@ Appl. Gen. Topol. 16, no. 2(2015), 127-139doi:10.4995/agt.2015.3257
c© AGT, UPV, 2015

Rational criterion for testing the density of

additive subgroups of Rn and Cn

Mohamed Elghaoui
a
and Adlene Ayadi

b

a
University of Carthage, Faculty of sciences of Bizerte, Tunisia. (elghaoui.mohamed@yahoo.fr)

b
University of Gafsa, Faculty of sciences of Gafsa, Tunisia. (adlenesoo@yahoo.com)

Abstract

In this paper, we give an explicit criterion to decide the density of

finitely generated additive subgroups of R
n

and C
n

.

2010 MSC: 47A06; 47A99.

Keywords: dense; additive group; rationally independent; Kronecker.

1. Introduction

It is a classical result of Kronecker that the additive group Z + αZ, α ∈ R,
is dense in R whenever α is irrational. In higher dimensions, this is generalized
as follows: Zn + Z[θ1, . . . , θn]

T is dense in Rn if and only if 1, θ1, . . . , θn are
rationally independent (see [6] and [5]). For general generated additive groups

of the form H =
p∑

k=1

Zuk, where p ≥ 1 and uk ∈ Kn (K = R or C), a criterion
for the density was given by Waldschmidt ([7], see proposition 2.1 for the real
case). However, the use of this theorem in higher dimension or with a large
number of generators is more difficult. So the main aim of this paper is to give
an explicit arithmetic way for checking the density of any finitely generated
additive subgroup of Cn and Rn, which may be used in a future algorithm.
This criterion can be used as a tool to characterize the density of any orbit
given by the natural action of any abelian linear or affine group on Kn (see for
example, [1], [2], [3] and [4]).

Received 5 September 2014 – Accepted 27 July 2015

http://dx.doi.org/10.4995/agt.2015.3257


M. Elghaoui and A. Ayadi

2. Preliminaries

First of all, let us introduce the following proposition which characterizes
the density of additive subgroups Zu1 + · · · + Zup of Rn.

Proposition 2.1 ([7], Proposition 4.3, Chapter II). Let H = Zu1 + · · · + Zup
with uk ∈ Rn, k = 1, . . . , p. Then H is dense in Rn if and only if for every
(s1, . . . , sp) ∈ Zp\{0}:

rank

[
u1 . . . . . . up
s1 . . . . . . sp

]
= n + 1.

If p = n + 1 and (u1, . . . , un) is a basis of R
n, the additive group H =

p∑
i=1

Zui,

where un+1 =
n∑

i=1

θiui is isomorphic (by a linear map) to Z
n + Z[θ1, . . . , θn]

T .

In this case, proposition 2.1 becomes explicit and have the following form: H
is dense in Rn if and only if 1, θ1, . . . , θn are rationally independent.

Now, for the general case, if H is dense in Rn then p ≥ n + 1 and the vector

space

p∑

k=1

Ruk is equal to R
n (proposition 2.1). The last condition means that

a basis of Rn can be extracted from the set of vectors uk, k = 1, . . . , p.

So let us assume here and after that this basis is (u1, . . . , un) and that
p ≥ n + 1.

With these assumptions, the rank condition in proposition 2.1 becomes:

(2.1) rank




1 0 . . . 0 αn+1,1 . . . αp,1

0
...

...
...

...
...

...
...

...
... 0

...
...

...
0 . . . 0 1 αn+1,n . . . αp,n
s1 . . . . . . sn sn+1 . . . sp



= n + 1

where the scalars αk,i are the coordinates of uk, k = n + 1, . . . , p in the basis
(u1, . . . , un), i.e.

uk =

n∑

i=1

αk,iui for all k = n + 1, . . . , p

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 128



Rational criterion for testing the density of additive subgroups

Simplifying further (2.1) using elementary row operations, we get:

rank




1 0 . . . 0 αn+1,1 . . . αp,1

0
...

...
...

...
...

...
...

...
... 0

...
...

