The following is an operation $*$ on $\mathbb R$.

$$x*y= \min\{x+1, y+1\}$$

Explain whether or not $\mathbb R$ has an identity element with respect to $*$.

This question also had two other parts, asking to explain whether it is commutative and associative. I said that it is commutative but not associative.

As for the identity element, I am confused because $\mathbb R$ goes to $-\infty$ so won't we always be able to find an element that is smaller than $x+1$?

I feel like there cannot be a neutral element because the operation $*$ is defined on $\mathbb R$ and I feel like you will always be able to find a smaller element than $x+1$.

Plus, the neutral element has to be unique so how could this operation have a unique neutral element?

Thank you for any help!

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    $\begingroup$ What is $x*x{}$? $\endgroup$ – Angina Seng Sep 20 '17 at 16:56
  • $\begingroup$ But that isn't unique for the operation? $\endgroup$ – JxxYsde3 Sep 20 '17 at 16:57
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    $\begingroup$ It's true, though. $\endgroup$ – Professor Vector Sep 20 '17 at 17:05
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    $\begingroup$ It is equivalent to you operation, JxxYsde3. I simply thought looking at what the operation is actually doing might be helpful. $\endgroup$ – amWhy Sep 20 '17 at 17:07
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    $\begingroup$ The operation isn't interesting, I bet it isn't even associative. And there isn't a neutral element, naturally, exactly for the reason you doubt it: it would mean $x*e=x$ for all $x$, but if $x<e$, you have $x*e=x+1\neq x$. $\endgroup$ – Professor Vector Sep 20 '17 at 17:11

The operation $x*y = \min(x+1, y+1)$, for $x, y \in \mathbb R$, is equivalent to the following:

$$x*y = \min(x, y) + 1$$

The neutral element, say $e \in \mathbb R$ must be such that $e*x = x*e = x$, that is $\min(e, x) + 1 = \min(x, e) +1 = x$, for all $x \in \mathbb R$.

But let's just pick a specific $x_1 \in \mathbb R$ to text. $\min(e, x_1) + 1 = \min(x_1, e) = x_1 \iff e = x_1 - 1$.

Let's try another distinct element, $x_2 \neq x_1$ in $\mathbb R$. We need $e' \in \mathbb R$ such that $e'*x_2 = x_2*e' = x_2 \iff e'= x_2 - 1$.

But we see that $e'\neq e$, because $e = x_1 -1$, and $e'= x_2 - 1$, and since $x_1\neq x_2$, we know that $e' \neq e$.

So we see quickly, that there is a different "neutral element" dependent on and defined for, each distinct element $x\in \mathbb R$.

So no, there does not exist a unique neutral (identity) element $e$ in $\mathbb R$ such that for all $x \in \mathbb R$, $e*x = x*e = x$.

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    $\begingroup$ I'm surprised by this terminology: if a neutral element exists, it is unique. If we had two neutral elements, $e$ and $e'$, we'd have $e=e*e'=e'$ by definition. $\endgroup$ – Professor Vector Sep 20 '17 at 17:30
  • $\begingroup$ My point being, @Professor, I am showing that using the property of a neutral element e with respect to a set (real numbers) given the operation $*$, that property being that there exists one element, $e$ unique in that for all $x \in \mathbb e* x = x*e = x$ fails, because for each specific $x \in \mathbb R$ the satisfying "e" applies is specific and valid for that specific $x$; i.e., there is no "universal, unique identity $e\in \mathbb R$" that serves as an identity for each and every $x \in \mathbb R$. $\endgroup$ – amWhy Sep 20 '17 at 17:46
  • $\begingroup$ Kind of like: we have that $$\forall x \in \mathbb R, \exists! e\in \mathbb R (e*x = x*e = x)$$ But we also have that $$\not \exists e! \in \mathbb R, \forall x \in \mathbb R(e*x= x*e = x)$$ And because of the latter statement, no such neutral (identity) element e exists for the operation $*$, on the Real numbers. $\endgroup$ – amWhy Sep 20 '17 at 17:53
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    $\begingroup$ Iron logic, I'd say. =D. $\endgroup$ – Professor Vector Sep 20 '17 at 18:06

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