Question

Consider

$G=\{m+n\sqrt{2}:m,n\in \mathbb{Z}\}$

as a subgroup of the additive group ℝ.
Which of the following statements is/are TRUE?

1. 𝐺 is a cyclic subgroup of ℝ under addition
2. 𝐺 ∩ 𝐼 is non-empty for every non-empty open interval 𝐼 ⊆ ℝ
3. 𝐺 is a closed subset of ℝ
4. 𝐺 is isomorphic to the group ℤ × ℤ, where the group operation in ℤ × ℤ is defined by $\left({𝑚}_1,{𝑛}_1\right)+\left({𝑚}_2,{𝑛}_2\right)=\left({𝑚}_1+{𝑚}_2,{𝑛}_1+{𝑛}_2\right)$

Answer:
Option-2 $\to$ 𝐺 ∩ 𝐼 is non-empty for every non-empty open interval 𝐼 ⊆ ℝ
Option-4 $\to$ 𝐺 is isomorphic to the group ℤ × ℤ, where the group operation in ℤ × ℤ is defined by $\left({𝑚}_1,{𝑛}_1\right)+\left({𝑚}_2,{𝑛}_2\right)=\left({𝑚}_1+{𝑚}_2,{𝑛}_1+{𝑛}_2\right)$

Solution :-

Option (1)

A subgroup G of $\left(ℝ,+\right)$ is cyclic if there exists an element $g\in G$ such that every element of G can be written as an integer multiple of g, i.e., there exists $g\in G$ such that:

Now we check if G is cyclic:

• Assume G is cyclic, meaning there exists some $g_{}\in G$ such that $G=\{n{g}_{}:n\in \mathbb{Z}\}\mathrm{.}$
• If $g_{}=a+b\sqrt{2}$ (for some integers $a,b\in \mathbb{Z}$ ), then every element in G should be an integer multiple of $g_{}i\mathrm{.}e\mathrm{.},m+n\sqrt{2}$ must be expressible as $k\left(a+b\sqrt{2}\right)$ for some integer k.

This would imply that:

which simplifies to the system of equations:

This suggest that G is cyclic if and only if every m and n are scalar multiple of a and b.
i.e., every elements of G is of the form $ka+kb\sqrt{2}$ , where $k\in \mathbb{Z}$ but not every elements of G is of this form. We have a contradiction, so our assumption is wrong.

However, no single pair of integers a and b can generate all possible integer pairs (m, n). This is because $\sqrt{2}$ is irrational, and the integers m and n cannot always be proportional to a single generator $a+b\sqrt{2}$ .

There is no single element $g_{}\in G$ such that every element of G can be written as an integer multiple of $g_{}$ . Therefore, G is not a cyclic subgroup of $\left(ℝ,+\right)$ .

So, the option (1) is incorrect.

Option (2)

Given that $G=\{m+n\sqrt{2}:m,n\in \mathbb{Z}\}\mathrm{.}$
The set G contains numbers of the form $m+n\sqrt{2}$ , where m and n are integers. This means that every element of G is a linear combination of 1 and $\sqrt{2}$ , with integer coefficients.

Also, 1 and $\sqrt{2}$ are linearly independent over $\mathbb{Q}\left(rationalnumbers\right)\mathrm{.}$ Since 1 and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$ , we can apply kronecker’s theorem.

Kronecker’s theorem guarantees that for any real number $x\in ℝ$ and for any small $ϵ>0$ , there exist integers m and n such that:

This tells us that we can always find an element of the form $m+n\sqrt{2}$ (which belongs to G) that is arbitrary close to any real number $x\mathrm{.}$ Therefore, the set G can approximate any real number as closely we want.

Conclusion:

By Kronecker’s Approximation Theorem, since 1 and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$ , the set $G=\{m+n\sqrt{2}:m,n\in \mathbb{Z}\}$ is dense in $ℝ\mathrm{.}$ This means that for any real number $x$ and any $ϵ>0$ , there exist integers m and n such that $\mid x-\left(m+n\sqrt{2}\right)\mid <ϵ\mathrm{.}$

In summary, the density of G in $ℝ$ follows directly from the fact that 1 and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$ , and Kronecker’s theorem guarantees that integer linear combination of such numbers can approximate any real number arbitrarily closely.

Because of G is dense in $ℝ$ , for any non-empty open interval $I\subseteq ℝ$ , $G\cap I$ non-empty.

So, the option (2) is correct.

Option (3)

Because of G is dense in $ℝ$ , the derived set of G is $ℝ$ $\left(i\mathrm{.}e\mathrm{.},G^{\prime }=ℝ\right)\mathrm{.}$
But $ℝ⊈G$ and hence $G^{\prime }⊈G\mathrm{.}$

FACT : Any set G is closed if and only if the derived set of G is subset of G ( $i\mathrm{.}e\mathrm{.},G^{\prime }\subseteq G$ ). Therefore, G is not closed subset of $ℝ\mathrm{.}$

So, the option (3) is incorrect.

Option (4)

To show the isomorphism, we need to prove two things:

1. Homomorphism : There exists a function $\varphi :G\to \mathbb{Z}\times \mathbb{Z}$ that preserves the group operation.
2. Bijectiveness : The function & is both injective (one-to-one) and surjective (onto), meaning it has an inverse.

Step 1: Define the Isomorphism

Let’s define a map $\varphi :G\to \mathbb{Z}\times \mathbb{Z}$ as follows:

This map takes an element $g=m+n\sqrt{2}\in G$ and sends it to the ordered pair $\left(m,n\right)\in \mathbb{Z}\times \mathbb{Z}$ .

Step 2: Check that is a Homomorphism

We need to check that $\varphi$ preserves the group operation, i.e., for any two elements $g_{}=m_1+n_1\sqrt{2}\in G$ and $g_{}=m_2+n_2\sqrt{2}\in G$ , we have:

• First, calculate $g_{}+g_{}$ :

• Now apply $\varphi$ to the sum:

• On thr other hand, applying $\varphi$ to $g_{}$ and $g_{}$ separately gives:

Adding these

Since $\varphi \left(g_{}+g_{}\right)=\varphi \left(g_{}\right)+\varphi \left(g_{}\right),$ the map $\varphi$ preserves the group operation and is a homomorphism.

Step 3: Check that $\varphi$ is Bijective

Injective(one-one):

To check injectivity, suppose $\varphi \left(g_{}\right)=\varphi \left(g_{}\right)\mathrm{.}$ This means that:

If $\left(m_1,n_1\right)=\left(m_2,n_2\right),$ then it must be that $m_1=m_2$ and $n_1=n_2,$ which implies that $g_{}=g_{}\mathrm{.}$ Therefore, $\varphi$ is injective.

Surjective(Onto):

For surjective, we need to show that for every $\left(m,n\right)\in \mathbb{Z}\times \mathbb{Z},$ there is a corresponding element $g\in G$ such that $\varphi (g)=\left(m,n\right)\mathrm{.}$

Given any $\left(m,n\right)\in \mathbb{Z}\times \mathbb{Z},$ we can take $g=m+n\sqrt{2}\in G\mathrm{.}$ Then:

Thus, $\varphi$ is surjective.

Step 4: Conclusion

Since $\varphi$ is a homomorphism and bijective, $\varphi$ is an isomorphism. Therefore, the group $G=\{m+n\sqrt{2}:m,n\in \mathbb{Z}\}$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}\mathrm{.}$