Question

Let$K$ be the set of all integer multiples of$\sqrt{2}$ , that is, all real numbers of the form $n\sqrt{2}$ with$n\in \mathrm{\mathbb{Z}}$ . Show that$K$ satisfies Axioms 1-5, but is not a ring.

Short answer

Hence, $S$ is not a ring, as$ab\notin K$ .

Step-by-step solution

Rings

A ringcan be defined as a non-empty set $\mathrm{ℝ}$whose elements deal with basically two operations: addition and multiplication, where elements belong to the real number $\mathrm{ℝ}$.

Proving Axioms

Let the assumed set be:

$K=\{n\sqrt{2}n\in \mathrm{\mathbb{Z}}\}$

Now, if $K$ is a subring of $\mathrm{\mathbb{Z}}$, then check the different axioms as:

1. If $a,b\in K$, then $a=n\sqrt{2},b=m\sqrt{2}\forall n,m\in \mathrm{\mathbb{Z}}$.

So, we have:

$\begin{array}{rcl}a+b& =& n\sqrt{2}+m\sqrt{2}\\ & =& \sqrt{2}\left(n+m\right),\left(n+m\right)\in K\end{array}$

.

2. $a+b=\left(n+m\right)\sqrt{2}=\left(m+n\right)\sqrt{2}=b+a$

3. If $c=k\sqrt{2}$, then:

$a+\left(b+c\right)=\left(n+m+k\right)\sqrt{2}=\left(a+b\right)+c$

4. Also, we have:

$0·\sqrt{2}\Rightarrow 0\in K$

.

5. And:

$\begin{array}{rcl}a+x& =& 0\\ & \Rightarrow & x=-n\sqrt{2}\end{array}$

6.$ab=n\sqrt{2}·m\sqrt{2}=2mn\notin K$

Here, we see that the set $K$satisfies all the above axioms 1-5, but it is not a ring as $ab\notin K$.