Prove that this ring is an integral domain based on newly defined binary operations "Define a new addition and multiplication on $\mathbb Z$ by the rules: $a(+) b = a + b – 1$ and $a(*) b = ab – (a + b) + 2$. Prove that with these new binary operations $\mathbb Z$ is an integral domain. You may assume that under these new operations $\mathbb Z$ is a ring."
I can show that $\mathbb Z$ is a commutative ring, I'm not sure how to find the identity element of $\mathbb Z$ to show that it's an integral domain.
Thanks in advance for any help. 
 A: First of all, we usually call the additive identity the "zero" and the multiplicitive identity just the "identity" for clarity. I believe you are looking for zero to show that the ring is an integral domain. To do this, you need to find some $x \in \mathbb{Z}$ such that for any $a \in \mathbb{Z}$, $a(+)x = a+x-1 = a$. You can do this by algebra. A similar method can be used to find the (multiplicitive) identity.
To show that the ring is an integral domain, you need to show that, if we denote the zero element by $z$,  $a\ast b = z \implies a = z $ or $b = z$.
A: This ring is constructed via a standard trick, which allows us to transport the structure of an arbitrary ring to any set with the same cardinality. 
Suppose $(A, +, *)$ is a ring. Let $f : A \to B$ be a bijection, where $B$ is an arbitrary set. Define operations on $B$ via
$$
x(+)y = f(f^{-1}(x) + f^{-1}(y)),
\qquad
x(*)y = f(f^{-1}(x) * f^{-1}(y)).
$$
Then it is immediate that $(B,(+),(*))$ is a ring, and $f$ an isomorphism.
In your case, $A = B = \Bbb{Z}$, and $f(x) = x+1$, and thus $f^{-1}(x) = x - 1$. In fact
$$
x(+)y = (x - 1) + (x-1) + 1 = x + y -1,
\quad
x(*)y = (x-1)(y-1) + 1 = x y -(x+y) +2.
$$
A: This exercise is best understood as a special case of the following trivial observation, which also explains how to come up with these rather exotic (ring) operations (which are, of course, useless, and this "exercise" is just an end in itsself).
Let $R$ be a ring with underlying set $|R|$. If there is a bijection $f : X \to |R|$ from a set $X$, then there is a unique ring $S$ with $|S|=X$ such that $f$ becomes an isomorphism of rings. Namely, one has $0_S = f^{-1}(0_R)$ and $s+t = f^{-1}(f(s)+f(t))$, the same with the multiplicative structure. In fact, the same works for arbitrary algebraic structures. Since $f$ is an isomorphism, every axiom or property of $R$ is inherited to $S$. For example, if $R$ is an integral domain, the same is true for $S$.
Now, take the bijection $\mathbb{Z} \to \mathbb{Z}$, $a \mapsto a-1$. The induced addition is $a+'b=(a-1)+(b-1)+1=a+b-1$, the zero is $1$, the multiplication is $a*'b = (a-1)*(b-1)+1=a*b-a-b+2$, the unit is $2$.
Thus, this ring is nothing else than the usual ring $\mathbb{Z}$, but with a different notation for its elements.