...
0 . . . 0 1 αn+1,n . . . αp,n

0 . . . . . . 0 sn+1 −
n∑

i=1

siαn+1,i . . . sp −
n∑

i=1

siαp,i




= n + 1

This condition is fulfilled if the last row is not null, which means that for every
(s1, . . . , sp) ∈ Zp\{0}, there is at least one integer k0 ∈ {n+1, . . . , p} such that
sk0 −

n∑
i=1

siαk0,i 6= 0, which gives rise to the following proposition:

Proposition 2.2. Let H = Zu1+· · ·+Zup, p ≥ n+1 and such that (u1, . . . , un)
is a basis of Rn with uk =

n∑
i=1

αk,iui, for every k = n + 1, . . . , p. Then H is

dense in Rn if and only if for every (s1, . . . , sp) ∈ Zp\{0}, there is at least one
integer k0 ∈ {n + 1, . . . , p} such that sk0 −

n∑
i=1

siαk0,i 6= 0.

Now, let us suppose that 1, αk,i1, . . . , αk,irk is the longest sequence extracted

from the list {1, αk,1, . . . , αk,n} which contains 1 and such that its elements are
rationally independent.

Set Ik := {i1, . . . , irk}.

• If Ik0 = {1, 2, . . . , n} for at least one integer k0 ∈ {n+1, . . . , p} then H is dense
in Rn. Indeed, otherwise, by proposition 2.2, there exists (s1, . . . , sp) ∈ Zp\{0}
such that for every k = n + 1, . . . , p

(2.2) sk −
n∑

i=1

siαk,i = 0

As Ik0 = {1, 2, . . . , n} then using equation 2.2 with k = k0, we get sk0 = 0 and
si = 0 for all i = 1, . . . , n. Using again this equation for the other values of
k ∈ {n+1, . . . , p}, we get sk = 0. Therefore si = 0 for every i = 1, . . . , p, which
leads to a contradiction since (s1, . . . , sp) ∈ Zp\{0}.

• If Ik = ∅ all the coordinates of the given vector uk are rational. Actually if
this condition is fulfilled for every k = n + 1, . . . , p then we have:

Proposition 2.3. If Ik = ∅ for every k = n + 1, . . . , p, then

p∑

j=1

Zuj is not

dense in Rn.

We need the following lemma:

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 129



M. Elghaoui and A. Ayadi

Lemma 2.4. Let (u1, . . . , un) be a basis of R
n, n ≥ 2. Then:

(i) The group
p∑

k=1

Zuk is closed in R
n, for any 1 ≤ p ≤ n.

(ii) The group
p∑

k=1

Ruk +
q∑

k=p+1

Zuk is closed in R
n, for any 1 ≤ p < q ≤ n.

Proof. let Φ : Rp −→
p∑

k=1

Ruk the natural isomorphism defined by

Φ(x1, . . . , xp) =

p∑

k=1

xkuk.

Then Φ is a homeomorphism and we have Φ(Zp) =

p∑

k=1

Zuk. Since Z
p is closed

in Rp, so

p∑

k=1

Zuk is closed in

p∑

k=1

Ruk and hence in R
n. A similar argument can

be used to the proof (ii) by considering Rp × Zq−p which is closed in Rq. �

Proof of proposition 2.3. If Ik = ∅ for every k = n + 1, . . . , p, then the coor-
dinates of every vector uk are rational. So there exist qk ∈ N∗ and pk,j ∈ Z
such that αk,j =

pk,j
qk

. Therefore, uk =
1
qk

n∑
j=1

pk,juj, for every k = n + 1, . . . , p.

Hence
p∑

j=1

Zuj ⊂ 1q
n∑

j=1

Zuj, where q = qn+1qn+2 . . . qp. By lemma 2.4,
1
q

n∑
j=1

Zuj

is closed in Rn, therefore

p∑

j=1

Zuj is not dense in R
n. �

• For a fixed k = n + 1, . . . , p, assume that Ik 6= ∅ and Ik 6= {1, 2, . . . , n}.
Then rewrite the scalars αk,j for every j /∈ Ik as a function of 1 and the scalars
{αk,i i ∈ Ik}. Thus there exist γ(k)j,i1, . . . , γ

