To demonstrate the associativity we need to show that x • (y • z) is the same thing as (x • y) • z. So consider

x • (y • z) = x + (y • z) + rx(y • z)

= x + y + z + ryz + rx(y + z + ryz)

= x + y + z + r(yz + xy + xz) + {r}^{2}xyz

and (x • y) • z = (x • y) + z + r(x • y)z

= x + y + rxy + z + r(x + y + rxy)z

= x + y + z + r(xy + xz + yz) + {r}^{2}xyz.

The two RHSs are equal, showing that the operation • is associative.

Firstly, suppose that x = −{r}^{−1}. Then

x • y = −\frac { 1 }{ r } + y + r \left( −\frac { 1 }{ r } \right) y = −\frac { 1 }{ r } + y − y = −\frac { 1 }{ r }.

Similarly y = −{r}^{−1} ⇒ x • y = −{r}^{−1}.

Secondly, suppose that x • y = −{r}^{−1}. Then

x + y + rxy = −{r}^{−1},

rx + ry + {r}^{2}xy + 1 = 0,

(rx + 1)(ry + 1) = 0,

⇒ either x = −{r}^{−1} or y = −{r}^{−1}.

Thus x • y = −{r}^{−1} ⇐⇒ (x = −{r}^{−1} or y = −{r}^{−1}).

If S = {real numbers \neq −{r}^{−1}}, then

(i) Associativity under • has already been shown.

(ii) If x and y belong to S, then x • y is a real number and, in view of the second result above, is \neq −{r}^{−1}. Thus x • y belongs to S and the set is closed under the operation •.

(iii) For any x belonging to S, x •0 = x+0+rx0 = x. Thus 0 is an identity element.

(iv) An inverse {x}^{−1} of x must satisfy x • {x}^{−1} = 0, i.e.

x + {x}^{−1} + rx{x}^{−1} = 0 ⇒ {x}^{−1} = − \frac { x }{ 1+rx } .

This is a real (finite) number since x \neq −{r}^{−1} and, further, {x}^{−1} \neq −{r}^{−1}, since if it were we could deduce that 1 = 0. Thus the set S contains an inverse for each of its elements.

These four results together show that S is a group under the operation •.