Set $S$ be the set of all real number except -1. Define $*$ on $S$ by
$$a*b=a+b+ab$$
a) Show that $*$ gives a binary operation on $S$.
Answer from my lecturer was;
Suppose that $a*b=a+b+ab=-1$. Then we obtain that $a=-1$ or $b=-1$. Therefore, $S$ is closed under *.
I not really clear how $a$ or $b$ be -1 since it is not belong to $S$ and how it be close? I need an explanation. Thanks in advanced.
$\endgroup$ 03 Answers
$\begingroup$The essence to prove $*$ a binary operation is to show that $*: S\times S\to S$ a map. In your question since $*$ is defined using multiplication and addition of $\mathbb{R}$ which are binary operations, we have $*:S\times S\to \mathbb{R}$ a map. As $S=\mathbb{R}\setminus \{-1\}$, it suffices to show that the range of $*$ is $S$. Suppose $a*b=-1$ and we see $a=-1$ or $b=-1$, a contradiction.
I guess your lecturer wrote $a*b=-1$ and you just mistook it...
$\endgroup$ 1 $\begingroup$Claim: $a+b+ab=-1$ if and only if $a=-1$ or $b=-1$
Proof:
$a+b+ab=-1$ iff $a+b+ab+1=0$ iff $(a+1)(b+1)=0$ iff $a=-1$ or $b=-1$
$\endgroup$ $\begingroup$It's a matter of logistics.
It's easy to show: if $a + b + ab = -1$ then either $a = -1$ or $b=-1$.
$[a+b +ab = -1 \implies a(1+b) = -(1+b) \implies 1+b = 0 \lor a = \frac{-(1+b)}{1+b} \implies b = -1 \lor a = -1]$
So what does this mean in terms of $*$ being a binary operation on S?
If $a,b \in S$ then $a\ne -1; b\ne -1$. So $a+b+ab \ne -1$. But $a+b+ab \in \mathbb R$, of course. So $a+b +ab \in \mathbb R \setminus \{-1\} = S$.
So if $a,b \in S$ then $a*b=a+b+ab \in S$. So $*$ is a binary operation.
$\endgroup$