(k)
j,irk

, tk,j ∈ Q such that

αk,j = tk,j +
∑

i∈Ik

γ
(k)
j,i αk,i

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 130



Rational criterion for testing the density of additive subgroups

We obtain:

uk =

n∑

j=1

αk,juj

=
∑

j∈Ik

αk,juj +
∑

j /∈Ik

(
tk,j +

∑

i∈Ik

γ
(k)
j,i αk,i

)
uj

=
∑

j∈Ik

αk,juj +
∑

i∈Ik

αk,i



∑

j /∈Ik

γ
(k)
j,i uj


 +

∑

j /∈Ik

tk,juj

=
∑

j∈Ik

αk,j


uj +

∑

i/∈Ik

γ
(k)
i,j ui


 +

∑

j /∈Ik

tk,juj

Let qk ∈ N∗ and m(k)i,j , pk,j ∈ Z such that tk,j =
pk,j
qk

and γ
(k)
i,j =

m
(k)
i,j

qk
. Therefore,

(2.3) qkuk =
∑

j∈Ik

αk,j


qkuj +

∑

i/∈Ik

m
(k)
i,j ui


 +

∑

j /∈Ik

pk,juj

Notice that the choice of the scalar qk is not unique as it can be replaced by a
positive multiple of it.

Set

u′k,j := qkuj +
∑

i/∈Ik

m
(k)
i,j ui

for every k = n + 1, . . . , p and j ∈ Ik. So

(2.4) qkuk =
∑

j∈Ik

αk,ju
′

k,j +
∑

j /∈Ik

pk,juj

For a fixed k, the family of vectors
(
u′k,j, j ∈ Ik

)
and (uj, j /∈ Ik) constitute

all together a basis of Rn since the obtained set is a result of transforming the
basis (u1, . . . , un) using elementary operations.

3. The main result: the real case

Now, assume that Ik 6= ∅ for at least one k and that Ik 6= {1, . . . , n} for
every k = n + 1, . . . , p.

Definition 3.1. We define MH to be the matrix of the coordinates of the
vectors u′k,j, j ∈ Ik and k = n + 1, . . . , p.

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 131



M. Elghaoui and A. Ayadi

The matrix MH is actually defined up to scaling of its columns which are the
vectors {u′k,j, j ∈ Ik ; k = n + 1, . . . , p}. Indeed, for a fixed k ∈ {n + 1, . . . , p},
the choice of qk in the definition of u

′

k,j is not unique as it can be replaced
by any positive multiple of it. However, the rank of the matrix MH does not
change with a particular choice of the set of vectors u′k,j.

Example 3.2. Let H = Zu1 + · · · + Zu7 with u1 = [1, 0, 0]T , u2 = [0, 1, 0]T ,
u3 = [0, 0, 1]

T , u4 = [1, 3
√
2, 2]T , u5 = [0,

√
2,
√
5]T , u6 = [2

√
2,
√
3, 1]T ,

u7 = [3,
√
2, 2

√
2]T . So n = 3 and p = 7.

Let k = 4. Then 1, 3
√
2 is the longest sequence which can be extracted from the

set {1, 1, 3
√
2, 2} (of the coordinates of u4 along with 1) such that its elements

are rationally independent. Since only α4,2 has been selected, so I4 = {2}.
Once α4,1 and α4,3 are written as a function of 1, 3

√
2, we get:

t4,1 = 1, t4,3 = 2, γ
(4)
1,2 = γ

(4)
3,2 = 0

Using the same procedure for the remaining values of k, we obtain:

I5 = {2, 3}, t5,1 = 0, γ(5)1,2 = γ
(5)
1,3 = 0

I6 = {1, 2}, t6,3 = 1, γ(6)3,1 = γ
(6)
3,2 = 0

I7 = {2}, t7,1 = 3, t7,3 = 0, γ(7)1,2 = 0, γ
(7)
3,2 = 2

We choose q4 = q5 = q6 = q7 = 1 so that pk,j = tk,j and m
(k)
i,j = γ

(k)
i,j for every

i ∈ Ik, j /∈ Ik and k = 4, 5, 6, 7.

The vectors u′k,j, j ∈ Ik, k = 4, 5, 6, 7 are:

u′4,2 = u2

u′5,2 = u2

u′5,3 = u3

u′6,1 = u1

u′6,2 = u2

u′7,2 = u2 + 2u3

So that MH is given by:

MH =




0 0 0 1 0 0
1 1 0 0 1 1
0 0 1 0 0 2




Theorem 3.3. Let H = Zu1 + · · · + Zup with uk ∈ Rn and MH defined as
above. Then H is dense in Rn if and only if rank(MH) = n.

