Question

Let$\mathit{K}$ be the set of all integer multiples of$\sqrt{\mathbf{2}}$ , that is, all real numbers of the form $\mathit{n}\sqrt{\mathbf{2}}$ with $\mathit{n}\mathbf{\in }\mathit{\mathbb{Z}}$. Show that $\mathit{K}$ 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 $\mathit{ℝ}$whose elements deal with basically two operations: addition and multiplication, where elements belong to the real number $\mathit{ℝ}$.

ProvingAxioms:

Let theassumed setbe:

$K=\left\{n\sqrt{2}|n\in \mathbb{Z}\right\}$

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

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

So, we have:

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

.

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

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

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

4) Also, we have:

$0\cdot \sqrt{2},\Rightarrow 0\in K$

.

5) And:

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

6)$ab=n\sqrt{2}\cdot m\sqrt{2}=2mn\notin K$

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