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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 •.

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