Identity element of the operation $x*y = \min\{x+1, y+1\}$ on $\mathbb R$? 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!
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 $x \in \mathbb R$, $e*x = x*e = x$.
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 $x \in \mathbb R$, $e*x = x*e = x$.