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Consider $A$ is a relation de fined on $R$ (real numbers) where $A = \{(a,b):|a-b|<4, a, b \in R\}$. Prove/disprove $A$ is transitive.

I know if $|a-b|<4$ and $|b-c|<4$, then, $|a-c|<4$ ; A is transitive. Can I directly prove it with any counter example such as for $a=6$, $b=3$ and $c=1$ this relation is not transitive because $|6-1|>4$.

Is this suitable for prove or disprove questions of relations?

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Your answer is correct. In general a relation $A$ is transitive if for all $a,b,c$ we have $(a,b)\in A$ and $(b,c)\in A$ implies $(a,c)\in A$. That means that if we can find just one counter example (such as $a=6$, $b=3$ and $c=1$) $A$ cannot be transitive.

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  • $\begingroup$ Now, it is clearer for me $\endgroup$ – user297634 Dec 20 '15 at 11:37
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Your answer is quite right. The question is basically "if $a$ is near to $b$, and $b$ is near to $c$, is $a$ near to $c$?"

The answer to that is clearly "not necessarily", as you say.

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