How to show distributivity in a ring, and what is wrong with my algebra? I am trying to show the following is a commutative ring with unity, however I am encountering a problem.
First, addition and multiplication are defined as:

$$a \oplus b=a+b-1$$$$a \odot b=ab-(a+b)+2$$

I have shown this "addition" is an abelian group.  The identity is $1$ and the negative is $2-a$.  Now, I am stuck on distributivity.  I found that:
$$a \odot(b \oplus c)$$
$$= a \odot (b+c-1)$$
$$=(a\odot b) + (a \odot c) - (a \odot 1)$$ 
$$=ab-(a+b)+2+ac-(a+c)+2-(a1-(a+1)+2)$$
$$=ab-a-b+2+ac-a-c+2-a+a+1-2$$
$$=ab+ac-2a-b-c+3$$
Why is this not equal to:
$$= a \odot (b+c+(-1))$$
$$=(a\odot b) + (a \odot c) + (a \odot (-1))$$
$$=ab-(a+b)+2+ac-(a+c)+2+a(-1)-(a+(-1))+2)$$
$$=ab-a-b+2+ac-a-c+2-a-a+1+2$$
$$=ab+ac-4a-b-c+7$$
However, the first equation is equal to
$$(a \odot b) \oplus (a \odot c)$$
$$=(ab-(a+b)+2) \oplus (ac-(a+c)+2)$$
$$=ab-(a+b)+2+ac-(a+c)+2-1$$
$$=ab-a-b+2+ac-a-c+2-1$$
$$=ab+ac-2a-b-c+3$$

So, why is the first equation not equal to the second, but equal to the third? Which one is correct?

 A: Note that $(a+1)\oplus (b+1)=(a+b)+1$: This makes us suspicious and we compute
$$\begin{align}(a+1)\odot(b+1)&=(a+1)(b+1)-(a+1)-(b+1)+2\\
&=ab+1.\end{align}$$
We conclude that $x\mapsto x+1$ maps the commutattive ring with unity $(\mathbb Z,+,\cdot)$ to the structure $(\mathbb Z,\oplus,\odot)$, which is therefore also a commutative ring with unity. (And we conclude that $0+1=1$ is its additive neutral and that $1+1=2$ is its unity. 
A: We want to show that $\odot$ distributes over $\oplus$. However, we can NOT assume that $\odot$ distributes over $+$.

For the first part, we have:
\begin{align*}
a \odot(b \oplus c)
&= a \odot (b + c - 1) \\
&= (a)(b + c - 1) - ((a) + (b + c - 1)) + 2 \\
&= (ab + ac - a) - (a + b + c - 1) + 2 \\
&= ab + ac - 2a - b - c + 3 \\
\end{align*}
For the last part, we have:
\begin{align*}
(a \odot b) \oplus (a \odot c)
&= (ab - (a + b) + 2) \oplus (ac - (a + c) + 2) \\
&= (ab - (a + b) + 2) + (ac - (a + c) + 2) - 1 \\
&= ab + ac - 2a - b - c + 3 \\
\end{align*}
which matches.
A: I'm assuming the domain is $\mathbb Z$, but the same argument would work for $\mathbb Q$, $\mathbb R$, or really any ring.
In general, if $f:X\to\mathbb Z$ is $1-1$ and onto, we can define operators on $X$:
$$x\oplus y = f^{-1}(f(x)+f(y))$$
$$x\odot y = f^{-1}(f(x)\cdot f(y))$$
In these cases, the additive identity is $f^{-1}(0)$.
In your case above,this $X=\mathbb Z$ and $f(x)=x-1$, $f^{-1}(n)=n+1$.
Now, with these definitions:
$$\begin{align}
x\odot(y\oplus z) &= x\odot f^{-1}(f(y)+f(z))\\
&= f^{-1}\left(f(x)\cdot(f(y)+f(z))\right)\\
&= f^{-1}(f(x)f(y)+f(x)f(z))\\
&= f^{-1}(f(x\odot y) + f(x\odot z))\\
&=(x\odot y)+(x\odot z)
\end{align}$$