We need the following lemmas for the proof of the theorem 3.3:

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 132



Rational criterion for testing the density of additive subgroups

Lemma 3.4. Let u1, . . . , un+1 ∈ Rn be such that (u1, . . . , un) is a basis of
Rn and un+1 =

n∑
i=1

αiui. Suppose that 1, αk1, . . . , αkr is the longest sequence

extracted from the list {1, α1, . . . , αn} which contains 1 and such that its ele-
ments are rationally independent. Then there exist q ∈ N∗, mk,1, . . . , mk,r ∈ Z
such that

∑

j∈I

Ru′j +
∑

j /∈I

Zuj ⊂
n+1∑

k=1

Zuk ⊂
∑

j∈I

Ru′j +
1

q

∑

j /∈I

Zuj

where u′j = quj +
∑
k /∈I

mk,juk for every j ∈ I and I = {kj, j = 1, . . . , r}.

Proof. Assume without loss of generality that kj = j, j = 1, . . . , r. In the
above discussion, we have introduced the vectors u′k,j when several vectors are

added to the basis (u1, . . . , un). But in this case, only one vector has been
added (p = n + 1), so we drop the index k from the definition of u′k,j, mk,j,
pk,j and Ik. Thus we have

qun+1 =
∑

j∈I

αju
′

j +
∑

j /∈I

pjuj

where u′j = quj +
∑
i/∈I

mi,jui. Moreover, let H :=

n+1∑

k=1

Zuk and

u′n+1 = qun+1 −
∑

j /∈I

pjuj =

r∑

j=1

αju
′

j.

Now, consider the vector space E of dimension r equipped with the basis B1 =
(u′1, . . . , u

′

r). The vector u
′

n+1 ∈ E and its coordinates with respect to the basis
B1 are [α1, . . . , αr]T . Moreover, since 1, α1, . . . , αr are rationally independent,
so for every (s1, . . . , sr+1) ∈ Zr+1\{0},

det




1 0 . . . 0 α1

0
...

...
...

...
...

...
... 0

...
0 . . . 0 1 αr
s1 . . . . . . sr sr+1



= sr+1 −

r∑

i=1

siαi 6= 0.

By applying proposition 2.1 to K′ :=
r∑

j=1

Zu′j + Zu
′

n+1, we get:

K′ = E

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 133



M. Elghaoui and A. Ayadi

On the other hand, E ⊕
(

n∑
k=r+1

Ruk

)
= Rn, so

r∑

j=1

Zu′j + Zu
′

n+1 +

n∑

k=r+1

Zuk =

r∑

j=1

Zu′j + Zu
′

n+1 +

n∑

k=r+1

Zuk

Using lemma 2.4,
n∑

k=r+1

Zuk is closed in R
n, thus:

r∑

j=1

Zu′j + Zu
′

n+1 +
n∑

k=r+1

Zuk = E +
n∑

k=r+1

Zuk

Finally, we have for every 1 ≤ j ≤ r:




u′n+1, u
′

j ∈
n+1∑
k=1

Zuk,

un+1, uj ∈ E + 1q
n∑

k=r+1

Zuk

So K′ ⊂
n+1∑
k=1

Zuk. Then

E +
n∑

k=r+1

Zuk = K′ +
n∑

k=r+1

Zuk ⊂
n+1∑

k=1

Zuk ⊂ E +
1

q

n∑

k=r+1

Zuk.

The proof is completed. �

Lemma 3.5. Let B = (u1, . . . , un) be a basis of Rn, n ≥ 2, and v1, . . . , vq ∈
n∑

i=1

Zui

with 1 ≤ q < n. Then the group
q∑

i=1

Rvi +

n∑

i=1

Zui is not dense in R
n.

Proof. Without loss of generality, we can assume that the vectors v1, . . . , vq are
linearly independent. So they can be completed to the basis B′ = (v1, . . . , vq,
vq+1, . . . , vn) using the basis B. We may also assume that vi = ui for every

i = q + 1, . . . , n. Since vi ∈
n∑

j=1

Zuj for every i = 1, . . . , n, it follows that,

through a change of basis, we have ui ∈
n∑

j=1

Qvj. So there exist p ∈ N∗ and

ni,j ∈ Z such that

ui =

n∑

j=1

ni,j
p

vj

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 134



Rational criterion for testing the density of additive subgroups

Hence ui ∈
1

p

n∑

j=1

Zvj for every i = 1, . . . , n. Therefore,

q∑

i=1

Rvi +

n∑

i=1

Zui ⊂
q∑

i=1

Rvi +
1

p

n∑

i=1

Zvi

=

q∑

i=1

Rvi +
1

p

n∑

i=q+1

Zvi

By lemma 2.4, the group

q∑

i=1

Rvi +
1

p

n∑

i=q+1

Zvi is closed in R
n, so the group

q∑

i=1

Rvi +

n∑

i=1

Zui is not dense in R
n. �

Proof of theorem 3.3. Let H :=
p∑

i=1

Zui. Suppose that H = R
n, and define

Hk :=
n∑

i=1

Zui + Zuk, for every k = n + 1, . . . , p. As

H ⊂
p∑

k=n+1

Hk

So, we have

(3.1)

