Prove or disprove |x-y| is a group

Question

I have to prove or disprove each four factors (closure, associativity, identity, inverse) of a set being a group.

The set I have is |x-y| in the natural numbers.

So far I have,

Closure: ∀x:∀y: |x-y|∈ N

Not closed. Use x=1 and y=1.

Then, |x-y|=|1-1|=0∉ N

Associativity: ∀x:∀y:∀z:||x-y|-z|= |x-|y-z||

Not associative. Use x=2, y=1, z=3.

Then, ||2-1|-3|≠|2-|1-3||.

2≠0.

From here, I am having a hard time figuring out how to prove/disprove identity and inverse.

Identity: Ǝe∈ N:∀x,y: |e-|x-y||=|x-y|

Inverse: ∀x:Ǝy: |x-y|=e

So you mean the set is $;\Bbb N;$ , and the operation on this set is $;|x-y|;$ ? — DonAntonio, Apr 30 '17 at 19:06
So @DonAntonio , the set includes all x,y in the Natural numbers. A group is a set equipped with a binary operation * and satisfies the four properties. In this case, x*y= |x-y|. — Katiee, Apr 30 '17 at 19:13
So for you $;0\neq\Bbb N;$ ? It can be either way, of course. — DonAntonio, Apr 30 '17 at 19:16
@DonAntonio it's a universal thing that $0\neq \mathbb N$ :) — , Apr 30 '17 at 19:24
@OpenBall Really??! Well, tell that to the thousands of mathematicians and among them logicians, that think otherwise...You can read here: https://math.stackexchange.com/questions/283/is-0-a-natural-number , or here: https://www.quora.com/Is-zero-a-natural-number-Why-or-why-not , or here: http://mathworld.wolfram.com/NaturalNumber.html . You can see that in many cases it is stated there's no general agreement on this, and some rather well known mathematicians ( say Bourbaki, Halmos, etc.) include zero as natural number, at least in some of their writings. Besides, you wrote $;0\neq\Bbb N;$ — DonAntonio, Apr 30 '17 at 19:29
@OpenBall ...where you wanted, apparently, to write $;0\notin\Bbb N;$ — DonAntonio, Apr 30 '17 at 19:29
@DonAntonio it was you who wrote $0\neq\mathbb N$. I hope the joke is clear now. — , Apr 30 '17 at 19:31
@OpenBall Oops, now I see...hehe. That was too subtle for my two working neurons. Thanks, got it. +1 — DonAntonio, Apr 30 '17 at 19:34

score 1 · Answer 1 · answered Apr 30 '17 at 19:17

For the identity you need to find (or not...) and element $\;e\in\Bbb N\;$ s.t. for all $\;n\in\Bbb N\;$ , we have that

$$|e-n|=|n-e|=n\implies e=0$$

But since you don't include zero as a natural number there is then no identity elements, and thus the last property doesn't simply apply.