to exist in the relation. And it does, because a and b happen to be the same number. But sets don't contain duplicates, so it either is transitive, or it cant be transitive if a and b are the same value.

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  • $\begingroup$ Why would the set need to contain duplicates? Transitivity only insists that $(1,2)$ (and, indeed $(1,1)$, if $a=b=c=1$) are in the set, not that they are in the set twice! $\endgroup$ – Milo Brandt Oct 28 '16 at 3:45
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    $\begingroup$ My confusion started with this video. youtu.be/LLHrSc4g0ZM?t=680 Where she says 1,1 and 1,2 require 1,2 to be in the set to be transitive. She says it isn't and moves on. $\endgroup$ – Grimchester Oct 28 '16 at 3:48
  • $\begingroup$ @naveendankal: Does not seem to be a duplicate. $\endgroup$ – Daniel McLaury Oct 28 '16 at 3:52
  • $\begingroup$ @Grimchester: There is indeed a mistake in the video. $\endgroup$ – Daniel McLaury Oct 28 '16 at 3:53
  • $\begingroup$ Yes I agree on the second look at both the question $\endgroup$ – naveen dankal Oct 28 '16 at 3:54
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A relation is transitive if whenever a~b and b~c we have a~c.

So when do we have both a~b and b~c?

Well, the only relations we have are 1~1 and 1~2. So if b~c then we must have b=1. So then we have a~1, which must mean that a=1. So the only two cases to check are $a=b=c=1$ and $a = b = 1$, $c=2$. That is:

  • Since 1~1 and 1~1, we must have 1~1 if the relation is transitive.
  • Since 1~1 and 1~2, we must have 1~2 if the relation is transitive.

Both of these hold, so the relation is indeed transitive.

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  • $\begingroup$ What about this: {(1,3),(1, 2),(2, 3),(2, 2)} .. it's transitive for (1,2)-(2,3) cause (1,3) exists. But (1,2)-(2,2) requires (1,2) to exist and it doesn't. So is the relation then not transitive, because it fails in 1 instance? $\endgroup$ – Grimchester Oct 28 '16 at 4:07
  • $\begingroup$ "But (1,2)-(2,2) requires (1,2) to exist and it doesn't" What do you mean it doesn't? You just said it did! $\endgroup$ – Daniel McLaury Oct 28 '16 at 4:08
  • $\begingroup$ Haha, oh wow.. /facepalm you're right! That mistake actually helped me solve a problem too :) $\endgroup$ – Grimchester Oct 28 '16 at 4:17
  • $\begingroup$ Am i correct in assuming if it fails for any one instance, the whole relation is not transitive? For Example. {(1,3),(1, 2),(2, 3),(2, 4)} $\endgroup$ – Grimchester Oct 28 '16 at 4:24
  • $\begingroup$ A relation is transitive if, whenever a~b and b~c, we have a~c. So this is non-transitive because 1~2 and 2~4 but we do not have 1~4. $\endgroup$ – Daniel McLaury Oct 28 '16 at 4:25

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