B. Properties of operations

1 If

$$ x \ast y = x + 2y +4 $$

then

$$ y \ast x = 2y + x + 4 $$

which is equal to $x \ast y$, so the operation $\ast$ is associative.

Because

$$ \begin{array}{r c l} x \ast (y \ast z) & = & x \ast (y + 2z + 4) \\[0.5em] & = & x + 2(y + 2z + 4) + 4 \\[0.5em] & = & x + 2y + 4z + 12 \end{array} $$

and

$$ \begin{array}{r c l} (x \ast y) \ast z & = & (x + 2y + 4) \ast z \\[0.5em] & = & (x + 2y + 4) + 2z + 4 \\[0.5em] & = & x + 2y + 2z + 8 \end{array} $$

which are not equal, the $\ast$ operation is not commutative.

Solving for the identity element $e$ in equation $x \ast e = x$, it can be seen that

$$ \phantom{,} \ x \ast e = x + 2e + 4 = x \ , $$

$$ 2e = -4 $$

which means that

$$ e = -2. $$

Substituting $e$ back to $x$ $+$ $2e$ $+$ $4$ $=$ $x$ shows that

$$ x + 2(-2) + 4 = x - 4 + 4 = x $$

so $\ast$ operation has an identity.

Because $\ast$ has an identity it’s worthwhile to look for an inverse of $\ast$. This means solving

$$ x \ast x^\prime = x + 2x^\prime + 4 = 2 = e $$

for $x^\prime$. Therefore,

$$ x^\prime = \dfrac{-x -2}{2} = - \dfrac{x}{2} - 1 $$

and substituting $x^\prime$ back to $x$ $+$ $2x^\prime$ $+$ $4$ shows that

$$ \begin{array}{r c l} x \ast x^\prime & = & x + 2 \left(- \dfrac{x}{2} - 1\right) + 4 \\[0.5em] & = & x - x - 2 + 4 \\[0.5em] & = & 2, \end{array} $$

so it’s possible that inverse exists.

To be sure of this, $x^\prime \ast x$ also needs to be equal to newly found identity element $e$, so noticing that

$$ x^\prime \ast x \ = \ x^\prime + 2x + 4 \ = \ \left( - \dfrac{x}{2} - 1 \right) + 2x + 4 \ = \ - \dfrac{1}{2} x + 2x + 4 - 1 \ = \ - \dfrac{3}{2}x - 3 \ \neq \ 2, $$

$$ \begin{array}{r c l} x^\prime \ast x & = & x^\prime + 2x + 4 \\[1.0em] & = & \left( - \dfrac{x}{2} - 1 \right) + 2x + 4 \\[1.0em] & = & - \dfrac{1}{2} x + 2x + 4 - 1 \\[1.0em] & = & - \dfrac{3}{2}x - 3 \\[0.5em] & \neq & 2, \end{array} $$

which means that not all elements have inverse.

2 Checking for commutativity of

$$ x \ast y = x + 2y - xy $$

means evaluating

$$ y \ast x = y + 2x - yx = 2x + y - yx $$

Because $x \ast y$ $\neq$ $y \ast x$, the operation is not associate.

Moreover, since

$$ \begin{array}{r c l} x \ast (y \ast z) & = & x \ast (y + 2z - yz) \\[0.5em] & = & x + 2(y + 2z - yz) - x(y + 2z - yz) \\[0.5em] & = & x + 2y + 4z - 2yz - xy - 2xz + xyz \end{array} $$

$$ \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & x \ast (y + 2z - yz) \\[0.5em] = & x + 2(y + 2z - yz) - x(y + 2z - yz) \\[0.5em] = & x + 2y + 4z - 2yz - xy - 2xz + xyz \end{array} $$

and

$$ \begin{array}{r c l} (x \ast y) \ast z & = & (x + 2y - xy) \ast z \\[0.5em] & = & (x + 2y - xy) + 2z - (x + 2y - xy)z \\[0.5em] & = & x + 2y - xy + 2z - zx - 2yz + xyz \end{array} $$

$$ \begin{array}{c l} & (x \ast y) \ast z \\[0.5em] = & (x + 2y - xy) \ast z \\[0.5em] = & (x + 2y - xy) + 2z - (x + 2y - xy)z \\[0.5em] = & x + 2y - xy + 2z - zx - 2yz + xyz \end{array} $$

are not equal, operation $\ast$ does not commute.

