Question

Let $E$be the set of even integers with ordinary addition. Define a new multiplication $\ast$on $E$ by the rule "$a\ast b=ab/2$" (where the product on the right is ordinary multiplication). Prove that with these operations $E$is a

commutative ring with identity.

Short answer

It is proved that $E$ is a commutative ring with identity.

Step-by-step solution

Define a new definition of multiplication

Consider $E$is the set of integers. Here, the addition is defined as ordinary addition and multiplication:

$a\ast b=\frac{1}{2}ab$

Then, $E$ satisfies all conditions of the ring under ordinary addition. Therefore, $E$ is called as a commutative ring without identity.

Now, show these conditions under the new definition of multiplication.

Assume that$a,b,c\in E$ . Then,

1. $a\ast b=\frac{1}{2}ab\in E$

Thus, multiplication remains closed.

2.

$\begin{array}{rcl}a\ast \left(b\ast c\right)& =& a\ast \left(\frac{1}{2}bc\right)\\ & =& \frac{1}{4}\left(abc\right)\\ & =& \frac{1}{2}\left(\frac{1}{2}ab\right)c\\ & =& \left(a\ast b\right)\ast c\end{array}$

It shows that the multiplication is associative.

3.

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

And

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

The above two equations show that the multiplication holds distributive law.

Obtain that with these operations E  is a commutative ring with identity

Now, show the identity exists.

Consider that $a\ast e=a\Rightarrow \frac{1}{2}ae=a\Rightarrow e=2$

Thus, 2 = multiplicative identity.

Since the multiplicative is commutative,

$\begin{array}{rcl}a\ast b& =& \frac{1}{2}ab\\ & =& \frac{1}{2}ba\\ & =& b\ast a\end{array}$

Therefore this $E$ is a commutative ring with identity.

Hence, it is proved.