## Question:

Define a binary operation • on the set of real numbers by
x • y = x + y + rxy,
where r is a non-zero real number. Show that the operation • is associative.
Prove that $x • y = −{r}^{−1}$ if, and only if, $x = −{r}^{−1}$ or $y = −{r}^{−1}$. Hence prove that the set of all real numbers excluding $−{r}^{−1}$ forms a group under the operation •.

## Step-by-step

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 •.