Solving $x \ast e$ $=$ $x$ for $e$, means that

$$ \begin{array}{r c l} x \ast e & = & x + 2e - xe \\[0.5em] & = & x + e(2 - x) \\[0.5em] & = & x, \end{array} $$

$$ e = \dfrac{x - x}{2 - x} = 0. $$

Substituting $e$ $=$ $0$ back to $x \ast e$ shows that

$$ x + 2(0) - x(0) $$

indeed is $x$, so the operations has an identity.

Checking for inverse, the equation $x \ast x^\prime$ $=$ $e$ needs to be first solved for $x^\prime$. Because

$$ x \ast x^\prime = x + 2x^\prime - xx^\prime = 0 = e $$

and

$$ x + 2x^\prime - xx^\prime = x + x^\prime(2 - x) = 0 $$

so therefore

$$ x^\prime(2 - x) = -x $$

and hence

$$ x^\prime = - \dfrac{x}{2 - x}. $$

Substituting $x^\prime$ back to equation $x \ast x^\prime$ means that

$$ \begin{array}{r c l} x \ast x^\prime & = & x - 2 \dfrac{x}{2 - x} - x \dfrac{x}{2 - x} \\[1.0em] & = & x - \dfrac{2x - x}{2 - x} \\[1.0em] & = & x - x \\[1.0em] \end{array} $$

which is equal to $0$

$$ \begin{array}{r c l} x^\prime \ast x & = & -\dfrac{x}{2 - x} + 2x + \dfrac{x}{2 - x}x \\[0.5em] & = & 2x - \dfrac{x + x^2}{2 - x} \\[0.5em] & = & 0. \end{array} $$

Multiplying both sides of the equality leads to

$$ \begin{array}{r c l} (2 - x)2x - x + x^2 & = & 4x - 2x^2 - x + x^2 \\[0.5em] & = & -x^2 + 3x \\[0.5em] & = & x(-x + 3) \\[0.5em] & = & 0. \end{array} $$

Based on the zero product property the equality leads to two equations, either $x = 0$ or $-x + 3 = 0$, from where solving the latter leads to $x = 3$. Because $x$ is either $0$ or $3$, the $x \ast x^\prime$ is not unique, $x^\prime \ast x$ $\neq$ $x \ast x^\prime$ and the operation $\ast$ does not have inverses for all values in $\mathbb{R}$.

3 Because

$$ x \ast y = |x + y| = |y + x| = y \ast x, $$

the $\ast$ operation is commutative.

Since

$$ \begin{array}{r c l} x \ast (y \ast z) & = & x \ast (|y + z|) \\[0.5em] & = & |x + |y + z|| \\[0.5em] & = & ||x + y| + z| \\[0.5em] & = & (|x + y|) \ast z \\[0.5em] & = & (x \ast y) \ast z, \end{array} $$

the $\ast$ operation is associative.

Identity element exists if the equation can be solved for $e$ in $x \ast e = x$. Let

$$ x \ast e = x, $$

then the equation is equivalent to

$$ |x + e| = x, $$

so there’s two equations to be solved, $x + e$ $=$ $x$ and $-(x + e)$ $=$ $x$. Solution for the first is $e$ $=$ $x - x$ $=$ $0$ and for the second $e$ $=$ $x + x$ $=$ $2x$.

