Let $R$ be the set of all integers with alternative ring operations defined below. Show that $\Bbb Z$ is isomorphic to $R$.

For any integers $a,b$, define $a\oplus b=a + b – 1$ and $a\odot b=a + b – ab.$ Let $R$ be the ring of integers with these alternative operations. Show that $\Bbb Z$ is isomorphic to $R$.

What I thought about doing was showing $f(a + b) = a + b – 1$, meaning it isn’t isomorphic since it doesn’t equal $f(a)f(b)$. True?

Solutions Collecting From Web of "Let $R$ be the set of all integers with alternative ring operations defined below. Show that $\Bbb Z$ is isomorphic to $R$."

To establish an isomorphism, you would have to find a mapping $f:\Bbb Z \to R$ satisfying the definition of an isomorphism:

  1. $f(a+b)=f(a)\oplus f(b)$
  2. $f(ab)=f(a)\odot f(b)$
  3. $f$ one-to-one and onto (you could just find an inverse homomorphism)

You’re going to have to come up with a candidate for $f$.

My first hint would be to look at $a\odot 1$ and $a\odot 0$ and $a\oplus 1$. $0$ and $1$ are “special” in $R$, and if you sort out what they are doing in the new operation, you can deduce the right $f$ to pick.

You officially need to first show that $R$ is indeed a ring, before you can go about showing anything involving $R$ is a ring homomorphism. However, if you have a candidate bijection $f$ with another (known) ring, in this case $\Bbb Z$, which means it satisfies $f(x+y)=f(x)\oplus f(y)$ and $f(x\times y)=f(x)\odot f(y)$ as indicated in the answer by rschwieb, then you can cheat a bit because $f$ translates all the operations of $\Bbb Z$ into those of$~R$, and their properties come with them. For instance checking the left distributive law in $R$ can be done as
$$
\begin{aligned}
a\odot(b\oplus c)
&=f(x)\odot(f(y)\oplus f(z))
=f(x)\odot(f(y+z))
=f(x\times(y+z))\\
&=f(x\times y+x\times z)
=f(x\times y)\oplus f(x\times z)\\
&=(f(x)\odot f(y))\oplus(f(x)\odot f(z))
=(a\odot b)\oplus(a\odot c),
\end{aligned}
$$
where $x,y,z\in\Bbb Z$ are such that $f(x)=a,f(y)=b,f(z)=c$ (which is well defined since $f$ is a bijection). While it looks a bit complicated, it just transfers the responsability for the distributive law from $R$ to $\Bbb Z$ (for which it is used in the middle of the computation). So nothing is really going on. The other axioms can be checked similarly without effort.

This assumes you have a candidate $f$, but also that you know the neutral elements for $\oplus$ and for $\odot$, which must be chosen as $f(0)$ and $f(1)$ respectively (it is part of the requirement for homomorphisms). You can easily deduce from the definitions of $\oplus$ and $\odot$ which these neutral elements are. Once you got them, you know $f(0)$ and $f(1)$, and other values follow, like $f(2)$ which must be $f(1+1)=f(1)\oplus f(1)$. You will see it is downhill from there.

Well OP choose $f(x) = -x + 1$. To prove the isomorphism, we shall show $$f(a*_1b) = f(a)*_2f(b)$$ under addition and multiplication. Thus, we have for $a, b \in \mathbb{Z}$,

$$f(a)*_2f(b) = (-a + 1) + (-b + 1) – 1 = -(a + b) + 1 = f(a*_1b)$$ and
$$f(a)*_2f(b) = (-a + 1) + (-b + 1) – (ab-(a+b)+1) = -(a + b) + 2 – ab + (a + b) – 1 = -ab + 1 = f(a*_1b)$$

As you can see OP it is all about finding a function by trial-and-error, finding out if it is operation preserving, and performing simple high-school algebra. In case you are unsure you can check if $f(e_G) = e_H$ for example.