Question

Define a new addition and multiplication on$\mathrm{\mathbb{Q}}$ by$r\oplus s=r+s+1$

and$r\ominus s=rs+r+s$

Prove that with these new operations $\mathrm{\mathbb{Q}}$ is a commutative ring with identity. Is

it an integral domain?

Short answer

It is proved that the given ring is an integral domain.

Step-by-step solution

Define multiplication and addition

Firstly, define addition on $\mathrm{\mathbb{Q}}$ by

$r\oplus s=r+s+1$

Also, define multiplication on $\mathrm{\mathbb{Q}}$ by

$r\ominus s=rs+r+s$

Let’s consider that

$r,s,t\in \mathrm{\mathbb{Q}}$

Obtaining with addition operation if  ℚ  is a commutative ring with identity or not

1. Finding addition remains closed

As $r\oplus s\in \mathrm{\mathbb{Q}}$

Thus, the addition remains closed.

2. Associative property:

$\begin{array}{rcl}r\oplus \left(s+t\right)& =& r\oplus \left(s+t+1\right)\\ & =& r+s+t+1+1\\ & =& \left(r+s++1\right)+t+1\\ & =& \left(r\oplus s\right)+t+1\\ & =& \left(r\oplus s\right)\oplus t\end{array}$

Therefore, the addition is associative.

1. Commutative property:

$\begin{array}{rcl}r\oplus s& =& r+s+1\\ & =& s+r+1\\ & =& s\oplus t\end{array}$

Thus, the addition is commutative.

2. Let’s consider that

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

Therefore, it is additive. Additive$e=-1\in \mathrm{\mathbb{Q}}$ .

3. Consider$x\in \mathrm{\mathbb{Q}}$ . Then,

$\begin{array}{rcl}r\oplus x& =& r+x+1\\ & =& -1\Rightarrow x=-2-r\in \mathrm{\mathbb{Q}}\end{array}$

The inverse additive also lies in $\mathrm{\mathbb{Q}}$.

Obtaining with multiplication operation if ℚ  is a commutative ring with identity or not 

4. As given,

$r⊝s=rs+r+s\in \mathrm{\mathbb{Q}}$

That means the multiplication remains closed.

5. Multiplication associated property:

$\begin{array}{rcl}r⊝\left(s⊝t\right)& =& r⊝\left(st+s+t\right)\\ & =& r\left(st+s+t\right)+r+st+s+t\\ & =& rst+rs+rt+st+r+s+t\\ & =& \left(rs+r+s\right)t+\left(rs+r+s\right)t\\ & =& \left(r⊝s\right)t+r⊝s+t\\ & =& \left(r⊝s\right)⊝t\end{array}$

Thus, this is a multiplication associative.

6. Associative law:

$\begin{array}{rcl}r⊝\left(s\oplus t\right)& =& r⊝\left(s+t+1\right)\\ & =& r\left(s+t+1\right)r+s+t+1\\ & =& rs+r+s+rt+r+t+1\\ & =& r⊝s+r⊝t+1\\ & =& \left(r⊝s\right)\oplus \left(r⊝t\right)\end{array}$

And,

$\begin{array}{rcl}\left(r\oplus s\right)⊝t& =& \left(r+s+1\right)t\\ & =& \left(r+s+1\right)t+r+s+1+t\\ & =& rt+t+t+st+s+t+1\\ & =& r⊝t+s⊝t+1\\ & =& \left(r⊝t\right)\oplus \left(s⊝t\right)\end{array}$

Here, the associated law is held under the new definition of multiplication and addition.

Therefore it is proved that with these new operations that $\mathrm{\mathbb{Q}}$ is a commutative ring with identity.

Obtaining integral domain of  

To obtain that it is commutative,

$\begin{array}{rcl}r⊝s& =& rs+r+s\\ & =& sr+s+r\\ & =& s⊝r\end{array}$

This multiplication shows that it is commutative multiplication.

Now, show that the existence of unity

Assume that,

$r⊝a=r\Rightarrow ra+r+a=r\Rightarrow a\left(r+1\right)=0$

As r is chosen arbitrarily and $a=0$ is the multiplicative identity of the ring,

Consider

$r⊝s=-1$

Here -1 is the additive identity

Then,

$\begin{array}{rcl}rs+r+s& =& -1\Rightarrow r\left(s+1\right)\\ & =& -\left(s+1\right)\end{array}$

The result of these are $r=-1$ or $s=-1$

Therefore, it is proved that the given ring is an integral domain.