p∑

k=n+1

Hk = R
n

On the other hand, by lemma 3.4, we have

∑

j∈Ik

Ru′k,j +
∑

j /∈Ik

Zuj ⊂ Hk ⊂
∑

j∈Ik

Ru′j,k +
1

qk

∑

j /∈Ik

Zuj

where for every j ∈ Ik, u′k,j = qkuj +
∑
i/∈Ik

m
(k)
i,j ui, with m

(k)
i,j ∈ Z and qk ∈ N∗.

It follows that

p∑

k=n+1

Hk ⊂
p∑

k=n+1



∑

j∈Ik

Ru′k,j +
∑

j /∈Ik

1

qk
Zuj




⊂
p∑

k=n+1



∑

j∈Ik

Ru′k,j


 +

p∑

k=n+1



∑

j /∈Ik

1

qk
Zuj




c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 135



M. Elghaoui and A. Ayadi

Let q = qn+1 . . . qp, the last formula then simplifies to

p∑

k=n+1

Hk ⊂
p∑

k=n+1



∑

j∈Ik

Ru′k,j


 + 1

q

p∑

k=n+1



∑

j /∈Ik

Zuj




⊂
p∑

k=n+1



∑

j∈Ik

Ru′k,j


 + 1

q

n∑

j=1

Zuj

and by equation 3.1, we have

(3.2)

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 + 1

q

n∑

j=1

Zuj = R
n

Suppose that

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 6= Rn, then we can extract a maximal set of

independent vectors {v1, . . . , vm} with m < n from the set of vectors {u′k,j, j ∈

Ik, k = n + 1, . . . , p}. As u′k,j ∈
1

q

n∑

j=1

Zuj so vi ∈
1

q

n∑

j=1

Zuj for every i =

1, . . . , m. Using lemma 3.5,

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 + 1

q

n∑

j=1

Zuj =

m∑

j=1

Rvj +
1

q

n∑

j=1

Zuj

is not dense in Rn, this leads to a contradiction with equation 3.2. So

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 = Rn.

Since

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 is the span of the columns of the matrix MH so

rank(MH) = n.

Conversely, suppose rank(MH) = n, i.e.

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 = Rn. By lemma 3.4,

we have: for every k = n + 1, . . . , p

∑

j∈Ik

Ru′k,j ⊂
∑

j∈Ik

Ru′k,j +
∑

j /∈Ik

Zuj ⊂ Hk

So

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 ⊂

p∑

k=n+1

Hk

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 136



Rational criterion for testing the density of additive subgroups

As Hk ⊂ H, so
p∑

k=n+1

Hk ⊂ H. Thus

R
n =

p∑

k=n+1



∑

j∈Ik

Ru′k,j


 ⊂ H

It follows that H = Rn. �

Example 3.6. Let H = Zu1 + · · ·+ Zu7, where u1 = [1, 0, 0]T , u2 = [0, 1, 0]T ,
u3 = [0, 0, 1]

T , u4 = [1,
√
2, 1]T , u5 = [0, 1,

√
3]T , u6 = [

√
2,
√
3, 1]T , u7 =

[1,
√
2,
√
2]T . So n = 3 and p = 7. The sets Ik, k = 4, . . . , 7 are:

I4 = {2}, I5 = {3}, I6 = {1, 2}, I7 = {2}.

We obtain:

u′4,2 = u2

u′5,3 = u3

u′6,1 = u1

u′6,2 = u2

u′7,2 = u2 + u3

So that:

MH =




0 0 1 0 0
1 0 0 1 1
0 1 0 0 1




Since rank(MH) = 3 then H is dense in R
3.

Now, let us summarize the approach to follow in order to test the density of a

given additive group H =

p∑

k=1

Zuk of R
n:

(1) If p ≤ n or
p∑

k=1

Ruk 6= Rn, then H is not dense in Rn.

