Is $|x-y| \le 7$ symetric relation?

Question

to prove $(x-y)$ divisible by $3$ as symmetric we use this method: let $xRy$ belongs to $A$ $$ \begin{align} (x-y)&=3k \\ -(x-y)&=-3k \\ (y-x)&=3(-k) \end{align} $$ therefore $yRx$ also belongs to $A$.

Now how do I prove $|x-y|\le 7$ as symmetric? I'm not able to understand. Or is it symmetric because of mod function and we don't use this method?

Answer 1

Let $S = \{ (x,y) \in \mathbb{R}^2 \mid |x-y| \le 7 \}$.

Since $|x-y| = |y-x|$, if $(x,y) \in S$ then $(y,x) \in S$. Hence, symmetric.

Answer 2

Yes, it is symmetric. I think the most fruitful way to look at it would be to do the definitions literally.

If $|a-b| \le 7$ does that mean $|b-a| \le 7$.

And then answer is: as $|a-b| = |b-a|$ then whatever is true about $|a-b|$ is true about $|b-a|$ as they are the same thing. So yes. If $|a-b| \le 7$ then $|b-a| =|a-b| \le 7$.

....

If you really want to plod through but pound it close with a sledgehammer.

Case 1) $a \ge b$ and if $k = a-b$ then $k \ge 0$. So $|a-b| =|k| = k$. If $k \le 7$ then $|b-a| = |-k| = k$ and $k \le 7$ so that is symmetric if $a \ge b$.

Case 2) $a < b$ and if $j = b-a > 0$ then $|a-b| = |-j| = j$. If $j \le 7$ then $|b-a| =|j| = j \le 7$ so that is symmetric of $b < a$.

So for all $a,b$..... if $|a-b| \le 7$ then $|b-a| = |a-b| \le 7$.

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The less fruitful way would be:

if an operation $*$ is commutative then the relation $a*b$ has property $X$ will always be symmetric. This is because $a*b = b*a$ so whatever property $a*b$ has then $b*a$ will have it as it is the same thing.

And the operation $a*b := |a-b|$ is commutative.