Checking that $x \ast e$ $=$ $e \ast x$ $=$ $x$ for the identities means that for $x \ast e$ $>$ $0$,

$$ x \ast e = |x + 0| = x + 0 = x = 0 + x = |0 + x| = e \ast x, $$

$$ \begin{array}{r c l} x \ast e & = & |x + 0| \\[0.5em] & = & x + 0 \\[0.5em] & = & x \\[0.5em] & = & 0 + x \\[0.5em] & = & |0 + x| \\[0.5em] & = & e \ast x, \end{array} $$

and for $x \ast r$ $\leq$ $0$,

$$ x \ast e = |x + 2x| = x + 2x = 3x \neq x \neq 2x + x = |2x + x| = e \ast x, $$

$$ \begin{array}{r c l} x \ast e & = & |x + 2x| \\[0.5em] & = & x + 2x \\[0.5em] & = & 3x \\[0.5em] & \neq & x \\[0.5em] & \neq & 2x + x \\[0.5em] & = & |2x + x| \\[0.5em] & = & e \ast x, \end{array} $$

from which it can be claimed that the latter identity element does not hold, but the first does. Assuming that the identity is true for when $x \ast r$ $\leq$ $0$, the previous equation becomes equal to the first and it holds. Because of this, letting $e$ be equal to $0$, the identity is unique and exits for the operation $\ast$.

If the inverse for $\ast$ exists, then the equation $x \ast x^\prime$ $=$ $e$ can be solved for $x^\prime$. Then, because

$$ x \ast x^\prime = |x + x^\prime| = e, $$

so if $|x + x^\prime|$ $>$ $0$ then

$$ e = x + x^\prime $$

or if $|x + x^\prime|$ $\leq$ $0$ then

$$ e = -(x + x^\prime) = -x - x^\prime. $$

Solving these cases for $x^\prime$ means that

$$ x^\prime = e - x = 0 - x = -x, $$

and

$$ x^\prime = -e - x = 0 - x = -x, $$

so substituting $x^\prime$ to equation $x$ $\ast$ $x^\prime$ $=$ $e$ means that

$$ |x + x^\prime| = |x - x| = |0| = 0, $$

and same for equation $x^\prime$ $\ast$ $x$ $=$ $e$ means that

$$ |x^\prime + x| = |-x + x| = |0| = 0, $$

which means that the inverse exists, namely $x^\prime$ $=$ $-x$, because $x \ast x^\prime$ leads to $e$.

4 Operation $x \ast y$ $=$ $|x - y|$.

Commutativity: If $x \ast y$ $=$ $y \ast x$ the operation is commutative. When $x \ast y$ $>$ $0$, $x \ast y$ $=$ $x - y$ and when $x \ast y$ $\leq$ $0$, $x \ast y$ $=$ $-(x - y)$ $=$ $-x + y$. Similarly, when the operation is $y \ast x$, the equations are $y - x$ and $-y + x$. Now, when $x \ast y$ and $y \ast x$ are both strictly greater than $0$, $x \ast y$ $=$ $|x - y|$ $=$ $x - y$ $\neq$ $y - x$ $=$ $|y - x|$ = $y \ast x$, and when $x \ast y$ and $y \ast x$ are less or equal to $0$, $x \ast y$ $=$ $|x - y|$ $=$ $-(x - y)$ $=$ $-x + y$ $\neq$ $-y + x$ $=$ $-(y - x)$ $=$ $|y - x|$, so the operation $\ast$ does not commute.

Associativity: The operations is associative if $x \ast (y \ast z)$ $=$ $(x \ast y) \ast z$. Rewriting the left hand-side as

$$ x \ast (y \ast z) = x \ast |y - z| = |x - |y - z|| $$

shows that the parentheses lead to nested absolute value equation. Solving the outer absolute value first means that

$$ x \ast (y \ast z) = |x - |y - z|| $$

is either $x - |y - z|$ or $-(x - |y - z|)$ $=$ $-x + |y - z|$. Moreover, the inner absolute means that both of these possible solutions can have two different solutions, namely:

$$ \begin{array}{ c c c c c c c c } (1): & x - (y - z), & (2): & x + (y - z), & (3): & -x + (y - z), & (4): & -x - (y - z). \\[0.5em] & \| & & \| & & \| & & \| & \\[0.5em] (1): & x - y + z, & (2): & x + y - z, & (3): & -x + y - z, & (4): & -x - y + z. \end{array} $$

$$ \begin{array}{ c r c c r } (1): & x - (y - z) & = & (1): & x - y + z, \\ (2): & x + (y - z) & = & (2): & x + y - z, \\ (3): & -x + (y - z) & = & (3): & -x + y - z, \\ (4): & -x - (y - z) & = & (4): & -x - y + z. \end{array} $$

