I have a question in the following exercise:
Let $*$ be the binary operation defined on $\Bbb R$ by: $$x*y=xy+(x^2-1)(y^2-1)$$$\forall x,y\in \Bbb R$.
Show that $*$ has a neutral and determine it explicitly.
I have already proven that $*$ is commutative, so it is only enough to prove that $ * $ has neutral on one side. So we look for $e$ such that $x*e=x$ for all $x\in \Bbb R$.
Developing this we have:
$$x*e=xe+(x^2-1)(e^2-1)=x \iff x(e-1)+(x^2-1)(e-1)(e+1)=0$$$$\iff (e-1)(x+(x^2-1)(e+1))=0\iff e-1=0 \,\, \text{ or } (x^2-1)(e+1)+x=0$$$$\iff e=1 \, \, \text{ or } e=\dfrac{1-x-x^2}{x^2-1}$$
Then the neutral element would be 1 in case $x^2-1=0$ and $\dfrac{1-x-x^2}{x^2-1}$ in another case.
Is this correct?
$\endgroup$ 13 Answers
$\begingroup$Other approach
If $ e $ is a neutral element, then necessarily
$$0 * e = 0$$
or
$$-(e^2-1)=0$$
So$$e=\pm 1$$
We just need to check that $ e =1$ satisfies$$(\forall x\in \Bbb R)\;\; e * x = x$$
$\endgroup$ 1 $\begingroup$As you wrote, we are looking for $e$ such that $x*e = x$ for all $x \in \Bbb R$.
You don't have to find $e$ in terms of $x$. You need one real number $e$ that works for all values of $x$.
Since the value you are looking for should work with every $x$, you should look for a value of $x$ that simplifies the finding of such $e$.
Look what happens if we choose $x=1$. Then $e$ should satisfy $1*e=1$. That is $1\cdot e+(1^2-1)(e^2-1)=1$ that reduces to $e=1$. Once you find that $e=1$ works for $x=1$, just verify that $e=1$ works for all $x$:
$x*1 = x\cdot 1 + (x^2-1)(1^2-1) = x$
$\endgroup$ 1 $\begingroup$There can be only one neutral element. And it follows from your computations that $e=1$ will work.
$\endgroup$ 1
