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On $\Bbb N=\{0,1,2,...\}$ we define the operation $\otimes$ by $m\otimes n= |m-n|$. Are there any identity element?

I came up with two identity elements. $e= 0$ and $e=2m$ for an element $m$ in the set, but how is that possible? I thought on a operation $\otimes$ that it would always be one identity element?

Sebastiano
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Erika
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  • You could always add a formal copy of a neutral element to any operation, and you'd get two neutral elements. But you'd lose properties for this law. However, in your case, 2m has no meaning, since it depends on m ! – Anthony Dec 31 '20 at 17:22
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    the identity element should be the same for all elements – J. W. Tanner Dec 31 '20 at 17:22
  • So an identity element can´t depend on the element? – Erika Dec 31 '20 at 17:25
  • @Erika No, it can't depend on the element. It needs to be the same for all elements. –  Dec 31 '20 at 17:38
  • @AnthonySaint-Criq When you add a formal copy of a neutral element, you have to define what $e\cdot e'$ and $e'\cdot e$ are and there is no choice that makes both $e$ and $e'$ two sided neutral elements when $e\neq e'$. – Christoph Dec 31 '20 at 17:42
  • @Christoph oh yes I am mistaken, I went a bit fast, sorry ! – Anthony Jan 01 '21 at 08:19

2 Answers2

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The definition of the identity element is $$\exists e, \forall x, e*x=x*e=x$$

and not $$\forall x, \exists e, e*x=x*e=x$$

TheSilverDoe
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It's not possible. If a set with binary operation has an identity element then it must be unique. No matter if associativity, commutativity or other things hold.

If $1, 1'$ are identity elements then $1\cdot 1' = 1 = 1'$ from the properties of identity elements.

Identity element can't depend on other elements. There is however a different but similar notion of local identity element, which depends on an element chosen. We say that $e$ is a local identity element for $x$ if $e\cdot x = x\cdot e = x$. We may denote $e$ by $e(x)$ to indicate that $e$ depends on $x$.

I saw this one used in some texts about theory of quasigroups, without going into details of what a quasigroup is.

Here, $0$ and $2m$ are all possible local identities for $m$, but only $0$ is the identity element.

Jakobian
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  • @Aaron I'm sorry too. I haven't given my response a second thought and might come up as a bit harsh. – Jakobian Dec 31 '20 at 17:56
  • If someone had downvoted a reasonable and correct answer because they had misread it, I would probably not respond in a gracious manner, no worries. – Aaron Dec 31 '20 at 17:59