The operation is xy=x+y-xy Not xy=x-y-xy

Thanks for the correction! Let's re-evaluate the operation $x * y = x + y - xy$.

Step 1: Determine if the operation is commutative. An operation is commutative if $x * y = y * x$ for all $x, y \in \mathbb{Z}$. Given $x * y = x + y - xy$. Now, let's find $y * x$: $y * x = y + x - yx$ Since addition and multiplication of integers are commutative ($x+y = y+x$ and $xy = yx$), we can see that: $x + y - xy = y + x - yx$ Therefore, $x * y = y * x$. The operation is commutative.

Step 2: Find the identity element. An identity element $e \in \mathbb{Z}$ for the operation $*$ must satisfy $x * e = x$ for all $x \in \mathbb{Z}$. Using the definition of the operation: $x + e - xe = x$ Subtract $x$ from both sides: $e - xe = 0$ Factor out $e$: $e(1 - x) = 0$ For this equation to hold for all $x \in \mathbb{Z}$, the value of $e$ must be $0$. If $e = 0$, then $0(1 - x) = 0$, which simplifies to $0 = 0$. This is true for all integers $x$. Therefore, the identity element is $\boxed{e = 0}$.

Step 3: Find the inverse of $x$. An inverse element $x^{-1}$ for an element $x$ exists if $x * x^{-1} = e$, where $e$ is the identity element. We found $e = 0$. So, we need to solve $x * x^{-1} = 0$: $x + x^{-1} - xx^{-1} = 0$ Factor out $x^{-1}$ from the terms containing it: $x + x^{-1}(1 - x) = 0$ $x^{-1}(1 - x) = -x$ If $1 - x \neq 0$ (i.e., $x \neq 1$), we can solve for $x^{-1}$: $x^{-1} = \frac{-x}{1 - x} = \frac{x}{x - 1}$ For $x^{-1}$ to be an inverse in $\mathbb{Z}$, it must be an integer. Let's test some values: • If $x = 0$, $x^{-1} = \frac{0}{0 - 1} = 0$. (This is an integer). • If $x = 2$, $x^{-1} = \frac{2}{2 - 1} = 2$. (This is an integer). • If $x = 3$, $x^{-1} = \frac{3}{3 - 1} = \frac{3}{2}$. (This is not an integer). • If $x = -1$, $x^{-1} = \frac{-1}{-1 - 1} = \frac{-1}{-2} = \frac{1}{2}$. (This is not an integer).

Also, consider the case when $x = 1$: If $x = 1$, the equation $x + x^{-1} - xx^{-1} = 0$ becomes: $1 + x^{-1} - 1 \cdot x^{-1} = 0$ $1 + x^{-1} - x^{-1} = 0$ $1 = 0$ This is a contradiction, which means that $x=1$ does not have an inverse.

Since $x^{-1}$ is not an integer for all $x \in \mathbb{Z}$ (e.g., for $x=3$) and does not exist for $x=1$, the inverse of $x$ does not exist for all $x \in \mathbb{Z}$.

Summary of findings: • The operation is commutative. • The identity element is $\boxed{e = 0}$. • The inverse of $x$ does not exist for all $x \in \mathbb{Z}$.

What's next?