# Is the relation $x \geq 2y$ transitive?

I am trying to understand if the relation $x \geq 2y$ is transitive. I think the answer is no for the following reasons. Can someone please let me know if I am correct or incorrect. If incorrect, why am I incorrect.

Definition of transitive: If $\left ( x,y \right )$ are elements of the relation, and $\left ( y,z \right )$ are elements of the relation, then this implies that $\left ( x,z \right )$ are elements of the relation.

Here is a counter example: $\left ( 4,2 \right )$ are elements of the relation because $4 \geq 2\times 2$. $\left ( 2,1 \right )$ are also elements of the relation because $2 \geq 2\times 1$. But, $\left ( 4,1 \right )$ are not elements of the relation because $4$ is not greater than or equal to $2 * 1$.

• But 4 is greater than or equal to $2 \cdot 1 = 2$. – Julia Jackson Aug 6 '15 at 13:46
• At the very end of the question you claim that $4$ isn't greater than (or equal to) $2$. Do you see something wrong with this? – Git Gud Aug 6 '15 at 13:46
• Ok. Now I feel stupid. I guess it's a case where I'm not seeing the forest through all the trees. Thanks for answering. – user3457456 Aug 6 '15 at 13:58

$$x\ge2y\land y\ge2z\implies x\ge2y\land 2y\ge4z\implies x\ge 4z\implies x\ge2z.$$

Update: The last implication requires $z\ge0$.

• This assumes, of course, that $x,y,z \geq 0$ – Omnomnomnom Aug 6 '15 at 13:58
• @Omnomnomnom: no, why ? – Yves Daoust Aug 6 '15 at 14:00
• Take $x = -3$, $z = -1$. We have $x \geq 4z$, but not $x \geq 2z$. – Omnomnomnom Aug 6 '15 at 14:00
• Yep, stupid me, we need $4z\ge2z$. (But $x$ and $y$ can be negative.) – Yves Daoust Aug 6 '15 at 14:02
• @Crostul your comment is good! Use \neg rather than \not. – Omnomnomnom Aug 6 '15 at 14:02

What set is your relation defined on? If it's for positive numbers, then it is transitive, as Yves Daoust showed.

If the numbers can be negative, we could have: $$x = -3, y =-\frac{7}{4}, z = -1$$ Then $x = -3 \geq-3.5 =2y$, $y =-\frac{7}{4} \geq -2 = 2z$, but $x = -3 < -2 = 2z$.

• It only occurs because of $z<0$. – Yves Daoust Aug 6 '15 at 14:07
• It seems to work if $x$ is non-negative. Suppose $x\geq 0$, and we have that $x\geq 2y$ and $y\geq 2z$, and $z < 0$. Then $x \geq 0 > 2z$. – Santeri Aug 6 '15 at 14:31
• The original problem I was working on was defined on the naturals but I am very appreciative that you showed how the relation is not transitive if it includes negative numbers. – user3457456 Aug 6 '15 at 14:42
• @Santeri: the transitivity can only be falsified when $z<0$, irrespective of the signs of $x$ and $y$. – Yves Daoust Aug 6 '15 at 15:38
• I agree. I thought you meant 'if and only if' with your first comment. It's a necessary but not sufficient condition. – Santeri Aug 6 '15 at 15:49

Your relation is not transitive, since it fails to be transitive when you work with negative numbers. As an example, $$-4 \ge 2 \cdot (-2) \mbox{ and } -2 \ge 2 \cdot (-1)$$ but $$-4 \ge 2 \cdot (-1) \mbox{ is false}$$