On the set $R-\{-1\}$ define the operations $a\oplus b = a + b + ab$ and $a \times b = 0$. Determine if $\big(R-\{-1\}, \oplus,\times\big)$ is a ring. Is it a commutative ring with unity?
Using the definition of a ring I know I must prove:
- $(R, \oplus)$ forms an abelian group
- $\times$ is associative on $R$
- The distributive law holds
Its been about two years since I have taken a proofs related class so I am really struggling with how to prove these things… For part 1 I know I must show that $\oplus$ is associative on $R$, $R$ contains the identity for $\oplus$ and every element has an inverse for $R$.
I think the major issue I am having is I don't even know what $R-\{-1\}$ means or is supposed to look like.
Best Answer
$\Bbb R -${-1} = {$x \in \Bbb R | x \neq -1$}
We start by factoring $a+b+ab$ into $(1+a)(1+b)-1$
1) Show (R,$\oplus$) is a group
a) closure
Now to show that $a \oplus b$ is closed, we can start by saying that we know $\Bbb R$ is closed under addition and multiplication. Then we just need to show that for $a,b \in \Bbb R - ${-1}, that $a \oplus b \in \Bbb R -${-1}
Let's use proof by contradiction. So suppose that $a+b+ab=-1$. Then $(1+a)(1+b)-1 = -1$
$(1+a)(1+b)=0$
But then either $a=-1$ or $b=-1$, a contradiction.
So $a \oplus b \in \Bbb R -${-1} shows closure under $\oplus$
b) associative
($a \oplus b$) $\oplus$ c = $(a+b+ab) \oplus c$ = $a+b+ab + c +(a+b+ab)c$
= $a+b+ab+c+ac+bc+abc$ = $a + (b+c+bc)+a(b+c+bc)$
= $a \oplus (b+c+bc)$ = $a \oplus (b \oplus c)$
c) identity element
Let $e=0$. Then $a \oplus e = a+0+a(0) = a$ and $e \oplus a = 0+a+(0)a = a$
d) inverses
Suppose $(1+a)(1+b)-1 = 0$. Then $(1+a)(1+b)=1$ and for any $a$, $a^{-1}$ is such that $(1+b)=\frac{1}{1+a} \to b= \frac {1}{1+a} -1$ which is defined for all $\Bbb R$ except {-1}
*(R,$\oplus$) is abelian
Since addition and multiplication are commutative, then $a+b+ab=b+a+ba$
Hence, $a \oplus b = b \oplus a$
This shows the group is abelian
That's part 1