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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:

  1. $(R, \oplus)$ forms an abelian group
  2. $\times$ is associative on $R$
  3. 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.

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  • $\begingroup$ These operations are a little confusing. On the left, you mean + as defined in this new way, and on the right, you mean + in the ring operation. Then you define every product to be 0, yes? $\endgroup$ – Alfred Yerger Apr 5 '16 at 1:01
  • $\begingroup$ Sorry, I fixed it so it should be easier to read, and yes every product will be 0. $\endgroup$ – p.l Apr 5 '16 at 1:05
  • $\begingroup$ @p.l. Is $R$ here the real numbers, or some arbitrary ring? (If the real numbers, use $\mathbb{R}$ to avoid confusion.) $\endgroup$ – Ted Apr 5 '16 at 4:25
  • $\begingroup$ It is worth mentioning that many people consider multiplicative identity to be part of the definition of a ring. See, for example, this discussion on meta. Or Wikipedia articles on rings and rngs. (For this reason, it is very good that you have included the definition you are working with in your post.) $\endgroup$ – Martin Sleziak Apr 5 '16 at 5:25
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$\Bbb R -${-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

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  • $\begingroup$ Thank you so much, Is there anything I can be doing to get better at this? the book is absolutely no help and I spent 5 hours working on this and couldn't figure it out at all... $\endgroup$ – p.l Apr 5 '16 at 4:34
  • $\begingroup$ This is the final class I need for the math major but honestly at this point I'm pretty convinced there is nothing I can do to pass this class... $\endgroup$ – p.l Apr 5 '16 at 4:36
  • $\begingroup$ Experience. The main reason I got this problem so fast is...I've seen it before. I didn't nail it my first time either. Also, be diligent about the order you take steps. Write every move line by line. Do not try and do several things at once. Notice how carefully I hold left/right multiplication in part (2) until I get the thing into a space where I know commutativity holds. I'd say those are the biggest errors new people make. Also if you have problems over $\Bbb R$ $\Bbb Q $ or $\Bbb N $, before you waste too much time. $\endgroup$ – Phillip Hamilton Apr 5 '16 at 11:27
  • $\begingroup$ ...look for easy factorizations. Historically the problems these guys were working on were solutions to polynomials, so a lot of interesting results were found around looking for roots. With abstraction, it's sometimes easy to lose sight of that. Besides that, don't get discouraged. I promise you my first abstract algebra course went equally terribly. I retook it upon entering grad school. $\endgroup$ – Phillip Hamilton Apr 5 '16 at 11:28
  • $\begingroup$ As a start to improving your performance, you should change the R's in your question to $\Bbb R $'s. I've seen this problem so I knew what you meant. But step 1 is for you to be rigorous about your notation. A ring R is not the same thing as the set of real numbers $\Bbb R$ $\endgroup$ – Phillip Hamilton Apr 5 '16 at 11:32

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