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$\def\R{\mathrel R}$A relation $R$ is transitive if $a\R b$ and $b\R a$ implies $a\R c$

because it is never $a\R b$ and $b\R a$, it is always true because no matter what $a\R c$ is, if the LHS is false, the statement is always true.

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Your definition of 'transitive' has a typo. – TonyK Nov 22 '12 at 22:03
Transitivity of a relation $R$ on a set $S$ means that for all $a, b, c \in S$, IF $\,a\,R\,b\,$ AND $\,b\,R\,c$, then necessarily $\,a\,R\,c$. – amWhy Nov 22 '12 at 22:41

No. It's easy to prove: you just have to find a counterexample. Take

$$a = -9, ~ b = 3, ~ c = -1$$

  • $aRb~$ since $~-9 + 3 \cdot 3 = 0$

  • $bRc~$ since $~3 + 3(-1) = 0$

  • but $aRc~$ is false: $~-9 + 3\cdot (-1) \not= 0$

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Though it depends, on which set $R$ is supposed to be a relation. On $\mathbb Z\setminus 3\mathbb Z$ relation $R$ is transitive. – Hagen von Eitzen Nov 22 '12 at 23:08

No. You have $xRy \iff y = -\frac{1}{3} x$. So $9R(-3)$ and $(-3)R1$ but you do not have $9R1$.

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No. A relation $\R$ is transitive iff $a\R b$ and $b\R c$ imply $a \R c$. For your $R$, suppose $a \R b$ and $b \R c$ hold, then we have $a + 3b = 0$, and $b + 3c = 0$, then $$ a + 3c = -3b + 3c = -4b$$ That is $a\R c$ holds iff $b = 0$. So $\R$ isn't transitive as for example $-3\R 1$ and $1\R -\frac 13$, but $-3 \not\R -\frac 13$.

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