Doing the same for $(x \ast y) \ast z$ leads to similar results, that is four different possible solutions:

$$ \begin{array}{ c c c c c c c c } (1^\prime): & (x - y) - z, & (2^\prime): & -(x - y) - z, & (3^\prime): & -(x - y) + z, & (4^\prime): & (x - y) + z. \\[0.5em] & \| & & \| & & \| & & \| & \\[0.5em] (1^\prime): & x - y - z, & (2^\prime): & -x + y - z, & (3^\prime): & -x + y + z, & (4^\prime): & x - y + z. \end{array} $$

From these we can see that only $(3)$ $=$ $(3^\prime)$, that is, the $\ast$ operation is associative only when the outer absolute value is less or equal to $0$ and the inner absolute value strictly greater than $0$. For the operation to be commutative, all of the cases $(1), (2), (3)$ and $(4)$ should equal. This does not happen so the $\ast$ is not associative.

Identity: Solving for currently not known identity element $e$ in the equation $x \ast e$ $=$ $|x - e|$ = $x$ leads again to the two cases: $x - e$ $=$ $x$ or $-x + e$ $=$ $x$. Solving the first results in $e$ $=$ $0$ and the second in $e$ $=$ $2x$, so there’s no unique identity element. Therefore $\ast$ does not have identity.

Inverses: …

5 Because $x \ast y$ $=$ $xy + 1$ and $y \ast x$ $=$ $yx + 1$ that is equal to $xy + 1$, because multiplication is commutative, then $x \ast y$ is commutative. Since $x \ast (y \ast z)$ $=$ $x \ast (yz + 1)$ $=$ $x(yz + 1) + 1$ and $(x \ast y) \ast z$ $=$ $(xy + 1) \ast z$ $=$ $(xy + 1)z + 1$, the operations are not equal, so $\ast$ is not commutative. Solving $x \ast e$ $=$ $xe + 1$ $=$ $x$, for $e$, shows that $e$ $=$ $1 - \frac{1}{x}$. When substituting $e$ back to original equation it can be seen that

$$ x \ast e = xe + 1 = x \left(1 - \dfrac{1}{x} \right) + 1 = x - 1 + 1 = x, $$

$$ \begin{array}{ c l } & x \ast e \\[0.5em] = & xe + 1 \\[0.5em] = & x \left(1 - \dfrac{1}{x} \right) + 1 \\[0.5em] = & x - 1 + 1 \\[0.5em] = & x, \end{array} $$

so the identity exits, i.e. $e = 1 - \frac{1}{x}$. Assuming this, the inverse seems to exist for $x \ast x^\prime$ $=$ $e$. Because

$$ x \ast x^\prime = xx^\prime + 1 = 1 - \dfrac{1}{x} = e, $$

and solving $x^\prime$ means dividing through both sides of the equals signs, it can be shown that

$$ x^\prime + \dfrac{1}{x} = \dfrac{1}{x} - \dfrac{1}{x} \left( \dfrac{x}{1} \right) = \dfrac{1}{x} - 1, $$

$$ x^\prime = \dfrac{1}{x} - 1 - \dfrac{1}{x} = -1. $$

Because $x \ast x^\prime$ $=$ $xx^\prime - 1$ $=$ $x(-1) = -1$, then $x$ $=$ $1$, but $x^\prime x - 1$ $=$ $-x - 1$ $=$ $-1$, that is $x$ $=$ $0$, the inverse does not exist for the element $\ast$.

