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As we knows that Transitive relation R is (a, b) € R and (b, c) € R implies (a, c) € R

So we can suppose that let a be 1 and b be 2 and c be 1 Then, (a, b) € R means (1,2) € R and (b, c) € R means (2,1) € R implies (a, c) € R means (1,1) € R

Then it must be transitive .... But why isn't? Plz plz tell me fastly

Anton Vrdoljak
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5 Dots
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2 Answers2

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The relation isn't transitive because while $(2, 1), (1, 2)$ are in the relation, $(2, 2)$ is not.

Niki Di Giano
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In order to be a transitive relation on some set X, R has to satisfy: $$a\text{R}b \quad \text{and} \quad b\text{R}c \quad \Rightarrow \quad a\text{R}c, \quad \text{for each} \quad a,b,c \in X.$$

In your example, we can take $a=2, b=1, c= 2$, and it is obvious that $a$ is in relation with $b$ and that $b$ is in relation with $c$. But, it is also obvious that $a$ is not in relation with $c$, so your relation is not transitive.

Anton Vrdoljak
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  • Hey buddy but we can also take a =1, b = 2 and c = 1,then we gotta our tran. Relation – 5 Dots Mar 09 '20 at 05:01
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    To @5Dots : an accent is on for each $a,b,c \in X$... – Anton Vrdoljak Mar 09 '20 at 07:35
  • The constraint must hold for all $a$, $b$, and $c$. Thus the relation is not transitive if there exists a counter example: where $(a R b)\wedge (b R c) \wedge \neg(a R c)$. $$(2 R 1)\wedge(1 R 2)\wedge\neg(1 R 1)$$ – Graham Kemp Mar 13 '20 at 04:12