I have the following exercise, it is required to check if

the following equation defines an operation on the set of integers, and if so, check if it is associative, commutative and identity element.

$m * n = mn + 1$

I have taken $m = 2, n = 3$

$2*3 = (2 \cdot 3) + 1 = 6 + 1 = 7 \in \mathbb{Z}$

so that equation defines a valid operation.

now I check if it is associative, commutative, identity:

**i) associative.**

if $m=2, n=3, o=4$

I have to prove that

$(m*n)*o = m*(n*o)$

checking the left side:

$\begin{align}(2*3)*4 &= ((2 \cdot 3) + 1)*4 \\ &= 7 * 4 \\ &= (7 \cdot 4) + 1 \\ &= 28 + 1 \\ &= 29\end{align}$

checking the right side:

$\begin{align}2 * (3*4) &= 2 * ((3 \cdot 4) + 1) \\ &= 2* (12+1) \\ &= 2 * 13 \\ &= (2 \cdot 13) + 1 \\ &= 26 + 1 \\ &= 27 \end{align}$

therefore it is not associative because:

$(m*n)*o \ne m*(n*o)$

**ii) commutative:**

if $m = -2, n = 7$

we have to prove if:

$m*n = n * m$

checking the left side:

$\begin{align} (-2)*7 &= (-2 \cdot 7) + 1 \\ &= -14 + 1 \\ &= - 13 \end{align}$

checking the right side:

$\begin{align} 7 * (-2) &= (7 \cdot -2) + 1 \\ &= -14 + 1 \\ &= -13\end{align}$

therefore it is commutative because:

$m*n = n*m$

**iii) identity element**

I cite the following:

so I have taken $e = 1$ because it is considered inside a product in $\mathbb{Z}$.0 is an identity element for addition of integers, and 1 is an identity element for multiplication of integers.

if $m = 2$,

we have to prove if:

$m*e = e * m = m$

checking the left side:

$\begin{align} m * e &= (m \cdot e) + 1 \\ &= (m \cdot 1) + 1 \\ &= (2 \cdot 1) +1 \\ &= 2+1 \\ &= 3 \end{align}$

checking the right side:

$\begin{align} e * m &= (e \cdot m) + 1 \\ &= (1 \cdot m) + 1 \\ &= (1 \cdot 2) + 1 \\ &= 2+1 \\ &= 3 \end{align}$

I have also tried with $m = -2, m = 0$, and I have obtained that the identity element exists,

but, on my text it is written that there is not identity element. Please, can you tell me where is the error? Many thanks!