6 The operation $x \ast y$ $=$ $\max\{x,y\}$ is commutative because $x \ast y$ $=$ $\max\{x,y\}$ $=$ $\max\{y,x\}$ $=$ $y \ast x$. If $\ast$ is associative, then

$$ x \ast (y \ast z) = \max\{x,\max\{y,z\}\} = \max\{\max\{x,y\},z\} = (x \ast y) \ast z. $$

$$ \begin{array}{r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} \\[0.5em] & = & \max\{\max\{x,y\},z\} \\[0.5em] & = & (x \ast y) \ast z. \end{array} $$

To prove this, solving the following six inequalities need to be tested¹^,² for both $x \ast (y \ast z)$ and $(x \ast y) \ast z$:

$$ \begin{array}{ l } (1) \ x \leq y \leq z: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & z \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & z \end{array} % \\[1.5em] (2) \ x \leq z \leq y: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & y \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & y \end{array} % \\[1.5em] (3) \ y \leq x \leq z: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & z \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & z \end{array} % \\[1.5em] (4) \ y \leq z \leq x: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & x \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & x \end{array} % \\[1.5em] (5) \ z \leq x \leq y: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & y \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & y \end{array} % \\[1.5em] (6) \ z \leq y \leq x: \\[1.0em] \qquad \begin{array}{r c r c l} x \ast (y \ast z) & = & \max\{x,\max\{y,z\}\} & = & x \\[0.5em] (x \ast y) \ast z & = & \max\{\max\{x,y\},z\} & = & x \end{array} % \\[1.5em] \end{array} $$

$$ \begin{array}{ l } (1) \ x \leq y \leq z: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & z \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & z \end{array} % \\[1.5em] (2) \ x \leq z \leq y: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & y \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & y \end{array} % \\[1.5em] (3) \ y \leq x \leq z: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & z \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & z \end{array} % \\[1.5em] (4) \ y \leq z \leq x: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & x \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & x \end{array} % \\[1.5em] (5) \ z \leq x \leq y: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & y \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & y \end{array} % \\[1.5em] (6) \ z \leq y \leq x: \\[1.5em] \qquad \begin{array}{c l} & x \ast (y \ast z) \\[0.5em] = & \max\{x,\max\{y,z\}\} \\[0.5em] = & x \\[1.5em] & (x \ast y) \ast z \\[0.5em] = & \max\{\max\{x,y\},z\} \\[0.5em] = & x \end{array} % \\[1.5em] \end{array} $$

Thus, it can be seen that operation $\ast$ is associative, because no matter what the order of operation is, $\ast$ always is the largest element of $x$, $y$ and $z$. Identity of the operation, if it exists, means that $x \ast e$ $=$ $\max\{x,e\}$ $=$ $x$. Noticing that when $e$ $=$ $-\infty$, for all $x$, $\max\{x,e\}$ $=$ $x$, and the same holds for $\max\{e,x\}$, so $x \ast e$ $=$ $e \ast x$ $=$ $x$, and the identify element of $\max$ is $-\infty$. Finding out the inverse of $\max$ would require that there is a unique inverse element for each $x$. But, for example,

$$ \max\{1,1\} = 1 = \max{1,0} $$

so there are element(s) for which the requirement does not hold and therefore $\ast$ does not have inverse element.

7 Since $x \ast y$ $=$ $\frac{xy}{x + y + 1}$ $=$ $\frac{yx}{y + x + 1}$ $=$ $y \ast x$, the $\ast$ operation is commutative. The operation $\ast$ is associative because,

