Finding symmetry in a relation

Question

Given a relation R such that xRy =|x-y|≤3, show that R is an equivalence relation.

Okay, so for reflexivity, I simply stated that the absolute value of x-x is less than or equal to 3. (0 < 3.)

I'm unsure about symmetry. Show that yRx, which would mean that |y-x|≤3.

So possibly that means that |y-x|≤-3, I think. I don't know. I'm stuck.

Answer 1

To show symmetry, you need to show that if $xRy$, then $yRx$. $xRy$ means that $\left| x - y\right|\leq 3$, and since $\left|z\right| = \left|-z\right|$, we have $\left|x - y\right| = \left|y - x\right|\leq 3$, but this last inequality is precisely $yRx$.

Transitivity: show that if $xRy$ and $yRz$, then $xRz$. As an example, take $x = 1$, $y = 2$, and $z = 3$. Then $\left|1 - 2\right| = 1\leq 3$, so that $1R2$, and $\left|2 - 3\right| = 1\leq 3$, so that $2R3$. We see that $\left|1 - 3\right| = 2\leq 3$, so that $1R3$. Thus transitivity is verified for $x = 1$, $y = 2$, and $z = 3$. What happens if $x = 0$, $y = 3$, and $z = 6$?

Answer 2

You are almost right. Given the relation $xRy$ defined as $|x-y| \leq 3$, you are correct that $|x-x| = 0 \leq 3$, and so $xRx$.

For symmetry, given $xRy$, how would you show $yRx$? One way is $|x-y| = |-1(y-x)| = |-1| |y-x| = |y-x|$, this shows that $|x-y| = |y-x|$. So if $xRy$, then $yRx$.