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 •.

Question

Accepted Answer

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)
[latex]= x + y + z + r(yz + xy + xz) + {r}^{2}xyz[/latex]
and (x • y) • z = (x • y) + z + r(x • y)z
= x + y + rxy + z + r(x + y + rxy)z
[latex]= x + y + z + r(xy + xz + yz) + {r}^{2}xyz[/latex].
The two RHSs are equal, showing that the operation • is associative.
Firstly, suppose that [latex]x = −{r}^{−1}[/latex]. Then
[latex]x • y = −\frac { 1 }{ r } + y + r \left( −\frac { 1 }{ r } ight) y = −\frac { 1 }{ r } + y − y = −\frac { 1 }{ r }[/latex].
Similarly [latex]y = −{r}^{−1} ⇒ x • y = −{r}^{−1}[/latex].
Secondly, suppose that [latex]x • y = −{r}^{−1}[/latex]. Then
[latex]x + y + rxy = −{r}^{−1}[/latex],
[latex]rx + ry + {r}^{2}xy + 1 = 0[/latex],
(rx + 1)(ry + 1) = 0,
⇒ either [latex]x = −{r}^{−1} or y = −{r}^{−1}[/latex].
Thus [latex]x • y = −{r}^{−1} ⇐⇒ (x = −{r}^{−1} or y = −{r}^{−1})[/latex].
If S = {real numbers [latex] eq −{r}^{−1}[/latex]}, 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 [latex] eq −{r}^{−1}[/latex]. 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 [latex]{x}^{−1}[/latex] of x must satisfy [latex]x • {x}^{−1} = 0[/latex], i.e.
[latex]x + {x}^{−1} + rx{x}^{−1} = 0 ⇒ {x}^{−1} = − \frac { x }{ 1+rx } [/latex].
This is a real (finite) number since [latex]x eq −{r}^{−1}[/latex] and, further, [latex]{x}^{−1} eq −{r}^{−1}[/latex], 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 •.

Question : Define a binary operation • on the set of real numbers by x ...

This Question has been Answered.

Already have an account?

Login