new addition and new multiplication x ⊕ y = x + y − 1, x ⊗ y = x + y − xy on set Z, prove the set Z equipped with these 2 new operation

Question

here says a new operation addition and new operation multiplication x ⊕ y = x + y − 1, x ⊗ y = x + y − xy on set Z,where the operations on the right hand side are ordinary addition and multiplication of integers. Prove that the set Z equipped with these two new operations ⊕, ⊗ is a ring. Does it have a multiplicative identity?

I tried to use the addition cancellation but I do not understand why or I got the wrong understanding for this question.

I assume the there is y and y' are 2 solution for equation x ⊕ y = x + y − 1, therefore: x ⊕ y = x + y − 1 x ⊕ y' = x + y − 1 and I get x ⊕ y = x ⊕ y' after the cancellation, then I get y=y', and it means the addictive is unique. So, can I say that the set Z equipped with new addiction x ⊕ y = x + y − 1? or I just go into a wrong way for solve this problem?

ask in advance, thank you

Answer 1

You will need to prove that all the ring axioms are true if, throughout the axioms, $+$ is replaced by $\oplus$ and $\times$ is replaced by $\otimes$. It may also be necessary to give a new meaning to $0$, $-x$ and $1$. Indeed, it may not be possible to give any meaning to $1$: this is what is meant by the question "does it have a multiplicative identity?".

Here is one example: the associative law for addition says $$x\oplus(y\oplus z)=(x\oplus y)\oplus z$$ for all $x,y,z$. To prove this is true, we have $$\eqalign{LHS &=x\oplus(y+z-1)\cr &=x+(y+z-1)-1\cr &=x+y+z-2\cr RHS &=(x+y-1)\oplus z\cr &=(x+y-1)+z-1\cr &=x+y+z-2\cr &=LHS\ .\cr}$$ See if you can do the other axioms for yourself.

Answer 2

Hint $ $ Any set bijection $\,h:R'\to R\,$ serves to transport the ring structure of $\,(R,+,*,0,1),$ to $\,(R',\oplus,\otimes,0',1') \,$ by defining the operations in $\,R'\,$ to make $\,h\,$ be a ring isomorphism, i.e.

$$\ \ \ \begin{align} &x \oplus y\, =\ h^{-1}(h(x) + h(y)),\quad 0' = h^{-1}(0)\\ &x \otimes y\, =\, h^{-1}(h(x)\, *\, h(y)),\quad 1' = h^{-1}(1)\end{align}$$ $\begin{align}\Longrightarrow\ \ \ \ \ \ \ \ h(x\oplus y)\ =\ h(x) + h(y), & \ \ \ \ \ \ \ \ \ \ \ \ \ h(x\otimes y)\ =\ \ \ h(x) * h(y)\\[.3em] {\rm e.g.}\ \ h(x)=1\!-\!x\!:\ \ 1\! -\! (x\!+\!y\!-\!1) =\, 1\!-\!x + 1\!-\!y, & \ \ \ \ 1\!-\!(x\!+\!y\!-\!xy) = (1\!-\!x)*(1\!-\!y)\end{align}$