$$ \begin{array}{ r c c c l} x \ast (y \ast z) & = & x \ast \left( \dfrac{yz}{y + z + 1} \right) \\[1.0em] & = & \dfrac{x \left( \dfrac{yz}{y + z + 1} \right)}{x + \left( \dfrac{yz}{y + z + 1} \right) + 1} \\[1.0em] & = & \dfrac{\dfrac{xyz}{y + z + 1}}{\dfrac{x(y + z + 1) + yz + (y + z + 1)}{y + z + 1}} \\[1.0em] & = & \dfrac{\dfrac{xyz}{y + z + 1}}{\dfrac{xy + xz + x + yz + y + z + 1}{y + z + 1}} \\[1.0em] & = & \dfrac{xyz(y + z + 1)}{y + z + 1(xy + xz + x + yz + y + z + 1)} \\[1.0em] & = & \dfrac{xyz}{xy + xz + x + yz + y + z + 1} \\[1.0em] & = & \dfrac{xyz}{xy + xz + yz + z + x + y + 1} \\[1.0em] & = & \dfrac{xyz(x + y + 1)}{x + y + 1(xy + xz + yz + z + x + y + 1)} \\[1.0em] & = & \dfrac{ \dfrac{xyz}{x + y + 1} }{ \dfrac{xy + xz + yz + z + x + y + 1}{x + y + 1} } \\[1.0em] & = & \dfrac{ \dfrac{xyz}{x + y + 1} }{ \dfrac{xy + z(x + y + 1) + (x + y + 1)}{x + y + 1} } \\[1.0em] & = & \dfrac{ \left( \dfrac{xy}{x + y + 1} \right) z }{ \left( \dfrac{xy}{x + y + 1} \right) + z + 1 } \\[1.0em] & = & \left( \dfrac{xy}{x + y + 1} \right) \ast z & = & (x \ast y) \ast z. \end{array} $$

$$ \begin{array}{ c l } & x \ast (y \ast z) \\[1.0em] = & x \ast \left( \dfrac{yz}{y + z + 1} \right) \\[1.0em] = & \dfrac{x \left( \dfrac{yz}{y + z + 1} \right)}{x + \left( \dfrac{yz}{y + z + 1} \right) + 1} \\[1.0em] = & \dfrac{\dfrac{xyz}{y + z + 1}}{\dfrac{x(y + z + 1) + yz + (y + z + 1)}{y + z + 1}} \\[1.0em] = & \dfrac{\dfrac{xyz}{y + z + 1}}{\dfrac{xy + xz + x + yz + y + z + 1}{y + z + 1}} \\[1.0em] = & \dfrac{xyz(y + z + 1)}{y + z + 1(xy + xz + x + yz + y + z + 1)} \\[1.0em] = & \dfrac{xyz}{xy + xz + x + yz + y + z + 1} \\[1.0em] = & \dfrac{xyz}{xy + xz + yz + z + x + y + 1} \\[1.0em] = & \dfrac{xyz(x + y + 1)}{x + y + 1(xy + xz + yz + z + x + y + 1)} \\[1.0em] = & \dfrac{ \dfrac{xyz}{x + y + 1} }{ \dfrac{xy + xz + yz + z + x + y + 1}{x + y + 1} } \\[1.0em] = & \dfrac{ \dfrac{xyz}{x + y + 1} }{ \dfrac{xy + z(x + y + 1) + (x + y + 1)}{x + y + 1} } \\[1.0em] = & \dfrac{ \left( \dfrac{xy}{x + y + 1} \right) z }{ \left( \dfrac{xy}{x + y + 1} \right) + z + 1 } \\[1.0em] = & \left( \dfrac{xy}{x + y + 1} \right) \ast z \\[1.0em] = & (x \ast y) \ast z. \end{array} $$

Moreover, solving $x \ast e$ $=$ $x$, for $e$, means that $x \ast e$ $=$ $xe$ $/$ $(x$ $+$ $e$ $+$ $1)$ $=$ $x$. Multiplying both sides with $(x$ $+$ $e$ $+$ $1)$ leads to $xe$ $=$ $x(x$ $+$ $e$ $+$ $1)$ $=$ $x^2$ $+$ $xe$ $+$ $x$, and dividing with $x$ means that $e$ $=$ $x$ $+$ $e$ $+$ $1$, from where it can be seen that the equation can not be solved, because $e$ can not be “separated” to none of the sides of the equality, so no identity element exists for the operation $\ast$. Because no indentity exists, no inverses exist either.