How can I show that $(\mathbb Q,\oplus,\cdot)$ is a commutative ring with identity where $\oplus$ and $\cdot$ are defined as, $a\oplus b=a+b-1$ and $a\cdot b=a+b$?

According to the book, an algebraic structure $(R,\oplus,\cdot)$ is called a ring if the following conditions are satisfied:

  1. $(R,\oplus)$ is an abelian group.
  2. Associativity of multiplication holds: $a\cdot(b\cdot c) = (a\cdot b)\cdot c$.
  3. The left distributive law $a\cdot(b\oplus c)=(a\cdot b)\oplus(a\cdot c)$ and the right distributive law $(b\oplus c)\cdot a=(b\cdot a)\oplus(c\cdot a)$ are satisfied by "$\oplus$" and "$\cdot$".

Though I was somehow able to prove first 2 conditions, the third condition is not getting satisfied. It's an "show that..." question, so the statement is definitely true. Can someone help me?

  • 1
    $\begingroup$ What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case? $\endgroup$ – Jonas Lenz Nov 29 '18 at 8:44
  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – José Carlos Santos Nov 29 '18 at 8:49
  • $\begingroup$ All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition. $\endgroup$ – P.Bendre Nov 29 '18 at 8:50
  • $\begingroup$ Can you please show your computations for distributivity? $\endgroup$ – Stockfish Nov 29 '18 at 9:05
  • $\begingroup$ for instance yes - you want to expand both sides and rearrange terms such that you get equality $\endgroup$ – Stockfish Nov 29 '18 at 9:07

It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.

Just note that $1$ is the neutral element for $\oplus$, and so it should satisfy $a\cdot 1=1$ for all $a\in R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)

But it does not: $a\cdot 1:=a+1\neq 1$, for any $a\neq 0$.

In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.


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