Is the relation $x+3y = 0$ transitive?

Question

$\def\R{\mathrel R}$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.

Answer 1

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: $~-9 + 3\cdot (-1) \not= 0$

Answer 2

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

Answer 3

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 $-3 \not\R -\frac 13$.