Is $\mathbb{R}\setminus \{-1\}$ a Semi group/Monoid/Group with $a\cdot b=a+b+ab$?

Question

Is $\mathbb{R}\setminus \{-1\}$ a Semi group/Monoid/Group Where $a\cdot b=a+b+ab$?

So we have to show: 1. that the operation is closed and associativity, then to find an identity element, and then if there is an inverse for all elements.

Associativity: $$(a\cdot b)\cdot c=(a+b+ab)\cdot c=$$

$$= a+b+c+ac+ab+abc=a+(b+c+bc)+ac+bc+abc=a\cdot (b \cdot c)$$

But I am now sure about closure, I need to show that for all $a,b\in \mathbb{R}\setminus \{-1\} a+b+ab\neq -1$

I have started to look at cases, for $a,b>0$ it is true, for $a,b<0$ it is true too, but in mix signs it is getting difficult to determine

P.S $$a+b+ab=-1\iff a(1+b)+b=-1\iff a(1+b)+b+1=0\iff (a+1)(1+b)=0\iff a,b = -1$$

So it is closed

Answer 1

For closure, fix $a$ and solve $$a+x+ax=-1$$ $$a+x(1+a)=-1$$ $$x= \frac{1+a}{-1-a} = -1$$

which is not possible since $x \in \mathbb{R} \setminus\{- 1\}$. So $a \cdot b$ is never equal to $-1$, and hence you have closedness.

To find the identity element, solve $a\cdot x = a$. $$a+x+ax = a$$ $$(a+1)x = 0$$ $$x=0$$ (recall $a \neq -1$)

Finally, for an inverse, solve $a\cdot x = 0$ (I saw your comment so I'm going to do it instead of leaving an exercise): $$a+x+ax = 0$$ $$x=\frac{-a}{a+1}$$ which is well defined since $a \neq -1$. New exercise: verify that $a \cdot \frac{-a}{a+1} = 0$.

Answer 2

Hint Adding 1 to both sides of the defining equation gives $$a \cdot b + 1 = ab + a + b + 1 = (a + 1)(b + 1).$$ So, in terms of $\phi(x) = x + 1$, the multiplication rule satisfies $$\phi(a \cdot b) = \phi(a) \phi(b) .$$ Since $\phi$ is invertible and $\phi(\Bbb R \setminus \{-1\}) = \Bbb R \setminus \{0\}$, that map identifies the given operation with multiplication on $\Bbb R \setminus \{0\}$.