Proving that $*$ is a binary operation on S

Question

Let $S$ be the set of all real numbers except $-1$. Define $*$ on $S$ by $$a*b=a+b+ab.$$

Goal: Show that $*$ gives a binary operation on S.

In order to prove that $*$ is a binary operation, I need to prove that $S$ is closed under $*$, so I tried to prove that $a+b+ab$ never equals $-1$. I cannot figure out, however, how to do this.

Answer 1

We do it by contradiction: Let $a,b\in S$, i.e., $a,b\neq -1$, but suppose that $a*b=-1$. This means that \begin{align*} a+b+ab&=-1\\ b+ab&=-1-a\\ b(1+a)&=-(1+a)\tag{$\star$} \end{align*} Since $a\neq -1$, then $(1+a)\neq 0$, so $1+a$ is invertible. We can thus cancel the term $(1+a)$ from both sides of equation $(\star)$ above and obtain $$b=-1$$ which contradicts the hypothesis that $b\neq-1$.

Therefore, for $a,b\neq -1$ we also have $a*b\neq -1$, i.e., $a*b\in S$.

Answer 2

Hint: Show that $a*b=(a+1)(b+1)-1$