Find the inverse with respect to the binary operation $a ∗ b = a + b + a^2 b^2$

Question

A binary operation on $\mathbb{R}$: $a * b = a + b + a^2 b^2$

The neutral element I found to be $0$. Then I need to find an invertible element having two distinct inverses. I don't know where to start for this question.

Answer 1

Hint: Suppose $a$ is fixed and you're trying to find those $b$ such that $a\ast b = 0$ and $b \ast a = 0$ (i.e. those $b$ that are inverse to $a$). As $\ast$ is commutative, you only need to check $a\ast b = 0$. Now note that $a + b + a^2b^2 = 0$ is a quadratic equation in $b$ (as long as $a \neq 0$); remember, we're thinking of $a$ as fixed. You can then determine the zeroes of this quadratic in terms of $a$. You just need to find a suitable choice of $a$ such that there are two distinct real zeroes.

Answer 2

You already found the identity for this operation to be $0$. So, given an $a$, you find an inverse $b$ by solving $a+b+a^2b^2=0$. You can rearrange this to

$$a^2b^2+b+a=0$$

This is a quadratic expression in $b$, with the leading coefficient $a^2$, linear coefficient $1$ and constant coefficient $a$. Solve for $b$ using the quadratic formula and you get up to two answers.

You get two distinct answers when the discriminant ($B^2-4AC$ for the usual quadratic equation) is positive.

Note that, since the operation is symmetric in its two parameters, that any left-inverse is also a right-inverse, and vice versa.

Answer 3

If $a = 0$, $b = 0$ is the unique inverse of $a$, since $0 = 0 * b = 0 + b + 0^2b^2 = b$ in this case.

If $a \ne 0$, $a * b = a + b + a^2b^2 = 0$ is a quadratic equation for $b$. It may be written as

$b^2 + \dfrac{1}{a^2}b + \dfrac{1}{a} = 0. \tag{1}$

Applying the quadratic formula to (1) yields

$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}), \tag{2}$

which may be simplified somewhat:

$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}) = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4a^3}{a^4}})$ $= \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1 - 4a^3}{a^4}}) = \dfrac{1}{2a^2}(-1 \pm \sqrt{1 - 4a^3}), \tag{5}$

which has one value, $b = (-1/2a^2)$, precisely when

$4a^3 = 1 \; \; \text{or} \; \; a = \dfrac{1}{\sqrt[3]{4}}; \tag{6}$

if $0 \ne 4a^3 < 1$, we obtain two real solutions for $b$, so there are two inversens for any such $a$; if $4a^3 > 1$, then the values of $b$ form a complex conjugate pair; their is no (real) inverse for such $a$. But any nonzero $a < (1 / \sqrt[3]{4})$ will have exactly two inverses in $\Bbb R$ for the operation $a * b = a + b + a^2b^2$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!