Question

Define a new addition and multiplication on Z by$a\oplus b=a+b-1$

and $a⊝b=ab-\left(a+b\right)+2$,

Prove that, with the new operations $\mathrm{\mathbb{Z}}$ is an integral domain.

Short answer

It is proved that $\mathrm{\mathbb{Z}}$ is an integral domain as there are no zero divisors.

Step-by-step solution

Define addition and multiplication

Consider the given equation of addition and multiplication,

$a\oplus b=a+b-1$and $a⊝b=ab-\left(a+b\right)+2$,

Define the addition on$\mathrm{\mathbb{Z}}$ ,

$a\oplus b=a+b-1$and

Defining multiplication on $\mathrm{\mathbb{Z}}$,

$a⊝b=ab-\left(a+b\right)+2$

Prove that addition remains closed

Now, prove that this is a ring.

Then, let’s consider$a,b,c\in \mathrm{\mathbb{Z}}$ .

1. $a\oplus b=a+b-1\in \mathrm{\mathbb{Z}}$, therefore the addition remains closed.

2. As it is known

$\begin{array}{rcl}a\oplus \left(b+c\right)& =& a\oplus \left(b+c-1\right)\\ & =& a+b+c-1-1\\ & =& \left(a+b-1\right)+c-1\\ & =& \left(a\oplus b\right)\oplus c\end{array}$

Thus, the addition in$\mathrm{\mathbb{Z}}$ is associated.

3. Let’s consider that, $a\oplus e=a$

Then, $a+e-1=a\Rightarrow e=1$

Therefore,$e=1$ is an additive identity.

4. As addition is commutative,

$\begin{array}{rcl}a\oplus b& =& a+b-1\\ & =& b+a-1\\ & =& b\oplus a\end{array}$

5. Consider that $a\oplus x=1$. Then,

$a+x-1=1\Rightarrow x=2-a$

Thus the $a\oplus x=1$ has solution in$\mathrm{\mathbb{Z}}$ .

Prove that multiplication remains closed

6.

The multiplication is defined as,

$a⊝b=ab-\left(a+b\right)+2$

The multiplication remains closed as$\mathrm{\mathbb{Z}}$ remains closed under the condition of addition and multiplication.

7. Check the property of multiplication associated property.

localid="1653747322008" $\begin{array}{rcl}a⊝\left(b⊝c\right)& =& a\ominus \left(bc-\left(b+c\right)+2\right)\\ & =& a\left(bc-\left(b+c\right)+2\right)-\left(a+\left(bc-\left(b+c\right)+2\right)\right)+2\\ & =& abc-ac-bc+2c-\left(ab-\left(a+b\right)+2+c\right)+2\\ & =& \left(ab-\left(a+b\right)+2\right)c-\left(\left(ab-\left(a+b\right)+2\right)+c\right)+2\\ & =& \left(a\ominus b\right)c-\left(a\ominus b+c\right)+2\\ & =& \left(a\ominus b\right)\ominus c\end{array}$

Thus the multiplication is associated.

8. Check if the multiplication is commutative or not.

$\begin{array}{rcl}a\ominus \left(b+c\right)& =& a\ominus \left(b+c-1\right)\\ & =& a\left(b+c-1\right)-\left(a+b+c-1\right)+2\\ & =& ab-\left(a+b\right)+2+ac-\left(a+c\right)+2-1\\ & =& \left(a\ominus b\right)+\left(a\ominus c\right)-1\\ & =& \left(a\ominus b\right)+\left(a\ominus c\right)\end{array}$

And

$\begin{array}{rcl}\left(a\oplus b\right)\ominus c& =& \left(a+b-1\right)\ominus c\\ & =& \left(a+b-1\right)c-\left(a+b-1+c\right)+2\\ & =& ac-\left(a+c\right)+2+bc-\left(b+c\right)+2-1\\ & =& a\ominus c+b\ominus c-1\\ & =& a\ominus c+b\ominus c\end{array}$

Finally,

$\begin{array}{rcl}a\ominus b& =& ab-\left(a+b\right)+2\\ & =& ba-\left(b+a\right)+2\\ & =& b\ominus a\end{array}$

Therefore, as shown above, the multiplication is commutative.

Let’s consider

$a\oplus e=a\Rightarrow a+e-1=a\Rightarrow e=1$

As 1 is an additive identity

Assume that

$$\begin{gathered} a\ominus b=1\Rightarrow ab-\left(a+b\right)+2=1 \\ \Rightarrow b\left(a-1\right)=a-1 \end{gathered}$$

When$a\ne 1$, then$b=1$. Also, when$b\ne 1$then$a=1$.

This shows that there are no zero divisors. Hence, it proved that $\mathrm{\mathbb{Z}}$is an integral domain.