Operator such as $-1$ is the identity element

Question

Short question
Do you know an operator such as $-1$ is the identity element ?

Long Question
This morning, I had a hard time with identity elements.
I'm pretty sure that the following isn't very rigourous, so please don't hesitate to comment !
I'm going to think in $\overline{\mathbb{R}}$, which means that $\infty$ and $-\infty$ are numbers like others real number. According to the addition, we can divide our real segment in two parts : $[-\infty, 0]$ and $[0, +\infty]$.
According to the multiplication, we can divide these two segments into four: $[-\infty, -1]$, $[-1, 0]$, $[0, 1]$ and $[1, +\infty]$.
So we can see $5$ key numbers : $-\infty, -1, 0, 1, \infty$.
The problem is that I can't find a function so that $-1$ is an identity element.
I hope the problem is not trivial ;)
We can picture the problem like that :
$$Function \to Identity\; element$$ $$max \to -\infty$$ $$??? \to -1$$ $$+ \to 0$$ $$\times \to 1$$ $$min \to \infty$$

@MikeMiller, out of curiosity, how did you come up with your own answer? — Théophile, Jun 19 '14 at 03:59
@Théophile, Yeah ;) but it makes sense since $[-\infty, -1]$ is sent on $[0, 1]$ and $[-1, 0]$ is sent on $[1, \infty]$. The symmetry I was looking for is still there ;) Please write an answer to show that the question is solved — Fabien, Jun 19 '14 at 04:00
@Théophile Well, the multiplicative identity is $1$. Instead of negating like you did I shifted downwards: mine's just $(a+2)(b+2)-2$. — , Jun 19 '14 at 04:02
@MikeMiller Thanks for the explanation. It's interesting to see how this generalizes in David's answer below. — Théophile, Jun 19 '14 at 04:52
@Théophile Looks like yours is a special case too, coming from $\phi(x)=-x$. — , Jun 19 '14 at 05:05

score 4 · Accepted Answer · answered Jun 19 '14 at 04:06

4

Take any bijection $\phi:{\Bbb R}\to{\Bbb R}$ and define an operation $$x*y=\phi\bigl(\phi^{-1}(x)\phi^{-1}(y)\bigr)\ .$$ Then the identity element for $*$ is $\phi(1)$, which could be anything you like, depending on your choice of $\phi$.

In particular, the choice $\phi(x)=x-2$ gives the example from the comment by @MikeMiller.

You could do the same thing with addition: $$x*y=\phi\bigl(\phi^{-1}(x)+\phi^{-1}(y)\bigr)\ .$$ and then the identity element would be $\phi(0)$.

answered Jun 19 '14 at 04:06

David

82,662

Thanks for the generalization ! Always a pleasure to read your answers. – Fabien Jun 19 '14 at 04:08
It's worth pointing out that this basically just means if we relabel the real numbers with the real numbers in some other way without changing the algebraic properties then of course whatever 1 gets relabeled as is the identity. The same could be said to be true of any set of the same cardinality. – jxnh Jun 20 '14 at 04:37

Operator such as $-1$ is the identity element

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