(2) Otherwise, compute the sets Ik, k = n + 1, . . . , p:
• If Ik = ∅ for every k = n + 1, . . . , p, then H is not dense in Rn.
• If there is an integer k0 ∈ {n + 1, . . . , p} such that Ik0 = {1, . . . , n},
then H is dense in Rn.

• If Ik 6= ∅ for at least one k and Ik 6= {1, . . . , n} for every k = n +
1, . . . , p, then compute the vectors u′k,j, j ∈ Ik, k = n + 1, . . . , p.

(3) Determine the matrix MH and its rank.
Then H is dense in Rn iff rank(MH) = n.

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 137



M. Elghaoui and A. Ayadi

4. The complex case: Density of additive subgroups of Cn

Let u1, . . . , up ∈ Cn, p ≥ 2n + 1. Suppose that (u1, . . . , u2n) is a basis of Cn
over R. Denote by ũk = [ℜ(uk), ℑ(uk)]T for every k = 1, . . . , p, where ℜ(w)
and ℑ(w) are respectively the real and the imaginary part of a vector w ∈ Cn.
We let H̃ = Zũ1 + · · · + ũp, so H̃ ⊂ R2n.

Theorem 4.1. Let H = Zu1 + · · · + Zup with uk ∈ Cn. Then H is dense in
Cn if and only if rank(M

H̃
) = 2n.

Proof. The proof results directly from theorem 3.3 and the fact that H = Cn

if and only if H̃ = R2n. �

Example 4.2. Let H = Zu1 + · · · + Zu8 with u1 = [1, 0]T , u2 = [0, 1]T ,
u3 = [i, 0]

T ,u4 = [0, i]
T , u5 = [1 + 2i, 5

√
3]T , u6 = [i

√
2,
√
3 + i]T , u7 =

[2
√
3 + i,

√
2 + 2i]T , u8 = [1 + 4i

√
2, 5i

√
3]T .

Then H̃ = Zu1 + · · · + Zu8, where ũ1 = [1, 0, 0, 0]T , ũ2 = [0, 1, 0, 0]T ,
ũ3 = [0, 0, 1, 0]

T , ũ4 = [0, 0, 0, 1]
T , ũ5 = [1, 5

√
3, 2, 0]T , ũ6 = [0,

√
3,
√
2, 1]T ,

ũ7 = [2
√
3,
√
2, 1, 2]T , ũ8 = [1, 0, 4

√
2, 5

√
3]T .

The sets Ik, k = 5, . . . , 8 are:
I5 = {2}, I6 = {2, 3}, I7 = {1, 2} and I8 = {3, 4}.

We obtain:

ũ′5,2 = ũ2

ũ′6,2 = ũ2

ũ′6,3 = ũ3

ũ′7,1 = ũ1

ũ′7,2 = ũ2

ũ′8,3 = ũ3

ũ′8,4 = ũ4

Then

M
H̃

=




0 0 0 1 0 0 0
1 1 0 0 1 0 0
0 0 1 0 0 1 0
0 0 0 0 0 0 1


 .

Since rank(MH̃) = 4 then H̃ is dense and so is H.

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 138



Rational criterion for testing the density of additive subgroups

Acknowledgements. This work is supported by the research unit: systèmes

dynamiques et combinatoire: 99UR15-15

References

[1] A. Ayadi and H. Marzougui, Dense orbits for abelian subgroups of GL(n, C), Foliations
2005, World Scientific, Hackensack, NJ, (2006), 47–69.

[2] A. Ayadi, H. Marzougui and E. Salhi, Hypercyclic abelian subgroups of GL(n, R), J.
Difference Equ. Appl. 18 (2012), 721–738.

[3] A. Ayadi, Hypercyclic abelian groups of affine maps on Cn, Canad. Math. Bull. 56
(2013), 477–490.

[4] N. S. Feldman, Hypercyclic tuples of operators and somewhere dense orbits, J. Math.
Anal. Appl. 346 (2008), 82–98.

[5] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth
Edition, Clarendon Press, Oxford, 1960.

[6] L. Kronecker, Nherungsweise ganzzahlige Auflsung linearer Gleichungen, Monats-

berichte Knigl. Preu. Akad. Wiss. Berlin, (1884), 1179–1193 and 1271–1299.
[7] M. Waldschmidt, Topologie des points rationnels, Cours de troisième Cycle, Université

P. et M. Curie (Paris VI), (1994/95).

c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 139